example problem solving for impulse

Impulse – problems and solutions

Second height (h 2 ) = 3.2 meter

v 2 = 2 g h

v t 2 = v o 2 + 2 g h

0 = v o 2 + 2 (-10)(3.2)

I = (0.005)(4 – (-6)) = (0.005)(4 + 6) = (0.005)(10) = 0.05 Newton second

Plus and minus sign indicates the opposite direction.

Impulse (I) = 0.12 Newton second.

9. A 2 kg object experiences a 4 N force for 5 seconds. Find the impulse. Solution: Impulse = FΔt = 4 × 5 = 20 N·s

15. An impulse of 10 N·s acts on a 2 kg object. What is the change in velocity? Solution: Δv = Impulse / mass = 10 / 2 = 5 m/s

21. An impulse of 15 N·s acts on a 5 kg object. What is the change in velocity? Solution: Δv = Impulse / mass = 15 / 5 = 3 m/s

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Impulse and Momentum – Physics Example Problem

Impulse and momentum are physical concepts that are easily seen from Newton’s Laws of Motion.

Start with this equation of motion for constant acceleration.

v = v 0 + at

where v = velocity v 0 = initial velocity a = acceleration t = time

If you rearrange the equation, you get

v – v 0 = at

Newton’s second law deals with force.

where F = force m = mass a = acceleration

solve this for a and get

Stick this into the velocity equation and get

v – v 0 = (F/m)t

Multiply both sides by m

mv – mv 0 = Ft

The left side of the equation deals with momentum (often denoted by a lower-case p) and the right side is impulse (often denoted by an upper-case letter J).

Mass times velocity is known as momentum and force applied over time is called impulse.

Impulse and Momentum Example Problem

Question: A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) What is the initial momentum of the mass? b) What is the final momentum of the mass? c) What was the force acting on the mass? d) What was the impulse acting on the mass?

Part a) What is the initial momentum?

Momentum is mass times velocity. Since the mass is at rest, the initial velocity is 0 m/s.

momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s

Part b) What is the final momentum?

After the force is finished acting on the mass, the velocity is 3 m/s.

momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s

Part c) What was the force acting on the mass?

From parts a and b, we know mv 0 = 0 kg⋅m/s and mv = 150 kg⋅m/s.

150 kg⋅m/s – 0 kg⋅m/s = Ft 150 kg⋅m/s = Ft

Since the force was in effect over 2 seconds, t = 2 s.

150 kg⋅m/s = F ⋅ 2s F = (15 kg⋅m/s) / 2 s F = 75 kg⋅m/s 2

Unit Fact: kg⋅m/s 2 can be denoted by the derived SI unit Newton (symbol N )

Part d) What was the impulse acting on the mass?

The impulse is the force multiplied by the time passed. It is also equal to the change in momentum over the same time period.

Ft = 75 N ⋅ 2 s Ft = 150 Ns or 150 kg⋅m/s

The impulse was 150 kg⋅m/s.

These problems are relatively simple as long as you keep your units straight. Impulse and momentum should have the same units: mass⋅velocity or force⋅time. Check your units when you check your answer.

Another possible way to cause errors is to confuse your vector directions. Velocity and Force are both vector quantities. In this example, the mass was pushed in the direction of the final velocity. If another force pushed in the opposite direction to slow down the mass, the force would have a negative value compared to the velocity vector.

If you found this helpful, check out other Physics Example Problems .

Related Posts

Impulse and Momentum

Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach. Timothy Treadwell, 2001

• Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
• Compute the momentum of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell.
• How fast would a 250 lb man have to run to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
• How fast would a 4000 lb car have to drive to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)

For the first month of its journey, the Mars Orbiter Spacecraft actually orbited the Earth. After a series of orbit raising maneuvers, the tiny main engine gave the spacecraft enough speed in the right direction to escape. For the next nine months the main engine was only used twice and only very briefly (less than a minute of total burn time) to correct the spacecraft's trajectory.

Nine months after leaving Earth orbit, the spacecraft arrived at Mars. To enter orbit around Mars, the spacecraft needed to slow down — a maneuver called orbit insertion. Since the main engine hadn't been used much in nine months, a quick little test was needed.

Press Release: September 22, 2014 Mars Orbiter Spacecraft's Main Liquid Engine Successfully Test Fired The 440 newton Liquid Apogee Motor (LAM) of India's Mars Orbiter Spacecraft, last fired on December 01, 2013, was successfully fired for a duration of 3.968 seconds at 1430 hrs IST today (September 22, 2014). This operation of the spacecraft's main liquid engine was also used for the spacecraft's trajectory correction and changed its velocity by 2.18 metre/second. With this successful test firing, Mars Orbiter Insertion (MOI) operation of the spacecraft is scheduled to be performed on the morning of September 24, 2014 at 07:17:32 hrs IST by firing the LAM along with eight smaller liquid engines for a duration of about 24 minutes. ISRO, 2014

• Using the press release, determine the mass of the Mars Orbiter Spacecraft at the time of this test firing.
• Why was the qualifying phrase "at the time of this test firing" added to the end of the previous question? Why does the mass of the spacecraft vary with time?

After all, the faster we go, the more difficult it is to avoid collisions with small objects and the more damage such a collision will wreak. Even if we are fortunate enough to miss all sizable objects, we can scarcely expect to miss the dust and individual atoms that are scattered throughout space. At two-tenths of the speed of light, dust and atoms might not do significant damage even in a voyage of 40 years, but the faster you go, the worse it is — space begins to become abrasive. When you begin to approach the speed of light, each hydrogen atom becomes a cosmic ray particle and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly the speed of light is a cosmic ray particle, and there is no difference if the ship strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza: "Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for the pitcher.") So 60,000 kilometers per second may be the practical speed limit for space travel. Isaac Asimov, 1987

The density of the interstellar medium is about one hydrogen atom per cubic centimeter. Imagine a 1,000 tonne, 4 by 6 meter, classroom-sized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri (the nearest solar system to our own) 4.243 light years away.

Kinematics:

• How long would it take our hypothetical spacecraft to complete its hypothetical journey?

Impulse-Momentum:

• Determine the momentum of our spacecraft.
• What mass of interstellar medium is swept up during the journey?
• What impulse does the interstellar medium deliver to the spacecraft?
• How does this impulse compare to the momentum of the spacecraft?

Work-Energy:

• Determine the kinetic energy of our spacecraft.
• What is the effective drag force of the interstellar medium during the journey?
• How much work does the interstellar medium do on the spacecraft?
• How does this work compare to the kinetic energy of the spacecraft?
• Write something.
• In older passenger cars, body panels were attached to a single frame around the perimeter, making them very rigid. This is known as body-over-frame construction. In newer cars, different body parts have stress-bearing elements within them and these parts are then welded to each other. This is known as unitized body construction. Repairing "unibody" cars after collision is comparatively difficult as stress (and thus damage) are distributed throughout the different parts. Why then are cars now built this way
• To escape from a horrible fire, two people are forced to jump from the third story of a burning building on to solid concrete. Which person is more likely to sustain serious injuries: the jumper who comes to an abrupt halt when he lands or the jumper who bounces after impact?
• What is the basic idea behind crash safety features in cars like seatbelts, airbags, crumple zones, etc. What quantities in the impulse-momentum theorem ( F ∆ t  =  m ∆ v ) change as a result of these features, how are they changed, and how does this result in increased safety during a crash?
• The phrase "roll with the punches" has its origin in boxing. What does it mean to roll with a punch (or ride a punch)? How does rolling reduce the severity of a punch?
• Is it possible for a motorcycle to have more momentum (that is, a momentum with a larger magnitude) than a train?
• When hit, the velocity of a 0.145 kg baseball changes from +20 m/s to −20 m/s. What is the magnitude of the impulse delivered by the bat to the ball?
• the ground is soft and the ball stops dead
• the ground is hard and the ball bounces straight up at 2.0 m/s
• Draw a free body diagram showing all the forces acting on the model rocket.
• the weight of the rocket
• the net force on the rocket while the engine was running
• the net impulse on the rocket while the engine was running
• the speed of the rocket when the engine stopped
• the height of the rocket above the ground when the engine stopped
• What is the acceleration of the rocket after the engine shut down?
• What maximum height above the ground did the rocket reach?
• What does the area under this curve represent?
• Calculate its cumulative value at 2 s intervals. Compile your results in a table like the one below.

• Sketch a graph of this quantity with respect to time.

statistical

• Use the given data to create a force-time graph.
• Determine the impulse acting on the projectile as a function of time.
• Compute the launch speed of the projectile.

Data adapted from Kampen, Kaczmarczik, and Rath; 2006 .

investigative

• Which plate have you chosen to work with?
• What is the speed of this plate ?
• What is the area of this plate?
• Is the plate you've chosen continental or oceanic?
• What is a typical density of this kind of crust?
• What is a typical thickness of this type of crust?
• Compute the momentum of the tectonic plate you've chosen from the data you've found. State your answer to the nearest order of magnitude (the nearest power of ten). Don't forget the unit.
• the thrust-time data used to generate the graph on the data sheet
• the propellant mass
• the mass after firing (i.e., the mass of the empty rocket)
• impulse provided by the motor
• fractional propellant mass loss (You need to determine the initial and final mass of the rocket. Assume that the loss of mass is directly proportional to the cumulative impulse. When the impulse is zero at the beginning, the mass loss is zero. When the impulse reaches its final value, the mass loss is 100%.)
• mass of the rocket
• speed of the rocket (Don't forget to include the force of gravity in your calculations.)
• acceleration of the rocket
• altitude of the rocket

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AP Physics 1 : Impulse and Momentum

Study concepts, example questions & explanations for ap physics 1, all ap physics 1 resources, example questions, example question #1 : impulse and momentum.

Since the collision is completely elastic, we know that both momentum and kinetic energy are conserved. We can write the following equations (initial momentum and energy of the second ball are neglected since it is not moving:

We can rewrite the second equation as:

Rearranging, we get:

Plug in our values for the initial and final velocities:

Example Question #2 : Impulse And Momentum

It does not matter whether the collision is elastic or inelastic (although it would be best to assume that it's inelastic). Momentum is conserved in either type of collision, and is the only value needed for our calculation. Since they come to a standstill, their momentums at the moment of collision are equal and opposite:

Plug in the given values from the question and solve:

Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:

There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:

Rearranging for final velocity, we get:

At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:

Example Question #4 : Impulse And Momentum

Impulse can be written as either of two popular expressions:

From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:

Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:

Canceling out mass and rearranging for final velocity, we get:

We know these variables, allowing us to solve for the velocity:

Plugging this value into the expression for impulse, we get:

Example Question #5 : Impulse And Momentum

Consider the following system:

To calculate the momentum of the block, we first need to know the velocity of the block. This can be found using the equation for the conservation of momentum:

If we assume that the final height is zero, we can eliminate initial kinetic energy and final potential energy, getting:

Substituting expressions for each term, we get:

Cancel out mass and rearrange to solve for velocity:

We can use the horizontal distance traveled and the angle of the slope to determine the initial height:

Now that we have the initial height, we can solve for final velocity:

Finally, we can now use the equation for momentum to solve the problem:

Example Question #6 : Impulse And Momentum

Momentum is always conserved. Equation for conservation of momentum: 

There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.

The equation for momentum is:

To maintain conservation of momentum, a new state must have the same momentum as a previous state:

Plug in known values and solve.

To solve this problem, we need to consider the change in the ball's momentum. To do so, we'll use the following equation.

Rearrange the above equation to solve for the average force.

Example Question #9 : Impulse And Momentum

Joe, of mass 90kg, jumps straight up. To do so, he bends his knees and produces an upwards force that results in a constant upward net force of 100N. If Joe experiences this force for 0.9s before leaving the ground, what is Joe's velocity immediately after he leaves the ground? 

To solve this problem we need to use the relationship between force and impulse, which is given by the following equation:

This equation represents that the rate of change of momentum with respect to time is equal to the net force that causes said change in momentum. Thus:

Which of the following explains why when we land on our feet, we instinctively bend our knees? Hint: think about the relationship between force, impulse, and time.

When we bend our knees we extend the time in which we apply the force that stops us, so our impulse is greater

When we bend our knees we extend the time in which we apply the force that stops us, so our impulse is smaller

By bending our knees we use a greater force to stop, which makes the impulse smaller

By bending our knees we extend the time it takes us to stop, which increases the impact force

By bending our knees we extend the time it takes us to stop, which diminishes the impact force

Note that the initial momentum does not depend on the impact force nor on how much time it takes to stop. The initial momentum depends on the velocity we have when we first hit the ground. This velocity is given by whatever happened before we hit the ground, which no longer concerns us since we only care about what happens from the moment we first hit the ground till the moment we stop. Yes, the time that passes for you to stop is very small, but it is impossible for it to be zero. So we have that the change in momentum (impulse) is a constant:

Remember that any change in momentum for a given mass occurs because its velocity changes. The velocity of the mass changes due to an acceleration and an acceleration is caused by a force. This gives us a relationship between force and impulse:

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8.1 Linear Momentum, Force, and Impulse

Section learning objectives.

By the end of this section, you will be able to do the following:

• Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem
• Describe Newton’s second law in terms of momentum
• Solve problems using the impulse-momentum theorem

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) calculate the mechanical energy of, power generated within, impulse applied to, and momentum of a physical system.

Section Key Terms

[BL] [OL] Review inertia and Newton’s laws of motion.

[AL] Start a discussion about movement and collision. Using the example of football players, point out that both the mass and the velocity of an object are important considerations in determining the impact of collisions. The direction as well as the magnitude of velocity is very important.

Momentum, Impulse, and the Impulse-Momentum Theorem

Linear momentum is the product of a system’s mass and its velocity . In equation form, linear momentum p is

You can see from the equation that momentum is directly proportional to the object’s mass ( m ) and velocity ( v ). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object.

Momentum is a vector and has the same direction as velocity v . Since mass is a scalar , when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.

Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.

Recall our study of Newton’s second law of motion ( F net = m a ). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum.

In equation form, this law is

where F net is the net external force, Δ p Δ p is the change in momentum, and Δ t Δ t is the change in time.

We can solve for Δ p Δ p by rearranging the equation

F net Δ t F net Δ t is known as impulse and this equation is known as the impulse-momentum theorem . From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time.

[OL] [AL] Explain that a large, fast-moving object has greater momentum than a smaller, slower object. This quality is called momentum.

[BL] [OL] Review the equation of Newton’s second law of motion. Point out the two different equations for the law.

Newton’s Second Law in Terms of Momentum

When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since Δ p = Δ ( m v ) Δ p = Δ ( m v ) . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. F net = m a F net = m a is actually derived from the equation:

For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of F net = m a F net = m a from

by substituting the definitions of acceleration and momentum.

The change in momentum Δ p Δ p is given by

If the mass of the system is constant, then

By substituting m Δ v m Δ v for Δ p Δ p , Newton’s second law of motion becomes

for a constant mass.

we can substitute to get the familiar equation

when the mass of the system is constant.

[BL] [OL] [AL] Show the two different forms of Newton’s second law and how one can be derived from the other.

Tips For Success

We just showed how F net = m a F net = m a applies only when the mass of the system is constant. An example of when this formula would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you can use Newton’s second law expressed in terms of momentum to account for the changing mass without having to know anything about the interaction force by the fuel on the rocket.

Hand Movement and Impulse

In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse.

• one tub filled with water
• Try catching a ball while giving with the ball, pulling your hands toward your body.
• Next, try catching a ball while keeping your hands still.
• Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop.
• After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive.
• Explain what happens in each case and why.
• a football player colliding with another, or a car moving at a constant velocity
• a car moving at a constant velocity, or an object moving in the projectile motion
• a car moving at a constant velocity, or a racket hitting a ball
• a football player colliding with another, or a racket hitting a ball

[OL] [AL] Discuss the impact one feels when one falls or jumps. List the factors that affect this impact.

Links To Physics

Engineering: saving lives using the concept of impulse.

Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2 ). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as Δ p = F net Δ t Δ p = F net Δ t .

Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force F net = Δ p Δ t , F net = Δ p Δ t , you can see how increasing Δ t Δ t while Δ p Δ p stays the same will decrease F net . This is another example of an inverse relationship. Similarly, a padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact.

Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision , especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident.

Grasp Check

You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem.

• Bending your knees increases the time of the impact, thus decreasing the force.
• Bending your knees decreases the time of the impact, thus decreasing the force.
• Bending your knees increases the time of the impact, thus increasing the force.
• Bending your knees decreases the time of the impact, thus increasing the force.

Solving Problems Using the Impulse-Momentum Theorem

Talk about the different strategies to be used while solving problems. Make sure that students know the assumptions made in each equation regarding certain quantities being constant or some quantities being negligible.

Worked Example

Calculating momentum: a football player and a football.

(a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s.

No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p . (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum:

To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation.

To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation.

The ratio of the player’s momentum to the ball’s momentum is

Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football.

Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams ( Figure 8.3 ) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds).

Recall that Newton’s second law stated in terms of momentum is

As noted above, when mass is constant, the change in momentum is given by

where v f is the final velocity and v i is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for Δ p Δ p , we can use F net = Δ p Δ t F net = Δ p Δ t to find the force.

To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above.

Now we can find the magnitude of the net external force using F net = Δ p Δ t F net = Δ p Δ t

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using F net = m a , but we would have had to do one more step. In this case, using momentum was a shortcut.

Practice Problems

• 0.5 kg ⋅ m/s
• 15 kg ⋅ m/s
• 50 kg ⋅ m/s

A 155-g baseball is incoming at a velocity of 25 m/s. The batter hits the ball as shown in the image. The outgoing baseball has a velocity of 20 m/s at the angle shown.

What is the magnitudde of the impulse acting on the ball during the hit?

• 2.68 kg⋅m/s.
• 5.42 kg⋅m/s.
• 6.05 kg⋅m/s.
• 8.11 kg⋅m/s.

Check Your Understanding

What is linear momentum?

• the sum of a system’s mass and its velocity
• the ratio of a system’s mass to its velocity
• the product of a system’s mass and its velocity
• the product of a system’s moment of inertia and its velocity

If an object’s mass is constant, what is its momentum proportional to?

• Its velocity
• Its displacement
• Its moment of inertia

What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is constant?

• F net = Δ v Δ m Δ t F net = Δ v Δ m Δ t
• F net = m Δ t Δ v F net = m Δ t Δ v
• F net = m Δ v Δ t F net = m Δ v Δ t
• F net = Δ m Δ v Δ t F net = Δ m Δ v Δ t

Give an example of a system whose mass is not constant.

• A spinning top
• A baseball flying through the air
• A rocket launched from Earth
• A block sliding on a frictionless inclined plane

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the assessment will help identify which objective is causing the problem and direct students to the relevant content.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates.

Access for free at https://openstax.org/books/physics/pages/1-introduction

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• Book title: Physics
• Publication date: Mar 26, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/physics/pages/1-introduction
• Section URL: https://openstax.org/books/physics/pages/8-1-linear-momentum-force-and-impulse

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Impulse Momentum Exam 1 and Problem Solutions

where ΔV=V₂-V₁=-3-4=-7m/s

I=m.ΔV=3.(-7)=-21kg.m/s

P₂=m.V₂=4kg.6m/s=24kg.m/s

ΔP=P₂+P₁ (vector addition)

ΔP²=P₂²+P₁²=m²(v₂²+v₁²)

ΔP=40kg.m/s

Impulse=change in momentum

I=ΔP=40kg.m/s

3. Find the impulse and force which make 12m/s change in the velocity of object having 16kg mass in 4 s.

F.Δt=ΔP=m.ΔV

F.4s=16kg.12m/s

F.Δt=Impulse=192kg.m/s

F.Δt=20.2/2+20.(6-2)+20.(10-6)/2

F.Δt=140kg.m/s

5. A ball having mass 500g hits wall with a10m/s velocity. Wall applies 4000 N force to the ball and it turns back with 8m/s velocity. Find the time of ball-wall contact.

F.Δt=ΔP=m.ΔV=m.(V₂-V₁)

-4000.Δt=0,5kg.(8-10)

Δt=0,00025s

Impulse and Momentum: Explanation and Examples

• The Albert Team
• Last Updated On: March 28, 2023

Now that you’ve learned about momentum, we can explore more complex situations where momentum changes and the conditions that cause momentum to change. In this post, we’ll explain the impulse equation, show you how to calculate impulse, and explore how impulse and momentum are related.

What We Review

Definition Of Impulse And Its Relation To Force

When a force is applied to an object, that force will impact the object’s motion. Impulse is a concept in physics we use to quantify the impact that force has on an object over a period of time. An everyday example of impulse is when a football player exerts a force on the quarterback to change his momentum. The impulse is equal to the product of the force the player exerts and the time they are in contact. 

How Are Impulse And Momentum Related?

Impulse and momentum are very closely related to each other. An impulse affects the object’s motion and thus causes the object’s momentum to change. How much the object’s momentum changes is exactly equal to the impulse the object experiences. 

How To Calculate Impulse

Equation for impulse.

…where:

• J is impulse

Impulse Units

The units for impulse come from those of force (newtons) and time (seconds). Since impulse is equal to the product of force and time, the units for impulse are \text{N}\cdot\text{s} . 

There is also an acceptable alternative unit for impulse. Since the amount of impulse that an object experiences is equal to its change in momentum, we can also use the units of momentum to measure impulse: \text{kg}\cdot\text{m/s} . 

Change In Momentum Formula

If you know the initial and final momentum of an object, you can calculate its change in momentum by finding the difference between those values. However, in many situations you may not directly know the object’s momentum and will instead be given the mass and velocity before and after the force is applied. As you learned before, momentum is the product of mass and velocity. Therefore, the change in momentum can be calculated as the product of mass and the change in velocity. 

• \Delta{p} is the change in momentum
• \Delta{v} is the change in velocity 

Impulse-Momentum Theorem 

Definition of the impulse-momentum theorem.

The impulse-momentum theorem states that the impulse an object experiences is equal to the object’s change in momentum. This theorem is represented by the equation:

If we substitute the equations for impulse and change in momentum, this theorem can also be expressed as:

Significance Of The Impulse-Momentum Theorem

The impulse-momentum theorem is an important concept in physics because it provides a relationship between force, time, and motion. This is extremely helpful for analyzing collisions between objects and understanding how manipulating the variables affects the outcome of a collision. 

One of the most important applications is in the car safety features that we enjoy today. In car accidents, cars experience a massive change in momentum as they go from a high speed to a full stop. The force of such a collision can be enormous and potentially fatal. The theorem tells us that if we can increase the amount of time it takes for the passenger’s momentum to become zero, we can reduce the amount of force the passenger experiences. For example, when an airbag deploys in an accident, it reduces the force on the person by slowing them down more gradually than if they were to hit the much harder dashboard. By increasing the time it takes to stop the person, the airbag decreases the force the passenger experiences and saves lives.  

Examples: Crash Testing

Let’s consider the following example of a crash test dummy with and without an airbag.

Crash Without an Airbag

Let’s try the following example. A 60\text{ kg} crash test dummy is traveling in a car moving at 30\text{ m/s} . The car collides with a wall and comes to rest. Without an airbag, the crash test dummy hits the steering wheel and comes to rest in 0.05\text{ s} . How much force does the dummy experience in the collision?

First, we will calculate the change in momentum: 

Now applying the impulse-momentum theorem:

Dividing by time produces:

This massive amount of force would almost certainly cause serious and potentially deadly injuries to a real passenger. 

Crash With an Airbag

Now the same crash test is conducted but with an airbag that brings the crash test dummy to rest in 0.3\text{ s} . With the airbag, how much force does the dummy experience? 

Here, the change in momentum is the same but the time is greater. We will apply the same method as before to calculate the force, but substitute the longer time of 0.3\text{ s} :

While that increase in time would be almost imperceptible to the human eye, the result is a significant reduction in the collision force. By harnessing the impulse-momentum theorem, engineers have designed safety features like airbags, seatbelts, and crumple zones that save thousands of lives by increasing the time and decreasing the collision force on car passengers. 

For more examples, check out this video from GPB Education.

In this post, we learned that impulse is the product of force and time and is equal to the change in momentum. We can apply the impulse-momentum theorem to analyze collisions and understand how increasing the time of impact decreases the force an object experiences. Like many concepts in physics, this theorem has important applications to everyday life. 

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Impulse momentum theorem  problems

These impulse momentum theorem problems will help you clearly see how to apply the impulse-momentum theorem. Study the solution we give carefully so you can tackle other similar problems.

Earlier in impulse and momentum , we saw that Ft = ▲mv

In many situations, the mass will not change. In this case, we can also rewrite the formula as Ft = m▲v  or Ft = m(final speed - initial speed)

Problem #1:

What average force is required to push a 20-kg stroller with your toddler in it for 5 seconds if the  weight of the toddler is 15 kg? Suppose that you can push with a speed of 1 m/s and the stroller is initially at rest.

You are pushing a total mass of 20 kg + 15 kg = 35 kg

Ft = m(final speed - initial speed)

F × 5 = 35 kg × (1 m/s - 0)

F × 5  = 35

Since 7 × 5 = 35, F = 7 newtons.

Interesting impulse momentum theorem problems

Problem #2:

An average force of 50 N is exerted on a 4-kg cart for 2 seconds.

a. What is the impulse?

b. What is the change in momentum?

c. What is the mass's change in velocity?

a. Impulse  =  F × t

Impulse  = 50 N × 2

Impulse  = 100 N.s

Impulse  = change in momentum

So, change in momentum  = impulse  = 100 N.s

The mass's change in velocity is ▲v

F × t = m▲v

100 N.s = 4 × ▲v

Since 100 = 4 × 25, ▲v = 25 m/s

Problem #3:

A soccer ball is heading toward a wall with a speed of 20 meters per second. After hitting the wall, the ball bounces back with a speed of 25 meters per second. The ball was in contact with the wall for 0.003 second. What is the average force the wall exerted on the ball?

F × t = m(final speed - initial speed)

The final speed is 25 meters per second

The initial speed is -20 meters per second. This speed is negative because the ball is going to the other direction.

The mass of a soccer ball is 0.45 kg

F × 0.003 = 0.45 (25 - -20)

F × 0.003 = 0.45 (45)

 F × 0.003 = 20.25

F = 20.25 / 0.003

Impulse and momentum

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• Physics Formulas

Impulse Formula

A car is travelling at full speed. It crashes into the barrier, because of the negligence of the driver.

Here the huge force is applied on the wall by the car in a very little time interval which we term as an impulse.

Impulse is the big force acting for a very small interval of time. It is represented by J⃗ J→ .

Impulse Formula is articulated as

• Force applied is given as F
• Time interval throughout which force is applied is given as t.

Impulse can also be articulated as the rate of change of momentum.

• Mass of the body is given as m
• The velocity with which the body is moving is given as v.

Velocity is articulated as

v=v f – v i

• Initial Velocity is given as v i
• Final Velocity is given as v f .

Therefore, the Impulsive force is articulated as

Impulse is articulated in Kgms -1 and Impulsive force is articulated in Newton(N).

Impulse-momentum formula

The Impulse-momentum formula is obtained from the impulse-momentum theorem which states that change in momentum of an object is equal to impulse applied on the object. The formula is given as follows:

As the SI unit of impulse and momentum are equal, it is given as Ns=kg.m.s -1

Solved Examples

Below are some problems on impulse:

Example 1: A batsman knocks back a ball straight in the direction towards the bowler without altering its initial speed of 12 m/s. If the mass of the ball is 0.15kg, calculate the impulse imparted to the ball?

Given: v i (Initial Velocity) = 12 m/s,

v f (Final Velocity) = -12m/s,

m (mass) = 0.15kg,

J (Impulse) = ?

Impulse is articulated as

J = mv f – mv i

= m(v f – v i )

= 0.15 Kg (-12 -12) m/s

Example 2: A golfer hits a ball of mass 45g at a speed of 40m/s. The golf club is in contact with the ball for 3 s. Compute the average force applied by the club on the ball?

m (Mass)= 0.045 kg,

v i (Initial Velocity) = 0,

v f (Final Velocity) = 40m/s,

Stay tuned with BYJU’S to learn more about other Physics related concepts.

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Mechanics: Momentum and Collisions

Calculator pad, version 2, momentum and collisions: problem set.

Determine the momentum of …

a. … an electron (m= 9.1 x10 -31 kg) moving at 2.18 x 10 6 m/s (as if it were in a Bohr orbit in the H atom). b. … a 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860 m/s. c. … a 110-kg professional fullback running across the line at 9.2 m/s. d. … a 360,000-kg passenger plane taxiing down a runway at 1.5 m/s

• Audio Guided Solution
• Show Answer

A bicycle has a momentum of 24 kg•m/s. What momentum would the bicycle have if it had …

a. … twice the mass and was moving at the same speed? b. … the same mass and was moving with twice the speed? c. … one-half the mass and was moving with twice the speed? d. … the same mass and was moving with one-half the speed? e. … three times the mass and was moving with one-half the speed? f. … three times the mass and was moving with twice the speed?

According to the Guinness Book of World Records, the fastest recorded baseball pitch was delivered by Nolan Ryan in 1974. The pitch was clocked at 100.9 mi/hr (45.0 m/s). Determine the impulse required to give a 0.145-kg baseball such a momentum.

Jerome plays middle linebacker for South's varsity football team. In a game against cross-town rival North, he delivered a hit to North's 82-kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s.

a. Determine the initial momentum of the running back. b. Determine the final momentum of the running back. c. Determine the momentum change of the running back. d. Determine the impulse delivered to the running back.

Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was stopped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.

a. Determine the momentum change of Kara's car. b. Determine the impulse experienced by Kara's car. c. Determine the magnitude of the force experienced by Kara's car.

An interesting story is often told of baseball star Johnny Bench when he was a rookie catcher in 1968. During a Spring Training game, he kept signaling to star pitcher Jim Maloney to throw a curve ball. Maloney continuously shook off Bench's signal, opting to throw fastballs instead. The rookie catcher walked to the mound and told the veteran Maloney that his fastball wasn't fast enough and that he should throw some curve balls. Bench again signaled for a curve. Maloney shook of the signal and threw a fastball. Before the ball reached the plate, Bench took off his glove; he then caught the pitch barehanded.

a. Determine the impulse required to stop a 0.145-kg baseball moving at 35.7 m/s (80.0 mi/hr). b. If this impulse is delivered to the ball in 0.020 seconds, then what is magnitude of the force acting between the bare hand and the ball?

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds.

a. Determine the force acting upon Logan's body. b. If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds. Determine what the force on his body would have been for such an abrupt collision.

NASA's Langley Research Center has been experimenting with the use of air bags to soften the landings of crew exploration vehicles (CEV) on land. What stopping time will be required in order to safely stop a 7250 kg CEV moving at 7.65 m/s with an average force of 426000 N (an average force of 6 Gs)?

In a study conducted by a University of Illinois researcher, the football team at Unity High School in Tolono, IL was equipped for an entire season with helmets containing accelerometers. Information about every impact in practice and in games was sent to a computer present on the sidelines. The study found that the average force on a top of the head impact was 1770 N and endured for 7.78 milliseconds. Using a head mass of 5.20 kg and presuming the head to be a free body , determine the velocity change experienced in such an impact.

Problem 10:

Cassie has just finished her session on the trampoline during PE. As she prepares to exit the trampoline, her vertical momentum is reduced by a series of three resistive impulses with the bounce mat. Just prior to this series of impulses, her 48.5-kg body is moving downward at 8.20 m/s. On the first impulse, Cassie experiences an average upward force of 230 N for 0.65 seconds. The second impulse of 112 N•s lasts for 0.41 seconds. The last impulse involves an average upward force of 116 N which cases a 84 kg•m/s momentum change. What vertical velocity does Cassie have after these three impulses?

Problem 11:

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails while moving at 19.8 m/s. The first guard rail delivered a resistive impulse of 5700 N•s. The second guard rail pushed against his car with a force of 79000 N for 0.12 seconds. The third guard rail collision lowered the car's velocity by 3.2 m/s. Determine the final velocity of the car.

Problem 12:

Mr. H ignites the enthusiasm of the class with a home-made cannon demonstration. The 1.27-kg cannon is loaded with a 54-gram tennis ball and placed on the floor. Mr. H adds the fuel, waits for its vapors to fill the reaction chamber and then brings a match nearby. The explosion stuns the crowd and propels the ball forward. A photogate measurement determines that the cannon recoiled backwards with a speed of 7.8 m/s. Determine the speed of the ball.

Problem 13:

An 82-kg male and a 48-kg female pair figure skating team are gliding across the ice at 7.4 m/s, preparing for a throw jump maneuver. The male skater tosses the female skater forward with a speed of 8.6 m/s. Determine the speed of the male skater immediately after the throw.

Problem 14:

A candy-filled piñata is hung from a tree for Matthew's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Hayden cracks it with a 0.54-kg stick moving at 4.8 m/s. The stick stops and the piñata undergoes a gentle swinging motion. Determine the swing speed of the piñata immediately after being cracked by the stick.

Problem 15:

During an in-class demonstration of momentum change and impulse, Mr. H asks Jerome (102 kg) and Michael (98 kg) to sit on a large 14-kg skate cart. Mr. H asks Suzie (44 kg) to sit on a second 14-kg skate cart. The two carts are placed on low friction boards in the hallway. Jerome pushes off of Suzie's cart. Measurements are made to determine that Suzie's cart acquired a post-impulse speed of 9.6 m/s. Determine the expected recoil speed of Jerome and Michael's cart.

Problem 16:

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sending her eastward with a speed of 4.64 m/s. Neglecting friction, determine the subsequent velocity of the boy.

Problem 17:

To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broom in an effort to scare the crow away, it pushes off the 670-gram feeder with a takeoff speed of 1.5 m/s. Determine the speed at which the feeder initially recoils backwards.

Problem 18:

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.

a. Determine the pre-collision momentum of the ball. b. Determine the post-collision momentum of the ball. c. Determine the momentum change of the ball. d. Determine the velocity change of the racket.

Problem 19:

Anna Litical and Noah Formula are doing The Cart and the Brick Lab. They drop a brick on a 2.6 kg cart moving at 28.2 cm/s. After the collision, the dropped brick and cart are moving together with a velocity of 15.7 cm/s. Determine the mass of the dropped brick.

Problem 20:

A Roller Derby exhibition recently came to town. They packed the gym for two consecutive weekend nights at South's field house. On Saturday evening, the 68-kg Anna Mosity was moving at 17 m/s when she collided with 76-kg Sandra Day O'Klobber who was moving forward at 12 m/s and directly in Anna's path. Anna jumped onto Sandra's back and the two continued moving together at the same speed. Determine their speed immediately after the collision.

Problem 21:

Ima Rilla Saari rushes to her car in order to hurry home and get dressed for work. Failing to realize the dangers of driving under slick and icy conditions, she collides her 940-kg Mazda Miata into the rear of a 2460-kg pick-up truck which was at rest at the light on Lake Avenue. Ima's pre-collision speed was 12.5 m/s. Determine the post-collision speed of the two entangled cars as they slide across the ice.

Problem 22:

In a goal line stand against New Greer Academy, South's linebackers Jerome (m=102 kg) and Michael (m=98 kg) meet the 84-kg halfback moving southward through the air at 6.4 m/s. Upon contact, Jerome and Michael are both moving at 3.6 m/s in the exact opposite direction. Determine the post-collision speed and direction of the collection of three players. Assume they move together after the collision.

Problem 23:

Hayden (m=24.3-kg) is gliding along the sidewalk on his skateboard at 8.6 ft/s. As he travels under a low-hanging branch of the tree where Matthew is sitting, he grabs the 4.5-kg bag of soccer balls from Matthew's hands. Determine the speed of Hayden immediately after grabbing the bag of soccer balls.

Problem 24:

Rex (m=86 kg) and Tex (92 kg) board the bumper cars at the local carnival. Rex is moving at a full speed of 2.05 m/s when he rear-ends Tex who is at rest in his path. Tex and his 125-kg car lunge forward at 1.40 m/s. Determine the post-collision speed of Rex and his 125-kg car.

Problem 25:

Abbey and Mia are in the basement playing pool. On Abbey's recent shot, the cue ball was moving east at 82 cm/s when it struck the slower 5-ball moving in the same direction at 24 cm/s. The 5-ball immediately speeds up to 52 cm/s. Determine the post-collision speed of the cue ball.

Problem 26:

Polly Ester and Ray Ahn are doing the Elastic Collision lab on a low-friction track. Cart A has a mass of 1.00 kg and is moving rightward at 27.6 cm/s prior to the collision with Cart B. Cart B has a mass of 0.50 kg and is moving leftward with a speed of 42.9 cm/s. After the magnetic repulsion of the two carts, Cart A is moving leftward at 10.1 cm/s. Determine the post-collision speed and direction of cart B.

Problem 27:

Bailey is on the tenth frame of her recent bowling competition and she needs to pick up the last pin for a spare and the first place trophy. She rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the collision. The pin went flying forward at 13.2 m/s. Determine the post-collision speed of the ball.

Problem 28:

Jack D. Ripper flipped out after missing a Must-Do-It question for the third time on his Minds On Physics assignment. Outraged by the futility of his efforts, he flings a 4.0-gram pencil across the room. The pencil lodges into a 221.0-gram Sponge Bob doll which is at rest on a countertop. Once in motion, the pencil/doll combination slide a distance of 11.9 cm across the countertop before stopping. The coefficient of friction between the doll and the countertop is 0.325. Determine the speed at which the pencil is moving prior to striking Sponge Bob.

Problem 29:

A physics student hurls a 315-gram ball directly into a 3.54-kg box which is at rest on a table top. The baseball strikes the box with a pre-impact speed of 54.1 m/s. The box is filled with towels to help absorb the blow  and effectively catch the ball. The coefficient of friction between the box and the table is 0.714. Determine the distance which the ball and box slide across the table after the collision.

Problem 30:

A 72-kg boy and a 48-kg girl, both wearing ice skates face each other at rest on a skating rink. The boy pushes the girl, sending her eastward with a speed of 6.8 m/s. When the impulse is completed, the boy and girl are a distance of 1.4 meters apart. Determine the distance of separation between the boy and the girl 5.0 seconds after the impulse is completed.

Problem 31:

The city police are in pursuit of Robin Banks after his recent holdup at the savings and loan. The high speed police chase ends at an intersection as a 2080-kg Ford Explorer (driven by Robin) traveling north at 32.6 m/s collides with a 18400-kg garbage truck moving east at 12.4 m/s. The Explorer and the garbage truck entangle together in the middle of the intersection and move as a single object. Determine the post-collision speed and direction of the two entangled vehicles.

Problem 32:

A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running west with a speed of 3.6 m/s. Assuming momentum conservation, determine the speed and direction of the two players immediately after the tackle.

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Ensemble of physics-informed neural networks for solving plane elasticity problems with examples

• Original Paper
• Published: 29 August 2024

Cite this article

• Aliki D. Mouratidou   ORCID: orcid.org/0000-0002-8382-1263 1 ,
• Georgios A. Drosopoulos 2 , 3 &
• Georgios E. Stavroulakis 1  

Two-dimensional (plane) elasticity equations in solid mechanics are solved numerically with the use of an ensemble of physics-informed neural networks (PINNs). The system of equations consists of the kinematic definitions, i.e. the strain–displacement relations, the equilibrium equations connecting a stress tensor with external loading forces and the isotropic constitutive relations for stress and strain tensors. Different boundary conditions for the strain tensor and displacements are considered. The proposed computational approach is based on principles of artificial intelligence and uses a developed open-source machine learning platform, scientific software Tensorflow, written in Python and Keras library, an application programming interface, intended for a deep learning. A deep learning is performed through training the physics-informed neural network model in order to fit the plain elasticity equations and given boundary conditions at collocation points. The numerical technique is tested on an example, where the exact solution is given. Two examples with plane stress problems are calculated with the proposed multi-PINN model. The numerical solution is compared with results obtained after using commercial finite element software. The numerical results have shown that an application of a multi-network approach is more beneficial in comparison with using a single PINN with many outputs. The derived results confirmed the efficiency of the introduced methodology. The proposed technique can be extended and applied to the structures with nonlinear material properties.

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The work of A.D.M. and G.E.S. has been supported by the Project Safe-Aorta, which was implemented in the framework of the Action “Flagship actions in interdisciplinary scientific fields with a special focus on the productive fabric”, through the National Recovery and Resilience Fund Greece 2.0 and funded by the European Union-NextGenerationEU (Project ID:TAEDR-0535983)

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School of Production Engineering and Management, Institute of Computational Mechanics and Optimization, Technical University of Crete, Kounoupidiana, 73100, Chania, Crete, Greece

Aliki D. Mouratidou & Georgios E. Stavroulakis

Discipline of Civil Engineering, School of Engineering and Computing, University of Central Lancashire, Preston campus, Preston, PR1 2HE, UK

Georgios A. Drosopoulos

Discipline of Civil Engineering, School of Engineering, University of KwaZulu-Natal, Durban campus, Durban, 4041, South Africa

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Mouratidou, A.D., Drosopoulos, G.A. & Stavroulakis, G.E. Ensemble of physics-informed neural networks for solving plane elasticity problems with examples. Acta Mech (2024). https://doi.org/10.1007/s00707-024-04053-3

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Received : 26 February 2024

Revised : 27 May 2024

Accepted : 31 July 2024

Published : 29 August 2024

DOI : https://doi.org/10.1007/s00707-024-04053-3

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The Solution to an Impulse Control Problem Motivated by Optimal Harvesting

We consider a stochastic impulse control problem that is motivated by applications such as the optimal exploitation of a natural resource. In particular, we consider a stochastic system whose uncontrolled state dynamics are modelled by a non-explosive positive linear diffusion. The control that can be applied to this system takes the form of one-sided impulsive action. The objective of the control problem is to maximise a discounted performance criterion that rewards the effect of control action but involves a fixed cost at each time of a control intervention. We derive the complete solution to this problem under general assumptions. It turns out that the solution can take four qualitatively different forms, several of which have not been observed in the literature. In two of the four cases, there exist only ε 𝜀 \varepsilon italic_ε -optimal control strategies. We also show that the boundary classification of 0 may play a critical role in the solution of the problem. Furthermore, we develop a way for establishing the strong solution to a stochastic impulse control problem’s optimally controlled SDE. Keywords : stochastic impulse control, linear diffusions, stochastic differential equations, optimal harvesting AMS 2020 subject classification : 93E20, 60J60, 60H10, 91B76

1 Introduction

We consider a stochastic dynamical system whose controlled state process is the strong solution to the SDE

where W 𝑊 W italic_W is a standard one-dimensional Brownian motion and ζ 𝜁 \zeta italic_ζ is a controlled càdlàg increasing piece-wise constant process. The objective of the optimisation problem is to maximise over all admissible processes ζ 𝜁 \zeta italic_ζ the performance criterion

where Δ ⁢ ζ t = ζ t − ζ t − Δ subscript 𝜁 𝑡 subscript 𝜁 𝑡 subscript 𝜁 limit-from 𝑡 \Delta\zeta_{t}=\zeta_{t}-\zeta_{t-} roman_Δ italic_ζ start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT = italic_ζ start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT - italic_ζ start_POSTSUBSCRIPT italic_t - end_POSTSUBSCRIPT , with the convention that ζ 0 − = 0 subscript 𝜁 limit-from 0 0 \zeta_{0-}=0 italic_ζ start_POSTSUBSCRIPT 0 - end_POSTSUBSCRIPT = 0 , and

Throughout the paper, we write 𝔼 x subscript 𝔼 𝑥 \operatorname{\mathbb{E}}_{x} blackboard_E start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT to denote expectation so that we account for the dependence of X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT on its initial value x 𝑥 x italic_x .

Stochastic impulse control problems arise in various fields. In the context of mathematical finance, economics and operations research, notable contributions include Harrison, Sellke and Tayor  [ 20 ] , Harrison and Taksar  [ 21 ] , Mundaca and Øksendal  [ 32 ] , Korn  [ 24 , 25 ] , Bielecki and Pliska  [ 9 ] , Cadenillas  [ 11 ] , Bar-Ilan, Sulem and Zanello  [ 7 ] , Bar-Ilan, Perry and Stadje  [ 6 ] , Ohnishi and Tsujimura  [ 33 ] , Cadenillas, Sarkar and Zapatero  [ 12 ] , LyVath, Mnif and Pham  [ 30 ] , and several references therein. Also, impulse control models motivated by the optimal management of a natural resource have been studied by Alvarez  [ 1 , 2 ] , Alvarez and Koskela  [ 4 ] and Alvarez and Lempa  [ 5 ] ; singular control versions of such models have been studied by Lungu and Øksendal  [ 29 ] , Framstad  [ 19 ] , Song, Stockbridge and Zhu  [ 40 ] , Alvarez and Hening  [ 3 ] and several references therein. In view of the wide range of applications, the general mathematical theory of stochastic impulse control is well-developed: apart from the contributions mentioned above, see also Richard  [ 39 ] , Stettner  [ 41 ] , Lepeltier and Marchal  [ 28 ] , Perthame  [ 36 ] , Egami  [ 18 ] , Davis, Guo and Wu  [ 16 ] , Djehiche, Hamadène and Hdhiri  [ 17 ] , Christensen  [ 13 ] , Helmes, Stockbridge and Zhu  [ 22 , 23 ] , Menaldi and Robin  [ 31 ] , Palczewski and Stettner  [ 35 ] , Christensen and Strauch  [ 14 ] , as well as the books by Bensoussan and Lions  [ 8 ] , Davis  [ 15 ] , Øksendal and Sulem  [ 34 ] , Pham  [ 37 ] , and several references therein.

The optimal management of a natural resource has motivated the problem that we study here. In this context, the state process X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT models the population density of a harvested species, while ζ t subscript 𝜁 𝑡 \zeta_{t} italic_ζ start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT is the cumulative amount of the species that has been harvested by time t 𝑡 t italic_t . The constant c > 0 𝑐 0 c>0 italic_c > 0 models a fixed cost associated with each harvesting cycle, while the function k 𝑘 k italic_k models the marginal profit arising from each harvest. Furthermore, the function h ℎ h italic_h models the utility of the harvested species, which could reflect the importance of the species to its associated ecosystem. Alternatively, the function h ℎ h italic_h can be used to model the revenue or cost associated with running the ecosystem. Relative to related references, such as the ones mentioned in the previous section, we generalise by considering (a) state-space discounting, (b) a state-dependent, rather than proportional, payoff associated with each harvest size, and (c) a running payoff such as the one modelled by the function h ℎ h italic_h . On the other hand, the assumptions that we make are of a rather similar nature.

In light of standard impulse control theory, a “ β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ ” strategy should be a prime candidate for an optimal one in the problem that we study here. Such a strategy is characterised by two points γ < β 𝛾 𝛽 \gamma<\beta italic_γ < italic_β in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ , which are both chosen by the controller, and can be described informally as follows. If the state process takes any value x ≥ β 𝑥 𝛽 x\geq\beta italic_x ≥ italic_β , then it is optimal for the controller to push it in an impulsive way down to level γ 𝛾 \gamma italic_γ . On the other hand, the controller should wait and take no action at all for as long as the state process takes values in the interval ] 0 , β [ ]0,\beta[ ] 0 , italic_β [ .

We show that a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy is indeed optimal, provided that a critical parameter x ¯ ¯ 𝑥 \underline{x} under¯ start_ARG italic_x end_ARG is finite and the fixed cost c 𝑐 c italic_c is sufficiently small (see Case I of Theorem  10 in Section  5 ). Otherwise, we show that only ε 𝜀 \varepsilon italic_ε -optimal strategies may exist (see Case II or Case IV of Theorem  10 ) or that never making an intervention may be optimal (see Case III of Theorem  10 ). The absence of an optimal strategy in Case IV of Theorem  10 is due to the relatively rapid growth of the function k 𝑘 k italic_k at infinity. It can therefore be eliminated if we make a suitable additional growth assumption. On the other hand, the absence of an optimal strategy in Case II of Theorem  10 is due to the nature of the problem that we solve.

The family of admissible controlled strategies that we consider do not allow for the state process to hit the boundary point 0 and be absorbed by it, which would amount to “switching off” the system. If we enlarged the set of admissible controls to allow for such a possibility and 0 were a natural boundary point, then we would face only the following difference: a β 𝛽 \beta italic_β -0 strategy would be optimal in Case II of Theorem  10 and we would not need to consider ε 𝜀 \varepsilon italic_ε -optimal strategies. On the other hand, the situation would be radically different if 0 were an entrance boundary point: in this case, β 𝛽 \beta italic_β -0 strategies would become an indispensable part of the optimal tactics. We discuss these observations more precisely in Remark  1 at the end of Section  5 . To the best of our knowledge, this is the first stochastic control problem in which the boundary classification of the problem’s state space has such a fundamental influence on the problem’s solution. We do not investigate this issue any further because this would require substantial extra analysis that would go beyond the scope of the present article.

The evolution of an impulse control problem’s state process is quite intuitive, provided that the corresponding uncontrolled dynamics are well-posed. For this reason, several references simply assume the existence of such processes. In the context of SDEs in ℝ d superscript ℝ 𝑑 \mathbb{R}^{d} blackboard_R start_POSTSUPERSCRIPT italic_d end_POSTSUPERSCRIPT , the state process of an impulse control problem can be derived by pasting together suitable strong solutions to the underlying uncontrolled SDE with random initial conditions (e.g., see Bensoussan and Lions  [ 8 , Section 6.1.1] ). In the context of general Markov processes, the classical construction of an impulse control strategy is substantially more technical and may involve countable products of canonical spaces (e.g., see Stettner  [ 41 ] and Lepeltier and Marchal  [ 28 ] ). If the uncontrolled state space process is a general Markov process with continuous sample paths, then comprehensive constructions of impulse control models have been derived by Helmes, Stockbridge and Zhu  [ 23 ] .

Impulse control problems with SDEs in ℝ d superscript ℝ 𝑑 \mathbb{R}^{d} blackboard_R start_POSTSUPERSCRIPT italic_d end_POSTSUPERSCRIPT can be formulated as in ( 1 )–( 3 ). In itself such a formulation is straightforward. Indeed, an SDE in ℝ d superscript ℝ 𝑑 \mathbb{R}^{d} blackboard_R start_POSTSUPERSCRIPT italic_d end_POSTSUPERSCRIPT such as ( 1 ) has a unique strong solution under suitable Lipschitz assumptions on b 𝑏 b italic_b and σ 𝜎 \sigma italic_σ for a wide class of controlled processes ζ 𝜁 \zeta italic_ζ (e.g., see Krylov  [ 26 , Theorem 2.5.7] ). On the other hand, a rigorous construction of an optimally controlled process ζ 𝜁 \zeta italic_ζ , such as a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy, is rather non-trivial. In the context of this paper, we construct a unique strong solution to the SDE ( 1 ) when the controlled process ζ 𝜁 \zeta italic_ζ is a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy (see Theorem  5 in Section  4 ). Despite the central role that such strategies play in stochastic impulse control, we are not aware of any such rigorous SDE result. Furthermore, this construction allows for a probabilistic derivation of the optimal expected discounted running reward as well as the optimal expected discounted reward from control exdenditure functionals (see ( 49 ) and ( 50 ) in Theorem  5 ). The construction that we make can most easily be adapted to derive the existence of strong solutions to optimally controlled SDEs that arise in other stochastic impulse control problems, even in dimensions higher than one.

The paper is organised as follows. Section  2 presents the precise formulation of the control problem that we solve, including all of the assumptions that we make. In Section  3 , we derive several results associated with a linear ODE that we need for the solution to the stochastic control problem we consider. In Section  4 , we prove that the SDE ( 1 ) has a unique strong solution when the controlled process ζ 𝜁 \zeta italic_ζ is a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy and we derive analytic expressions for certain associated functionals using probabilistic techniques. We derive the complete solution to the control problem that we consider in Section  5 . Finally, we present several examples illustrating the assumptions that we make and the results that we establish in Section  6 .

2 Formulation of the stochastic control problem

Fix a filtered probability space ( Ω , ℱ , ( ℱ t ) , ℙ ) Ω ℱ subscript ℱ 𝑡 ℙ \bigl{(}\Omega,\mathcal{F},(\mathcal{F}_{t}),\operatorname{\mathbb{P}}\bigr{)} ( roman_Ω , caligraphic_F , ( caligraphic_F start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ) , blackboard_P ) satisfying the usual conditions and carrying a standard one-dimensional ( ℱ t ) subscript ℱ 𝑡 (\mathcal{F}_{t}) ( caligraphic_F start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ) -Brownian motion W 𝑊 W italic_W . We consider a dynamical system, the uncontrolled stochastic dynamics of which are modelled by the SDE

and we make the following assumption.

Assumption 1

The functions b , σ : [ 0 , ∞ [ → ℝ b,\sigma:[0,\infty[\mbox{}\rightarrow\mathbb{R} italic_b , italic_σ : [ 0 , ∞ [ → blackboard_R are locally Lipschitz continuous and σ ⁢ ( x ) > 0 𝜎 𝑥 0 \sigma(x)>0 italic_σ ( italic_x ) > 0 for all x > 0 𝑥 0 x>0 italic_x > 0 .

This assumption implies that the scale function p 𝑝 p italic_p and the speed measure m 𝑚 m italic_m of the diffusion associated with the SDE ( 4 ), which are given by

are well-defined. Additionally, we make the following assumption on the boundary classification of the diffusion associated with ( 4 ).

Assumption 2

The boundary point 0 is inaccessible while the boundary point ∞ \infty ∞ is natural.

The state space of the linear diffusion associated with the SDE ( 4 ) is the interval ℐ = ] 0 , ∞ [ \mathcal{I}=\mbox{}]0,\infty[ caligraphic_I = ] 0 , ∞ [ . Recall that the boundary point p ∈ { 0 , ∞ } 𝑝 0 p\in\{0,\infty\} italic_p ∈ { 0 , ∞ } of ℐ ℐ \mathcal{I} caligraphic_I is called inaccessible if ℙ x ⁡ ( T p < ∞ ) = 0 subscript ℙ 𝑥 subscript 𝑇 𝑝 0 \operatorname{\mathbb{P}}_{x}(T_{p}<\infty)=0 blackboard_P start_POSTSUBSCRIPT italic_x end_POSTSUBSCRIPT ( italic_T start_POSTSUBSCRIPT italic_p end_POSTSUBSCRIPT < ∞ ) = 0 for all x ∈ ℐ 𝑥 ℐ x\in\mathcal{I} italic_x ∈ caligraphic_I and accessible otherwise. Furthermore, if the boundary p 𝑝 p italic_p is inaccessible, then it is natural if

and entrance otherwise, namely, if

(e.g., see Revuz and Yor  [ 38 , Definition VII.3.9] ). In these expressions, T y subscript 𝑇 𝑦 T_{y} italic_T start_POSTSUBSCRIPT italic_y end_POSTSUBSCRIPT is the first hitting time of the set { y } 𝑦 \{y\} { italic_y } , which is defined by

In Borodin and Salminen  [ 10 , II.1.6] , an inaccessible boundary point is called not-exit , while a natural (resp., entrance) boundary point is called natural (resp., entrance-not-exit ). Integral conditions for the classification of a boundary point p ∈ { 0 , ∞ } 𝑝 0 p\in\{0,\infty\} italic_p ∈ { 0 , ∞ } of ℐ ℐ \mathcal{I} caligraphic_I in terms of the scale function p 𝑝 p italic_p and the speed measure m 𝑚 m italic_m can be found in this reference.

We next consider the stochastic control problem defined by ( 1 )–( 3 ).

Definition 1

The family of all admissible controlled strategies is the set of all ( ℱ t ) subscript ℱ 𝑡 (\mathcal{F}_{t}) ( caligraphic_F start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ) -adapted càdlàg processes ζ 𝜁 \zeta italic_ζ with increasing and piece-wise constant sample paths such that the SDE ( 1 ) has a unique non-explosive strong solution and

Assumption 3

The discounting rate function r 𝑟 r italic_r is bounded and continuous. Also, there exists r 0 > 0 subscript 𝑟 0 0 r_{0}>0 italic_r start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT > 0 such that r ⁢ ( x ) ≥ r 0 𝑟 𝑥 subscript 𝑟 0 r(x)\geq r_{0} italic_r ( italic_x ) ≥ italic_r start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT for all x ≥ 0 𝑥 0 x\geq 0 italic_x ≥ 0 .

To complete the set of our assumptions, we consider the operator ℒ ℒ \mathscr{L} script_L acting on C 1 superscript 𝐶 1 C^{1} italic_C start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT functions with absolutely continuous first-order derivatives that is defined by

In the presence of Assumptions 1 , 2 and 3 , the second-order linear ODE ℒ ⁢ w ⁢ ( x ) = 0 ℒ 𝑤 𝑥 0 \mathscr{L}w(x)=0 script_L italic_w ( italic_x ) = 0 has two fundamental C 2 superscript 𝐶 2 C^{2} italic_C start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT solutions φ 𝜑 \varphi italic_φ and ψ 𝜓 \psi italic_ψ such that

If 0 is a natural boundary point, then

while, if 0 is an entrance boundary point, then

Symmetric results hold for the boundary point ∞ \infty ∞ (e.g., see Borodin and Salminen  [ 10 , II.10] ).

The functions φ 𝜑 \varphi italic_φ and ψ 𝜓 \psi italic_ψ admit the probabilistic representations

where Λ Λ \Lambda roman_Λ is defined by ( 3 ) with X 𝑋 X italic_X in place of X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT and T y subscript 𝑇 𝑦 T_{y} italic_T start_POSTSUBSCRIPT italic_y end_POSTSUBSCRIPT is defined by ( 7 ).

Furthermore, φ 𝜑 \varphi italic_φ and ψ 𝜓 \psi italic_ψ are such that

where C = φ ⁢ ( 1 ) ⁢ ψ ′ ⁢ ( 1 ) − φ ′ ⁢ ( 1 ) ⁢ ψ ⁢ ( 1 ) 𝐶 𝜑 1 superscript 𝜓 ′ 1 superscript 𝜑 ′ 1 𝜓 1 C=\varphi(1)\psi^{\prime}(1)-\varphi^{\prime}(1)\psi(1) italic_C = italic_φ ( 1 ) italic_ψ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ( 1 ) - italic_φ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ( 1 ) italic_ψ ( 1 ) and p 𝑝 p italic_p is the scale function defined by ( 5 ). To simplify the notation, we also define

Beyond involving standard integrability and growth assumptions, the conditions in the following assumption may appear involved. However, they are standard in the relevant literature and are satisfied by a wide range of problem data choices (see Examples  1 - 4 in Section  6 ).

Assumption 4

The following conditions hold true:

(ii) The function k 𝑘 k italic_k is absolutely continuous,

Furthermore,

(iii) If we define

then Θ Θ \Theta roman_Θ is continuous and there exists a constant ξ ∈ ] 0 , ∞ [ \xi\in\mbox{}]0,\infty[ italic_ξ ∈ ] 0 , ∞ [ such that the restriction of Θ / r Θ 𝑟 \Theta/r roman_Θ / italic_r in ] 0 , ξ [ ]0,\xi[ ] 0 , italic_ξ [ (resp., in ] ξ , ∞ [ ]\xi,\infty[ ] italic_ξ , ∞ [ ) is strictly increasing (resp., strictly decreasing).

3 Results associated with a linear ODE

Unless stated otherwise, the results in this section hold true if the coefficients of ( 4 ) satisfy the usual Engelbert and Schmidt conditions, rather than the stronger Assumption  1 , and the boundary points 0 0 , ∞ \infty ∞ are inaccessible. We start by recalling some standard results that we will need and can be found in, e.g., Lamberton and Zervos  [ 27 , Section 4] . Consider a Borel measurable function F : ] 0 , ∞ [ → ℝ F:\mbox{}]0,\infty[\rightarrow\mathbb{R} italic_F : ] 0 , ∞ [ → blackboard_R such that

where Λ Λ \Lambda roman_Λ is defined by ( 3 ) for X ζ = X superscript 𝑋 𝜁 𝑋 X^{\zeta}=X italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT = italic_X . This integrability condition is equivalent to

where Φ Φ \Phi roman_Φ and Ψ Ψ \Psi roman_Ψ are defined by ( 17 ). Given such a function F 𝐹 F italic_F , we define

The function R F subscript 𝑅 𝐹 R_{F} italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT admits the analytic presentation

ℒ subscript 𝑅 𝐹 𝐹 0 \mathscr{L}R_{F}+F=0 script_L italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT + italic_F = 0 . Furthermore,

Conversely, consider any function f : ] 0 , ∞ [ → ℝ f:\mbox{}]0,\infty[\rightarrow\mathbb{R} italic_f : ] 0 , ∞ [ → blackboard_R that is C 1 superscript 𝐶 1 C^{1} italic_C start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT with absolutely continuous first-order derivative and such that

Such a function is such that

Part (ii) of the following result will be important in appreciating the role that the boundary classification of 0 has on whether switching off the system might be optimal (see Remark  1 at the end of Section  5 ). In general, ( 27 ) is not true if 0 is an entrance boundary point (see ( 98 ) in Example  8 in Section  6 ).

Suppose that Assumptions  1 and  3 hold true. Also, suppose that the boundary points 0 and ∞ \infty ∞ of the diffusion associated with the SDE ( 4 ) are both inaccessible. Let F 𝐹 F italic_F be any Borel measurable function satisfying the equivalent integrability conditions ( 20 ) and ( 21 ), and consider the function R F subscript 𝑅 𝐹 R_{F} italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT defined by ( 22 ) and ( 23 ). The following statements hold true:

(i) Suppose that F 𝐹 F italic_F is bounded from below. If K 𝐾 K italic_K is any constant such that F ⁢ ( x ) / r ⁢ ( x ) ≥ K 𝐹 𝑥 𝑟 𝑥 𝐾 F(x)/r(x)\geq K italic_F ( italic_x ) / italic_r ( italic_x ) ≥ italic_K for all x > 0 𝑥 0 x>0 italic_x > 0 , then R F ⁢ ( x ) ≥ K subscript 𝑅 𝐹 𝑥 𝐾 R_{F}(x)\geq K italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT ( italic_x ) ≥ italic_K for all x > 0 𝑥 0 x>0 italic_x > 0 .

(ii) If 0 is a natural boundary point, then

Proof. Part (i) of the lemma follows immediately from the calculation

where we have used the definition ( 3 ) of Λ Λ \Lambda roman_Λ .

To establish part (ii) of the lemma suppose in what follows that 0 is a natural boundary point. Assuming that lim sup x ↓ 0 F ⁢ ( x ) / r ⁢ ( x ) ∈ ℝ subscript limit-supremum ↓ 𝑥 0 𝐹 𝑥 𝑟 𝑥 ℝ \limsup_{x\downarrow 0}F(x)/r(x)\in\mathbb{R} lim sup start_POSTSUBSCRIPT italic_x ↓ 0 end_POSTSUBSCRIPT italic_F ( italic_x ) / italic_r ( italic_x ) ∈ blackboard_R , fix any ε > 0 𝜀 0 \varepsilon>0 italic_ε > 0 and let x ε > 0 subscript 𝑥 𝜀 0 x_{\varepsilon}>0 italic_x start_POSTSUBSCRIPT italic_ε end_POSTSUBSCRIPT > 0 be any point such that

In view of ( 22 ), ( 23 ), the definition ( 3 ) of Λ Λ \Lambda roman_Λ and the second limit in ( 13 ), we can see that

Suppose that Assumptions  1 and  3 hold true, suppose that the boundary points 0 and ∞ \infty ∞ of the diffusion associated with the SDE ( 4 ) are both inaccessible and consider any Borel measurable function F 𝐹 F italic_F satisfying the equivalent integrability conditions ( 20 ) and ( 21 ). The function G F : ] 0 , ∞ [ → ℝ G_{F}:\mbox{}]0,\infty[\mbox{}\rightarrow\mathbb{R} italic_G start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT : ] 0 , ∞ [ → blackboard_R defined by

is such that

Furthermore, if the boundary point ∞ \infty ∞ is natural, then

Proof. We first note that the equality in ( 28 ) follows immediately from the definition ( 23 ) of R F subscript 𝑅 𝐹 R_{F} italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT and the identity ( 16 ). In view of ( 13 ) and ( 14 ), the assumption that the boundary point 0 is inaccessible implies that

This limit and the calculation

Similarly, the calculation

and the assumption that the boundary point ∞ \infty ∞ is inaccessible imply that

In view of ( 32 ) and the expression of G F subscript 𝐺 𝐹 G_{F} italic_G start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT on the right-hand side of ( 28 ), we can see that

These inequalities imply ( 29 ).

Combining these observations, we can see that

and ( 30 ) follows. □ □ \Box □

Suppose that Assumption  1 and  3 hold true. Also, suppose that the boundary points 0 and ∞ \infty ∞ of the diffusion associated with the SDE ( 4 ) are both inaccessible. Given any Borel measurable function F 𝐹 F italic_F satisfying the equivalent integrability conditions ( 20 ) and ( 21 ), if the boundary point 0 (resp., ∞ \infty ∞ ) is inaccessible, then

Furthermore, if there exists x † > 0 subscript 𝑥 † 0 x_{\dagger}>0 italic_x start_POSTSUBSCRIPT † end_POSTSUBSCRIPT > 0 ( resp., x † > 0 superscript 𝑥 † 0 x^{\dagger}>0 italic_x start_POSTSUPERSCRIPT † end_POSTSUPERSCRIPT > 0 ) such that the restriction of F / r 𝐹 𝑟 F/r italic_F / italic_r in ] 0 , x † [ ]0,x_{\dagger}[ ] 0 , italic_x start_POSTSUBSCRIPT † end_POSTSUBSCRIPT [ ( resp., ] x † , ∞ [ ]x^{\dagger},\infty[ ] italic_x start_POSTSUPERSCRIPT † end_POSTSUPERSCRIPT , ∞ [ ) is a monotone function, then

Proof. To establish the very first inequality in ( 35 ), we argue by contradiction. To this end, we assume that lim inf x ↓ 0 R F ′ ⁢ ( x ) / φ ′ ⁢ ( x ) > 0 subscript limit-infimum ↓ 𝑥 0 subscript superscript 𝑅 ′ 𝐹 𝑥 superscript 𝜑 ′ 𝑥 0 \liminf_{x\downarrow 0}R^{\prime}_{F}(x)/\varphi^{\prime}(x)>0 lim inf start_POSTSUBSCRIPT italic_x ↓ 0 end_POSTSUBSCRIPT italic_R start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT ( italic_x ) / italic_φ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ( italic_x ) > 0 , which implies that there exist ε > 0 𝜀 0 \varepsilon>0 italic_ε > 0 and x ε > 0 subscript 𝑥 𝜀 0 x_{\varepsilon}>0 italic_x start_POSTSUBSCRIPT italic_ε end_POSTSUBSCRIPT > 0 such that

To proceed further, we first note that ( 16 ) and the fact that ℒ ⁢ φ = ℒ ⁢ ψ = 0 ℒ 𝜑 ℒ 𝜓 0 \mathscr{L}\varphi=\mathscr{L}\psi=0 script_L italic_φ = script_L italic_ψ = 0 , where ℒ ℒ \mathscr{L} script_L is the differential operator defined by ( 9 ), imply that

In view of this observation and the definition ( 23 ) of R F subscript 𝑅 𝐹 R_{F} italic_R start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT , we can see that the function R F ′ / ψ ′ subscript superscript 𝑅 ′ 𝐹 superscript 𝜓 ′ R^{\prime}_{F}/\psi^{\prime} italic_R start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_F end_POSTSUBSCRIPT / italic_ψ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT is absolutely continuous with derivative

Now, suppose that there exists a point x † > 0 superscript 𝑥 † 0 x^{\dagger}>0 italic_x start_POSTSUPERSCRIPT † end_POSTSUPERSCRIPT > 0 such that F / r 𝐹 𝑟 F/r italic_F / italic_r is monotone in [ x † , ∞ [ [x^{\dagger},\infty[ [ italic_x start_POSTSUPERSCRIPT † end_POSTSUPERSCRIPT , ∞ [ . Given any points x 1 < x 2 subscript 𝑥 1 subscript 𝑥 2 x_{1}<x_{2} italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT in [ x † , ∞ [ [x^{\dagger},\infty[ [ italic_x start_POSTSUPERSCRIPT † end_POSTSUPERSCRIPT , ∞ [ , we use ( 32 ) to calculate

Finally, we can establish the other limit in ( 36 ) using symmetric arguments and ( 33 ). □ □ \Box □

Suppose that Assumption  1 and  3 hold true. Also, suppose that the boundary points 0 and ∞ \infty ∞ are both inaccessible. Given a function Θ Θ \Theta roman_Θ satisfying the conditions of Assumption  4 .(iii), as well as the equivalent integrability conditions ( 20 ) and ( 21 ), the following statements are true:

(i) There exists a unique x ¯ ∈ ] ξ , ∞ ] \underline{x}\in\mbox{}]\xi,\infty] under¯ start_ARG italic_x end_ARG ∈ ] italic_ξ , ∞ ] such that

where we adopt the convention ] ∞ , ∞ [ = ∅ ]\infty,\infty[\mbox{}=\emptyset ] ∞ , ∞ [ = ∅ .

In particular, this is the case if

(iii) If x ¯ < ∞ ¯ 𝑥 \underline{x}<\infty under¯ start_ARG italic_x end_ARG < ∞ and we define

with the usual convention that inf ∅ = ∞ infimum \inf\emptyset=\infty roman_inf ∅ = ∞ , then x ¯ > x ¯ ¯ 𝑥 ¯ 𝑥 \overline{x}>\underline{x} over¯ start_ARG italic_x end_ARG > under¯ start_ARG italic_x end_ARG ,

Proof. The limit in ( 40 ) follows from Lemma  3 and the assumption that Θ / r Θ 𝑟 \Theta/r roman_Θ / italic_r is strictly decreasing in ] ξ , ∞ [ ]\xi,\infty[ ] italic_ξ , ∞ [ . Using ( 32 ) and ( 38 ), we can see that

These expressions imply that

To establish the sufficient conditions in part (ii) of the lemma, we first use the integration by parts formula and ( 32 ) to observe that

The identity ( 32 ) implies that, given any constant K 𝐾 K italic_K ,

Θ 𝐾 𝑟 Q_{\Theta}=Q_{\Theta+Kr} italic_Q start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT = italic_Q start_POSTSUBSCRIPT roman_Θ + italic_K italic_r end_POSTSUBSCRIPT . If Θ / r Θ 𝑟 \Theta/r roman_Θ / italic_r satisfies the inequality in ( 42 ), then, for all K 𝐾 K italic_K such that

there exists η ( K ) ∈ ] ξ , ∞ [ \eta(K)\in\mbox{}]\xi,\infty[ italic_η ( italic_K ) ∈ ] italic_ξ , ∞ [ such that

It follows that

thanks to the fact that Q Θ subscript 𝑄 Θ Q_{\Theta} italic_Q start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT is strictly increasing in [ ξ , ∞ [ [\xi,\infty[ [ italic_ξ , ∞ [ .

On the other hand, we use ( 37 ) to calculate

In view of ( 36 ) and ( 47 ), these calculations and L’ Hôpital’s formula imply that

The implications in ( 45 ) follow from this analysis and the definition ( 43 ) of x ¯ ¯ 𝑥 \overline{x} over¯ start_ARG italic_x end_ARG . □ □ \Box □

4 The “ 𝜷 𝜷 \boldsymbol{\beta} bold_italic_β - 𝜸 𝜸 \boldsymbol{\gamma} bold_italic_γ ” strategy

In this section, we consider the β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy that is characterised by two points 0 < γ < β < ∞ 0 𝛾 𝛽 0<\gamma<\beta<\infty 0 < italic_γ < italic_β < ∞ and takes the following form. If the state process takes any value x ≥ β 𝑥 𝛽 x\geq\beta italic_x ≥ italic_β , the controller pushes it in an impulsive way down to the level γ 𝛾 \gamma italic_γ . For as long as the state process takes values inside the interval ] 0 , β [ ]0,\beta[ ] 0 , italic_β [ , the controller waits and takes no action. Accordingly, such a strategy is characterised by a controlled process ζ 𝜁 \zeta italic_ζ such that

where X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT is the associated solution to the SDE ( 1 ).

Suppose that Assumptions  1 and 3 hold true. Also, suppose that the boundary points 0 and ∞ \infty ∞ of the diffusion associated with the uncontrolled SDE ( 4 ) are both inaccessible. Given any points γ < β 𝛾 𝛽 \gamma<\beta italic_γ < italic_β in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ , there exists a controlled process ζ = ζ ⁢ ( β , γ ) 𝜁 𝜁 𝛽 𝛾 \zeta=\zeta(\beta,\gamma) italic_ζ = italic_ζ ( italic_β , italic_γ ) that is admissible in the sense of Definition  1 and is such that ( 48 ) holds true. Furthermore, given any x ∈ ] 0 , β [ x\in\mbox{}]0,\beta[ italic_x ∈ ] 0 , italic_β [ ,

Proof. We start with a recursive construction of the required process ζ 𝜁 \zeta italic_ζ and its associated solution to the SDE ( 4 ). To this end, we first consider any initial state x ∈ ] 0 , β [ x\in\mbox{}]0,\beta[ italic_x ∈ ] 0 , italic_β [ , we denote by X 1 superscript 𝑋 1 X^{1} italic_X start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT the solution to the uncontrolled SDE ( 4 ) and we define

subscript 𝜏 ℓ 𝑡 (\mathcal{F}_{\tau_{\ell}+t}) ( caligraphic_F start_POSTSUBSCRIPT italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT + italic_t end_POSTSUBSCRIPT ) -stopping time for all t ≥ 0 𝑡 0 t\geq 0 italic_t ≥ 0 ,

we can see that, on the event { τ ℓ < ∞ } subscript 𝜏 ℓ \{\tau_{\ell}<\infty\} { italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ } ,

where we have time changed the Lebesgue as well as the Itô integral (see Revuz and Yor  [ 38 , Propositions V.1.4, V.1.5] ). It follows that, if we define

Furthermore, we define

Also, we note that

Given the recursive construction we have just considered, we define

In view of ( 53 )–( 55 ), the process X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT given by ( 57 ) provides the unique solution to the SDE ( 1 ) for ζ 𝜁 \zeta italic_ζ being as in ( 57 ). Furthermore, these processes are such that ( 48 ) holds true. In the case that arises when the initial state x ≥ β 𝑥 𝛽 x\geq\beta italic_x ≥ italic_β , the only modification of the arguments above involves X 1 superscript 𝑋 1 X^{1} italic_X start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT being the solution to the uncontrolled SDE ( 4 ) for x = γ 𝑥 𝛾 x=\gamma italic_x = italic_γ and ζ 1 superscript 𝜁 1 \zeta^{1} italic_ζ start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT being the same as in ( 51 ) translated by adding the constant x − γ 𝑥 𝛾 x-\gamma italic_x - italic_γ to it.

ℓ 1 \widetilde{X}^{\ell+1} over~ start_ARG italic_X end_ARG start_POSTSUPERSCRIPT roman_ℓ + 1 end_POSTSUPERSCRIPT introduced at the beginning of the proof is independent of ℱ τ ℓ subscript ℱ subscript 𝜏 ℓ \mathcal{F}_{\tau_{\ell}} caligraphic_F start_POSTSUBSCRIPT italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT end_POSTSUBSCRIPT under the conditional probability measure ℙ ( ⋅ ∣ τ ℓ < ∞ ) \operatorname{\mathbb{P}}(\cdot\mid\tau_{\ell}<\infty) blackboard_P ( ⋅ ∣ italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ ) and its distribution under ℙ ( ⋅ ∣ τ ℓ < ∞ ) \operatorname{\mathbb{P}}(\cdot\mid\tau_{\ell}<\infty) blackboard_P ( ⋅ ∣ italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ ) is the same as the distribution of the solution X 𝑋 X italic_X to the uncontrolled SDE ( 4 ) with initial state X 0 = γ subscript 𝑋 0 𝛾 X_{0}=\gamma italic_X start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT = italic_γ under ℙ ℙ \operatorname{\mathbb{P}} blackboard_P . In particular,

\mathbb{R}_{+} blackboard_R start_POSTSUBSCRIPT + end_POSTSUBSCRIPT , where we denote by 𝔼 ℙ ( ⋅ ∣ τ ℓ < ∞ ) \operatorname{\mathbb{E}}^{\operatorname{\mathbb{P}}(\cdot\mid\tau_{\ell}<% \infty)} blackboard_E start_POSTSUPERSCRIPT blackboard_P ( ⋅ ∣ italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ ) end_POSTSUPERSCRIPT expectations computed under the conditional probability measure ℙ ( ⋅ ∣ τ ℓ < ∞ ) \operatorname{\mathbb{P}}(\cdot\mid\tau_{\ell}<\infty) blackboard_P ( ⋅ ∣ italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ ) . In view of these observations and the definition of conditional expectation,

To see this claim, we first note that the Radon-Nikodym derivative of ℙ ( ⋅ ∣ τ ℓ < ∞ ) \operatorname{\mathbb{P}}(\cdot\mid\tau_{\ell}<\infty) blackboard_P ( ⋅ ∣ italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT < ∞ ) with respect to ℙ ℙ \operatorname{\mathbb{P}} blackboard_P is given by

Given any event Γ ∈ ℱ τ ℓ Γ subscript ℱ subscript 𝜏 ℓ \Gamma\in\mathcal{F}_{\tau_{\ell}} roman_Γ ∈ caligraphic_F start_POSTSUBSCRIPT italic_τ start_POSTSUBSCRIPT roman_ℓ end_POSTSUBSCRIPT end_POSTSUBSCRIPT ,

and ( 58 ) follows.

In view of ( 52 )–( 58 ), we can see that

where Λ Λ \Lambda roman_Λ is defined by ( 3 ) with X 𝑋 X italic_X in place of X ζ superscript 𝑋 𝜁 X^{\zeta} italic_X start_POSTSUPERSCRIPT italic_ζ end_POSTSUPERSCRIPT and T β subscript 𝑇 𝛽 T_{\beta} italic_T start_POSTSUBSCRIPT italic_β end_POSTSUBSCRIPT is defined as in ( 7 ). Given any x ∈ ] 0 , β [ x\in\mbox{}]0,\beta[ italic_x ∈ ] 0 , italic_β [ , we iterate this result and use ( 15 ) to obtain

which establishes ( 50 ).

To show ( 49 ), we consider any x ∈ ] 0 , β [ x\in\mbox{}]0,\beta[ italic_x ∈ ] 0 , italic_β [ and we use ( 52 )–( 58 ) as well as ( 59 ) to derive the expression

Similarly, we can show that

Recalling the assumption that h ℎ h italic_h is bounded from below, we can use the monotone convergence theorem and these results to obtain

which proves ( 49 ). □ □ \Box □

5 The solution to the control problem

We will solve the control problem we have considered by deriving a C 1 superscript 𝐶 1 C^{1} italic_C start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT with absolutely continuous first-order derivative function w : ] 0 , ∞ [ → ℝ w:\mbox{}]0,\infty[\mbox{}\rightarrow\mathbb{R} italic_w : ] 0 , ∞ [ → blackboard_R that satisfies the HJB equation

Lebesgue-a.e. in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ . Given such a solution, the optimal strategy can be characterised as follows. The controller should wait and take no action for as long as the state process X 𝑋 X italic_X takes values in the interior of the set in which the ODE

is satisfied and should take immediate action with an impulse in the negative direction if the state process takes values in the set of all points x > 0 𝑥 0 x>0 italic_x > 0 such that

We first consider the possibility for a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy with γ < β 𝛾 𝛽 \gamma<\beta italic_γ < italic_β in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ to be optimal. The optimality of such a strategy is associated with a solution w 𝑤 w italic_w to the HJB equation ( 60 ) such that

To determine such a solution w 𝑤 w italic_w , we first consider the so-called “principle of smooth fit”, which requires that w ′ superscript 𝑤 ′ w^{\prime} italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT should be continuous, in particular, at the free-boundary point β 𝛽 \beta italic_β . This condition suggests the free-boundary equation

Next we consider the inequality

which is associated with impulsive action. For x = β 𝑥 𝛽 x=\beta italic_x = italic_β and z = β − u 𝑧 𝛽 𝑢 z=\beta-u italic_z = italic_β - italic_u , we can see that this implies that

This inequality and the identity

which follows from ( 62 ), can both be true if and only if the function u ↦ ∫ u β ( k ( s ) − w ′ ( s ) ) d s u\mapsto\int_{u}^{\beta}\bigr{(}k(s)-w^{\prime}(s)\bigr{)}\,\mathrm{d}s italic_u ↦ ∫ start_POSTSUBSCRIPT italic_u end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_β end_POSTSUPERSCRIPT ( italic_k ( italic_s ) - italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ( italic_s ) ) roman_d italic_s has a local maximum at γ 𝛾 \gamma italic_γ . This observation gives rise to the free-boundary condition

Every solution to ( 61 ) that can satisfy the so-called “transversality condition”, which is required for a solution w 𝑤 w italic_w to the HJB equation to identify with the control problem’s value function, is given by

for some constant A 𝐴 A italic_A , where R h subscript 𝑅 ℎ R_{h} italic_R start_POSTSUBSCRIPT italic_h end_POSTSUBSCRIPT is given by ( 22 ) and ( 23 ) for F = h 𝐹 ℎ F=h italic_F = italic_h . In view of the definition ( 19 ) of Θ Θ \Theta roman_Θ in Assumption  4 , the expression of R Θ subscript 𝑅 Θ R_{\Theta} italic_R start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT as in ( 23 ) and the representation ( 26 ), we can see that

\mathbb{R}_{+} blackboard_R start_POSTSUBSCRIPT + end_POSTSUBSCRIPT , thanks to the last condition in Assumption  4 .(ii) and ( 25 ). The identity ( 67 ) implies that ( 66 ) is equivalent to

Therefore, the solution to ( 61 ) that satisfies the boundary condition ( 63 ) is given by

Furthermore, the boundary conditions ( 65 ) and ( 64 ) are equivalent to

respectively, where

and G Θ subscript 𝐺 Θ G_{\Theta} italic_G start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT is defined by ( 28 ) in Lemma  2 .

The following result is about the solvability of the system of equations given by ( 70 ) for the unknowns γ 𝛾 \gamma italic_γ and β 𝛽 \beta italic_β . Note that Lemma  4 .(i) implies that a pair 0 ≤ γ < β < ∞ 0 𝛾 𝛽 0\leq\gamma<\beta<\infty 0 ≤ italic_γ < italic_β < ∞ satisfying the first equation in ( 70 ) might exist only if x ¯ < ∞ ¯ 𝑥 \underline{x}<\infty under¯ start_ARG italic_x end_ARG < ∞ .

Consider the stochastic control problem formulated in Section  2 and suppose that the point x ¯ ¯ 𝑥 \underline{x} under¯ start_ARG italic_x end_ARG introduced in Lemma  4 .(i) is finite. There exist a unique strictly decreasing function γ ⋆ : ] 0 , c ⋆ [ → ] 0 , x ¯ [ \gamma^{\star}:\mbox{}]0,c^{\star}[\mbox{}\rightarrow\mbox{}]0,\underline{x}[ italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT : ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ → ] 0 , under¯ start_ARG italic_x end_ARG [ and a unique strictly increasing function β ⋆ : ] 0 , c ⋆ [ → ] x ¯ , x ¯ [ \beta^{\star}:\mbox{}]0,c^{\star}[\mbox{}\rightarrow\mbox{}]\underline{x},% \overline{x}[ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT : ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ → ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ , where c ⋆ > 0 superscript 𝑐 ⋆ 0 c^{\star}>0 italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT > 0 is defined by ( 81 ) in the proof below and x ¯ ¯ 𝑥 \underline{x} under¯ start_ARG italic_x end_ARG , x ¯ ¯ 𝑥 \overline{x} over¯ start_ARG italic_x end_ARG are as in Lemma  4 , such that

for all c ∈ ] 0 , c ⋆ [ c\in\mbox{}]0,c^{\star}[ italic_c ∈ ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ . There exist no other points 0 < γ < β < ∞ 0 𝛾 𝛽 0<\gamma<\beta<\infty 0 < italic_γ < italic_β < ∞ satisfying the system of equations ( 70 ). The functions β ⋆ superscript 𝛽 ⋆ \beta^{\star} italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT and γ ⋆ superscript 𝛾 ⋆ \gamma^{\star} italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT are such that

Furthermore, c ⋆ < ∞ superscript 𝑐 ⋆ c^{\star}<\infty italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT < ∞ if and only if

Proof. In view of ( 40 ) and ( 43 ) in Lemma  4 , we can see that there exists a point γ ∈ ] 0 , β [ \gamma\in\mbox{}]0,\beta[ italic_γ ∈ ] 0 , italic_β [ such that the first equation in ( 70 ) holds true if and only if β ∈ ] x ¯ , x ¯ [ \beta\in\mbox{}]\underline{x},\overline{x}[ italic_β ∈ ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ , in which case, γ ∈ ] 0 , x ¯ [ \gamma\in\mbox{}]0,\underline{x}[ italic_γ ∈ ] 0 , under¯ start_ARG italic_x end_ARG [ . In particular, there exists a unique strictly decreasing function Γ : ] x ¯ , x ¯ [ → ] 0 , x ¯ [ \Gamma:\mbox{}]\underline{x},\overline{x}[\mbox{}\rightarrow\mbox{}]0,% \underline{x}[ roman_Γ : ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ → ] 0 , under¯ start_ARG italic_x end_ARG [ such that

It follows that the system of equations ( 70 ) has a unique solution γ < β 𝛾 𝛽 \gamma<\beta italic_γ < italic_β if and only if the equation

has a unique solution β ⋆ ( c ) ∈ ] x ¯ , x ¯ [ \beta^{\star}(c)\in\mbox{}]\underline{x},\overline{x}[ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) ∈ ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ . Using the first expression in ( 71 ), the identity in ( 76 ), the second of the inequalities in ( 77 ) and the fact that ψ 𝜓 \psi italic_ψ is strictly increasing, we calculate

Combining this result with the fact that

which follows from the first limit in ( 78 ), we can see that the equation F ⁢ ( Γ ⁢ ( β ) , β ) = − c 𝐹 Γ 𝛽 𝛽 𝑐 F\bigl{(}\Gamma(\beta),\beta\bigr{)}=-c italic_F ( roman_Γ ( italic_β ) , italic_β ) = - italic_c has a unique solution β ⋆ ( c ) ∈ ] x ¯ , x ¯ [ \beta^{\star}(c)\in\mbox{}]\underline{x},\overline{x}[ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) ∈ ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ if and only if

We conclude this part of the analysis by noting that the points β ⋆ ( c ) ∈ ] x ¯ , x ¯ [ \beta^{\star}(c)\in\mbox{}]\underline{x},\overline{x}[ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) ∈ ] under¯ start_ARG italic_x end_ARG , over¯ start_ARG italic_x end_ARG [ and γ ⋆ ( c ) := Γ ( β ⋆ ( c ) ) ∈ ] 0 , x ¯ [ \gamma^{\star}(c):=\Gamma\bigl{(}\beta^{\star}(c)\bigr{)}\in\mbox{}]0,% \underline{x}[ italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) := roman_Γ ( italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) ) ∈ ] 0 , under¯ start_ARG italic_x end_ARG [ provide the unique solution to the system of equations ( 70 ) if c ∈ ] 0 , c ⋆ [ c\in\mbox{}]0,c^{\star}[ italic_c ∈ ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ , while the system of equations ( 70 ) has no solution such that 0 < γ < β < ∞ 0 𝛾 𝛽 0<\gamma<\beta<\infty 0 < italic_γ < italic_β < ∞ if c ≥ c ⋆ 𝑐 superscript 𝑐 ⋆ c\geq c^{\star} italic_c ≥ italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT . In particular, the inequalities in ( 72 ) follow from the corresponding ones in ( 76 ).

The fact that the function β ↦ F ⁢ ( Γ ⁢ ( β ) , β ) maps-to 𝛽 𝐹 Γ 𝛽 𝛽 \beta\mapsto F\bigl{(}\Gamma(\beta),\beta\bigr{)} italic_β ↦ italic_F ( roman_Γ ( italic_β ) , italic_β ) is strictly decreasing, which we have established above, implies that the function c ↦ β ⋆ ⁢ ( c ) maps-to 𝑐 superscript 𝛽 ⋆ 𝑐 c\mapsto\beta^{\star}(c) italic_c ↦ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) is strictly increasing because β ⋆ ⁢ ( c ) superscript 𝛽 ⋆ 𝑐 \beta^{\star}(c) italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) is the unique solution to equation ( 79 ) for each c ∈ ] 0 , c ⋆ [ c\in\mbox{}]0,c^{\star}[ italic_c ∈ ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ . In turn, this result and the fact that Γ Γ \Gamma roman_Γ is strictly decreasing imply that the function γ ⋆ = Γ ∘ β ⋆ superscript 𝛾 ⋆ Γ superscript 𝛽 ⋆ \gamma^{\star}=\Gamma\circ\beta^{\star} italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT = roman_Γ ∘ italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT is strictly decreasing. The first limit in ( 74 ) follows immediately from ( 81 ). On the other hand, the second limit in ( 74 ) follows immediately from the the first limit in ( 74 ) and the second limit in ( 78 ). Furthermore, the identities in ( 73 ) follow from the first limit in ( 78 ) and ( 80 ).

To establish the equivalence of the inequality c ⋆ < ∞ superscript 𝑐 ⋆ c^{\star}<\infty italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT < ∞ with the condition in ( 75 ), we first use the first expression of F 𝐹 F italic_F in ( 71 ) and the definition ( 81 ) of c ⋆ superscript 𝑐 ⋆ c^{\star} italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT to observe that

We next use the second limit in ( 78 ) as well as Lemmas  2 and  4 . If x ¯ < ∞ ¯ 𝑥 \overline{x}<\infty over¯ start_ARG italic_x end_ARG < ∞ , then

and the proof is complete. □ □ \Box □

In light of ( 62 ), ( 69 ) and the previous lemma, we now establish the following result, which provides the solution to the HJB equation ( 60 ) identifying with the control problem’s value function when a β 𝛽 \beta italic_β - γ 𝛾 \gamma italic_γ strategy with γ < β 𝛾 𝛽 \gamma<\beta italic_γ < italic_β in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ is indeed optimal.

Consider the stochastic control problem formulated in Section  2 and suppose that the point x ¯ ¯ 𝑥 \underline{x} under¯ start_ARG italic_x end_ARG introduced in Lemma  4 .(i) is finite. Also, fix any c ∈ ] 0 , c ⋆ [ c\in\mbox{}]0,c^{\star}[ italic_c ∈ ] 0 , italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT [ , where c ⋆ > 0 superscript 𝑐 ⋆ 0 c^{\star}>0 italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT > 0 is as in Lemma  6 . The function w 𝑤 w italic_w defined by

where we write γ ⋆ superscript 𝛾 ⋆ \gamma^{\star} italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT and β ⋆ superscript 𝛽 ⋆ \beta^{\star} italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT in place of the points γ ⋆ ⁢ ( c ) superscript 𝛾 ⋆ 𝑐 \gamma^{\star}(c) italic_γ start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) and β ⋆ ⁢ ( c ) superscript 𝛽 ⋆ 𝑐 \beta^{\star}(c) italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ( italic_c ) given by Lemma  6 , is C 1 superscript 𝐶 1 C^{1} italic_C start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ and C 2 superscript 𝐶 2 C^{2} italic_C start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT in ] 0 , ∞ [ ∖ { β ⋆ } ]0,\infty[\mbox{}\setminus\{\beta^{\star}\} ] 0 , ∞ [ ∖ { italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT } . Furthermore, this function is a solution to the HJB equation ( 60 ) that is bounded from below.

Proof. The boundedness from below of w 𝑤 w italic_w follows immediately from Assumption  3 , the conditions in (i) and (ii) of Assumption  4 and Lemma  1 .(i).

By construction, we will establish all of the lemma’s other claims if we prove that

To this end, we use the first expression of w 𝑤 w italic_w in ( 69 ) and ( 72 ) to note that

The inequality ( 83 ) follows from this observation and the fact that

To show ( 84 ), we first use the expression

the definition ( 19 ) of Θ Θ \Theta roman_Θ in Assumption  4 and the first expression in ( 69 ) to calculate

where G Θ subscript 𝐺 Θ G_{\Theta} italic_G start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT is given by ( 28 ) in Lemma  2 for F = Θ 𝐹 Θ F=\Theta italic_F = roman_Θ . In view of the calculations

the inequalities ( 40 ) in Lemma  4 and the fact that β ⋆ > x ¯ superscript 𝛽 ⋆ ¯ 𝑥 \beta^{\star}>\underline{x} italic_β start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT > under¯ start_ARG italic_x end_ARG , we can see that

The second of these inequalities and the second expression of G Θ subscript 𝐺 Θ G_{\Theta} italic_G start_POSTSUBSCRIPT roman_Θ end_POSTSUBSCRIPT in ( 28 ) imply that

However, this result, ( 85 ) and the first inequality in ( 86 ) yield

and ( 84 ) follows. □ □ \Box □

To proceed further, we assume that the problem data is such that c ⋆ < ∞ superscript 𝑐 ⋆ c^{\star}<\infty italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT < ∞ , which is the case if and only if one of the two conditions of ( 75 ) in Lemma  6 holds true. In the first case, when x ¯ < ∞ ¯ 𝑥 \overline{x}<\infty over¯ start_ARG italic_x end_ARG < ∞ , the limits in ( 74 ) suggest the possibility for the function w 𝑤 w italic_w defined by ( 62 ) and ( 69 ) for γ = 0 𝛾 0 \gamma=0 italic_γ = 0 and some β > x ¯ 𝛽 ¯ 𝑥 \beta>\overline{x} italic_β > over¯ start_ARG italic_x end_ARG to provide a solution to the HJB equation ( 60 ) that identifies with the control problem’s value function. In this case, a free-boundary condition such as ( 65 ) is not relevant anymore and we are faced with only the free-boundary condition ( 64 ) with γ = 0 𝛾 0 \gamma=0 italic_γ = 0 , which is equivalent to the equation F ⁢ ( 0 , β ) = − c 𝐹 0 𝛽 𝑐 F(0,\beta)=-c italic_F ( 0 , italic_β ) = - italic_c , where F 𝐹 F italic_F is defined by ( 71 ).

Consider the stochastic control problem formulated in Section  2 and suppose that the point x ¯ ¯ 𝑥 \underline{x} under¯ start_ARG italic_x end_ARG introduced in Lemma  4 .(i) is finite. Also, suppose that the problem data is such that x ¯ < ∞ ¯ 𝑥 \overline{x}<\infty over¯ start_ARG italic_x end_ARG < ∞ , where x ¯ ¯ 𝑥 \overline{x} over¯ start_ARG italic_x end_ARG is defined by ( 43 ) in Lemma  4 . The following statements hold true:

(I) There exists c ∘ ∈ ] c ⋆ , ∞ ] c^{\circ}\in\mbox{}]c^{\star},\infty] italic_c start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT ∈ ] italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT , ∞ ] and a strictly increasing function β ∘ : [ c ⋆ , c ∘ [ → [ x ¯ , ∞ [ \beta^{\circ}:[c^{\star},c^{\circ}[\mbox{}\rightarrow[\overline{x},\infty[ italic_β start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT : [ italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT , italic_c start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT [ → [ over¯ start_ARG italic_x end_ARG , ∞ [ such that

where c ⋆ ∈ ] 0 , ∞ [ c^{\star}\in\mbox{}]0,\infty[ italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ∈ ] 0 , ∞ [ is as in Lemma  6 .

(III) Given any c ∈ [ c ⋆ , c ∘ [ c\in[c^{\star},c^{\circ}[ italic_c ∈ [ italic_c start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT , italic_c start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT [ , the function w 𝑤 w italic_w defined by

where we write β ∘ superscript 𝛽 \beta^{\circ} italic_β start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT in place of β ∘ ⁢ ( c ) superscript 𝛽 𝑐 \beta^{\circ}(c) italic_β start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT ( italic_c ) , is C 1 superscript 𝐶 1 C^{1} italic_C start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT in ] 0 , ∞ [ ]0,\infty[ ] 0 , ∞ [ and C 2 superscript 𝐶 2 C^{2} italic_C start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT in ] 0 , ∞ [ ∖ { β ∘ } ]0,\infty[\mbox{}\setminus\{\beta^{\circ}\} ] 0 , ∞ [ ∖ { italic_β start_POSTSUPERSCRIPT ∘ end_POSTSUPERSCRIPT } . Furthermore, this function is a solution to the HJB equation ( 60 ) that is bounded from below.

The second expression in ( 72 ) and the limits in ( 74 ) imply that

Part (I) of the lemma follows from this observation and the calculation