Journal of Applied Mathematics

Volume 2014 (2014), Article ID 127809, 13 pages

http://dx.doi.org/10.1155/2014/127809

Research Article

Quenching Time Optimal Control for Some Ordinary Differential Equations

Ping Lin

School of Mathematics & Statistics, Northeast Normal University, Changchun 130024, China

Received 23 January 2014; Accepted 28 February 2014; Published 10 April 2014

Academic Editor: Alberto Cabada

Copyright © 2014 Ping Lin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

This paper concerns time optimal control problems of three different ordinary differential equations in . Corresponding to certain initial data and controls, the solutions of the systems quench at finite time. The goal to control the systems is to minimize the quenching time. The purpose of this study is to obtain the existence and the Pontryagin maximum principle of optimal controls. The methods used in this paper adapt to more general and complex ordinary differential control systems with quenching property. We also wish that our results could be extended to the same issue for parabolic equations.

1. Introduction

In this paper, we study some quenching time optimal control problems of three different ordinary differential equations in . First of all, some notations will be introduced. We use and to stand for the Euclidean norm and the inner product of . For each matrix , we use and to denote its transposition and the operator norm, respectively. Let be a nontrivial matrix-value function in the space . Write for . For each given, we set

Each can be expressed as . Let

For each -function , its derivative will be written as

where with , .

The controlled systems under consideration are as follows:

Here, , , and , where

(i), , and ;(ii), , and ;(iii), , and .

Let

Define

Since, for each , is continuously differential over the domain , it is clear that, given , , and , the controlled system (4) has a unique solution. We denote this solution by and write for its maximal interval of existence.

It is shown in Section 2 that, given , , and , there exists a time with holding the property that

We say that the solution quenches at the finite time and is the quenching time of the solution .

The purpose of this paper is to study the existence and the Pontryagin maximum principle for the following time optimal control problems:

Because of (7), for each and , there exists a number in such that

which is called the optimal time for the problem . A control , in the set holding the property: , is called an optimal control, while the solution is called the optimal state corresponding to for the problem . We will simply write for the optimal state .

The main results of this paper are as follows.

Theorem 1. Given and , the problem has optimal controls.

Theorem 2. Let . Then, Pontryagin’s maximum principle holds for the problem . Namely, if is the optimal time, is an optimal control, and is the corresponding optimal state for the problem , then there is a nontrivial function in the space satisfying

Besides, it holds that

Theorem 3. Let . Then, Pontryagin’s maximum principle holds for the problem . Namely, if is the optimal time, is an optimal control, and is the corresponding optimal state for the problem , then there is a nontrivial function in the space satisfying

Besides, it holds that

It is worth mentioning that problem is more complicated than the other two problems since the target set of problem is more complicated than the target sets of the other two problems. Indeed, the target set of problem is , while the target sets for problem and problem are and , respectively. This is why we only obtain the existence but not Pontryagin’s maximum principle for problem .

The concept of quenching was first introduced by Kawarada in [1] for a nonlinear parabolic equation; it has more general sense than blowup. For instance, consider the equation , . It is obvious that the first quenching time of the solution for this equation is 1/24 at which . Then the solution can be extended, for instance, until . Further, the solution can be extended again until ; that is, the second quenching time is 3/8. This differs from the blowup ordinary differential equations, where the solutions tend to infinity at blowup time.

There have been many literatures concerning the properties of parabolic differential equations with quenching behavior (see [1, 2] and references therein). To the best of our knowledge, the study on quenching time optimal control problems has not been touched upon. In this paper, we focus on quenching time optimal control problems for ordinary differential equations with three particular vector fields , , in . Indeed, the methods used in this paper adapt to more general and complex ordinary differential control systems with quenching property. For instance, we can use the similar methods we use in this paper to consider the quenching time optimal problem for system (4), where , (we will give the details in the following section). We also wish that our results could be extended to the same issue for parabolic equations.

The differential systems whose solutions have the behavior of quenching arise in the study of chemical reactions. Quenching phenomena could describe the completion of a chemical reaction. Using catalysts to make a chemical reaction complete in the shortest time could be considered an optimal quenching time problem. It is also significant in the theory studies of the electric current transient phenomena in polarized ionic conductors. In certain cases, the quenching of a solution is desirable. Thus, it could be interesting to minimize the quenching time with the aid of controls in certain cases. It deserves to mention that the quenching time , with some control , can really be strictly less than . Here is an example. Consider the problem , where and for all . It can be directly checked that .

Because of the quenching behavior of solutions to system (4), the usual methods applied to solve the general optimal control problems (see, for instance, [3–7]) do not work for problem with and . We approach our main results by the following steps. First, we show some invariant properties for solutions of system (4). Next, we prove that, given and , the corresponding solution of system (4) quenches at finite time for each control . Then we give the quenching rate estimate for solutions of system (4). Furthermore, we obtain the following property. When a sequence of controls tends to a control in a suitable topology, the solutions with sufficiently large share a common interval of nonquenching with the solution . Finally, we use the above-mentioned results to verify our main theorems.

Since blowup could be regarded as a special quenching phenomenon, it deserves to mention the paper [8], where minimal blowup time optimal control problems of ordinary differential equations were studied. Inspired by the idea in [8], we develop new methods in this paper.

The rest of the paper is organized as follows. Section 2 presents some preliminary lemmas which supply some properties of solutions to controlled system (4). Section 3 proves the existence of optimal controls for problem with and . Section 4 verifies Theorems 2 and 3.

2. Preliminary Lemmas

In this section, we will introduce some properties of solutions to system (4), which will play important roles to prove our main results.

2.1. The Existence and Invariant Property

Consider the system

Since, for each , is continuously differential over the domain , it is clear that, given , , , and , system (15) has a unique solution. We denote this solution by and write for its maximal interval of existence.

Let , and . If ; then, by the continuity of the solution , we can find a positive number sufficiently small such that for all . Similarly, if , we can find a positive number sufficiently small such that for all . Given , , and , we will use the notation to denote such a subinterval of the interval . When , denotes the maximal time interval in which , while, when , denotes the maximal time interval in which . By the continuity of the solution again, it is clear from the existence theorem and the extension theorem of ordinary differential equations that is a left closed and right open interval, whose left end point is . Let , , , and .

We have the following lemma.

Lemma 4. Given , , and , then, for all ,

Proof. First, we claim the following property (A). Suppose that , , with , , and . Then, there is a positive number with such that the solution holds the following inequalities: and for all in the interval .

Indeed, since and , we can use the continuity of the solution to find a positive constant such that , , and for all . Hence, it follows from (15) with , (5), and the inequality that, for all ,

Here, we simply write for the solution . From the two inequalities mentioned above and the inequalities and , we get the property (A).

Now, we come back to prove the property (16).

By seeking a contradiction, suppose that there exist a time , initial data , a control , and a number pair with and such that the solution with and does not satisfy (16). Then we would find a number in the interval such that or . Write for and write for . We may as well assume that . Since is continuous over , we can use the definition of and and the inequality to derive the following properties: (a) and (b) corresponding to each with , there exists a number in such that ; (c) , for each . Write . Then, it follows from the properties (a), (c), and the definition of the interval that and . Consider the following system:

Making use of the property (A), we can choose a positive number sufficiently small such that

On the other hand, the function also satisfies the system (19) when . By the uniqueness of the solution to the system (19), we necessarily have for all . This, combined with the above-mentioned property (b), implies the existence of a number in such that . This contradicts to (20). Therefore, we have accomplished the proof of the property (16).

We can use the very similar argument of the proof of (16) to prove (17) (see [9] for the detailed proofs).

The following two lemmas play important roles in the proof of Theorem 1 with and and Theorem 3 (see [9] for the detailed proofs).

Let , , and ; we will use the notation to denote such a subinterval of the interval . When , denotes the maximal time interval in which , while, when , denotes the maximal time interval in which . It is clear that is a left closed and right open interval, whose left end point is . Let and .

We have the following lemma.

Lemma 5. Given , , and , then, for all ,

Let , , and ; we will use the notation to denote such a subinterval of the interval . When and , denotes the maximal time interval in which and , while, when and , denotes the maximal time interval in which and . It is clear that is a left closed and right open interval, whose left end point is . Let and .

We have the following lemma.

Lemma 6. Given , , and , then, for all ,

2.2. Quenching Property and Estimate of Quenching Rate

Lemma 7. For each and each , there exists a time in the interval holding the property that

Moreover, there exists a positive constant , independent of and , such that

Proof. Suppose that and . Let be the right end point of the interval . Thus, and (see the definition and the property of on page 3). We will prove Lemma 7 in the following two cases.

Case 1. and .

Step 1. This is to prove (23) in Case 1.

By property (16) in Lemma 4 and the definition of and , the solution with , , and holds the property that, for each ,

Then, from system (4) with , (5), and the inequality , it holds that

Here and throughout the proof, we simply write for .

Let be the solution to the following equation:

from which it is easy to check that

Furthermore, making use of (28) and (29), we can derive, from the comparison theorem of ordinary differential equations, the following inequality:

This, combined with (27) and (30), implies (23) in Case 1.

Step 2. This is to prove (24) in Case 1.

Indeed, by (27) and (28), it is clear that is monotonously increasing over the interval . This, combined with (27), implies that exists and .

Now, we claim that

By contradiction, if , then, by the continuity and the monotonicity of the solution , it holds that for and . Then, we can extend the solution and can find an interval with sufficiently small such that on the interval . However, and is the maximal interval in which . This is a contradiction. Thus, (32) holds.

On the other hand, by (27), it is clear that

from which and from (32), it holds that . This completes the proof of (24) in Case 1.

Step 3. This is to prove (25) in Case 1.

Since for each and is continuous over the interval , we can get (25) in Case 1.

Step  4. This is to prove (26) in Case 1.

By system (4) with , we conclude from (24), (27), and (5) that, for each ,

which implies (26) in Case 1.

Case  2. with and .

We can use the very similar argument as that of Case 1 to give a proof of Case 2 (see [9] for the detailed proofs).

Remark 8. Let and . Since , we can conclude from the definition of that, for each ,

The following two lemmas play important roles in the proof of Theorem 1 with and and Theorem 3 (see [9] for the detailed proofs).

Lemma 9. For each and each , there exists a time in the interval holding the property that

Moreover, there exists a positive constant , independent of and , such that

Lemma 10. For each and each , there exists a time in the interval holding the property that

Moreover, there exists a positive constant depending on , but independent of , such that, for each ,

2.3. Uniform Interval of Nonquenching

Let be a time interval. Given , and , we say that the solution does not quench on the interval if for each we have

Lemma 11. Suppose that and is a closed interval. Assume . Let and be an element and a bounded sequence in , respectively. Suppose that

and the solution does not quench on the interval . Then there is a natural number such that, for each with , the solution does not quench on the interval .

Using the lemmas we obtained in Sections 2.1 and 2.2, we can use the similar argument as we used in Lemma 2.2 of [8] to prove this lemma (see [9] for the detailed proofs.)

3. Existence of Optimal Control

Proof of Theorem 1. In this section, we will only give the proof of Theorem 1 in the case, where with and . We can use very similar arguments to prove Theorem 1 in the cases, where with , , , and .

Let with and . By Lemma 7 and Remark 8, it holds that, for each , . Moreover,

Thus

Then, we can utilize the definitions of to get a sequence of controls in the set holding the following properties. Each is a positive number;    and as ; for all .

For each , write simply for . Now, we will complete the proof by the following two steps.

Step 1. This is to prove .

Indeed, by Lemma 4 and Remark 8, we have

By system (4) with , it holds that, for each ,

from which and from (44) and (5), we get that, for each ,

Here and in what follows, is a positive constant independent of and , which may be different in different context. Then, by the property (2) held by sequence and Gronwall’s inequality, we derive, that for each ,

On the other hand, from system (4) with , (47), (26) in Lemma 7, and (5), it holds that, for each ,

If , then, from the property (2) held by sequence , we have the fact that the right side of the above inequality tends to as . This, together with the inequalities (48) and , implies that we can find a natural number and a positive number such that for all , which contradicts the property (3) held by sequence . This completes the proof of Step 1.

Step 2. This contains the existence of optimal control for the problem .

Fix such a number such that . It is clear that there exist a function in and a subsequence of the sequence , still denoted in the same way, such that

We extend the function by setting it to be zero on the interval and denote the extension by again. Obviously, this extended function is in the set .

Now, we will prove that is an optimal control for the problem . We will carry out its proof by the following two claims.

Claim One. By the definition of , it is obvious that the solution does not quench at any time in the interval .

Claim Two. . From the definition of , it holds that . By seeking a contradiction, suppose that . Then we would find a number with such that the solution does not quench on the interval . Thus, it follows from (49) and Lemma 11 that there exists a natural number such that, when , the solution does not quench on the interval . Thus, , when . Now, according to (26) in Lemma 7, we can easily verify that, for all ,

where is independent of . This, together with the property (2) held by , gives a positive constant independent of such that , for all . This contradicts with the property (3) by .

Thus, we have completed the proof of Theorem 1.

4. Proof of Pontryagin Maximum Principle

Proof of Theorem 2. We will only give the proof of Theorem 2 in the case, where with and . We can use very similar arguments to prove Theorem 2 in the case, where with and .

Let with and . We will prove the theorem in a series of steps as follows.

Step 1. This is to set up a penalty functional and to study the related properties.

Since the number is the optimal time for the problem , the solution , corresponding to each in the set , does not quench on , for any . Moreover, for all . Let be a fixed number such that . Write for the set . We introduce the Ekeland distance over the set by setting

Here and in what follows, stands for the Lebesgue measure of a measurable set in . Then forms a completed metric space (see [5], page 145). For each , we define a penalty functional by setting

We claim that is continuous over the space . Before moving forward to the proof of this claim, we make the following observation, which will be often used in what follows. Since , , and with , it holds that, for each with ,

Now, we come back to prove the above claim. Let be an element and let be a sequence in the space such that as . Then, it is clear that strongly in as . Throughout this proof, we will write simply for and, for each , we will write simply for ; is a constant independent of and , which may be different in different context. Since for all , it holds from the system (4) with that

From Lemmas 4 and 7 and Remark 8, we obtain that, for each ,

for all and for each ,

By the property that for all again, (55), and (56), it holds that

On the other hand, because is continuous over , we get and . This, together with (54), (55), (56), and (57), implies that, for each and ,

Now, we can apply Gronwall’s inequality to get that, as ,

Hence, we have proved the continuity of the functional .

Step 2. This is to apply the Ekeland variational principle.

It is clear that

Then, we can utilize Ekeland’s variational principle (see, for instance, [5], page 136-137) to find a control enjoying the following properties:

Let . By the variant of the Lyapunov theorem (see, for instance, [5], Chapter 4), we can get, corresponding to each , a measurable set in the interval such that and

Now, we construct the following spike function of with respect to by setting

It is obvious that the control is in . Write and simply for the solutions and , respectively. Clearly, they do not quench on the interval . Set . Then, for each , it holds that

Step 3. This is to show the uniform convergence of the family on the interval for .

It follows from (63) that as . Thus, we can use the same argument in the proof of (59) to get

On the other hand, making use of (62) and (63), it holds that, for all ,

Here, the function has the property for all .

Let be the unique solution to the following system:

Then, by (64) and (67), we get that, for each ,

Corresponding to each , write for the sum of the last two terms in the right hand of (68).

Clearly, it follows from (65) and (66) that uniformly on as .

Since is continuous and for all (see Lemma 4 and Remark 8), there exists a constant with such that for each . Then, it follows from (65) that, when is sufficiently small, we have for all . Hence, when is sufficiently small, it holds that, for each ,

On the other hand, from the continuity of and (65), we get that, when is sufficiently small,

Here and in what follows, is a constant independent of and , which may be different in different context. From the above inequality, (53), and (69), it follows that, when is sufficiently small,

from which and from (68), it holds from the Gronwall’s inequality that, when is sufficiently small,

Hence, it holds that uniformly on as , from which, we obtain that

Step 4. This is to get certain necessary conditions for the control .

By the second inequality of (61) and according to the definition of the functional , we can easily verify the following inequality:

This, together with (65) and (73), implies that

Let be the unique solution for the dual system:

Then, it follows from (75), (67), and (76) that the following inequality yields

Step 5. This is to obtain a uniform estimate for with sufficiently small.

Since solves (76), we see that

By Lemma 4 and Remark 8, it holds that

This, combined with (53), shows that, for each ,

Now we can apply the Gronwall’s inequality to get that, for all ,

On the other hand, according to the system satisfied by , it follows that, for every ,

namely, we have the equation as follows:

Then, making use of the above equation and by (81), we obtain

On the other hand, by Lemma 7 and the inequality , we get

where is independent of and . This, together with (84) and (5), implies that

where is independent of and .

Step 6. This contains the convergence of a subsequence of the family .

First of all, corresponding to each , we extend the function by setting to be on and setting to be 0 on denote the extended function by again. Clearly, this extended function is continuous on .

Now, we take a sequence of numbers from the interval such that (i) and (ii) . Corresponding to the number , we can take a sequence from the set such that and for all .

By (79), the sequence is uniformly bounded on the interval . This, together with (86), implies that is uniformly bounded on .

On the other hand, by system (4) with , we can use the similar argument we used to prove (47) to conclude that

where is independent of and .

Next, we will prove that the sequence is equicontinuous on .

Indeed, for each and in the interval with , it follows from (76), (53), and (86) that

from which and from (79), (85), and (87), it holds that

where is independent of , , and . This implies that the sequence is equicontinuous on . Hence, we can utilize the Arzela-Ascoli theorem and the standard diagonal argument to see that there exists a subsequence which is uniformly convergent on for each .

Let , . Then it holds that

Step 7. This is to extend the function over the interval .

Since all solutions, , , do not quench on every interval with and , we can use the same arguments in the proof of (59) to get

This, together with (86) and (90), yields the following inequality:

Since the constant is independent of , it holds that

On the other hand, because , we conclude from Lemma 7 that , from which and from (93), we obtain that . Now, we extend the function by setting it to be zero at the time and still denote the extension by . Clearly, this extended function is continuous on the interval and has the property .

Step  8. This is to verify that the function solves the system (10).

Clearly, the function holds the following property:

where . We first claim that

If (95) were not true, then there would exist a positive constant such that

On the other hand, we take a number with and . Because , we can find a number such that . Since , there is a natural number such that

Now, we can utilize Lemma 4 to get

which contradicts with (96). Therefore, we have proved (95).

Next, we claim that

Corresponding to each , we define a function by setting

It is clear that all functions, , , are measurable on the interval . We will first give an estimate on the sequence . Let . In the case that is such that , by (53), (79), (85), (86), and (87), we have the following estimate:

where is independent of and .

On the other hand, if is such that , then . This, together with (101), implies that

Then, we are going to show that the sequence is convergent for each . Indeed, corresponding to each , there exists a natural number such that when . Then, by (90), (91), and (102), we can apply Lebesgue dominated convergence theorem to get

from which, (99) follows immediately.

Now, let . Clearly, it holds that for sufficiently large. Thus, making use of (95) and (99), we can pass to the limit for in (94), where , to get

Since can be arbitrarily taken from in the above equation, we already hold the first equation of (10). On the other hand, we proved that in the end of Step 7. Hence, the function solves the system (10).

Step  9. This is to prove (11).

We can use the similar argument we used in the proof of Step 9 of Theorem 1.2 in [8] to prove (11).

Step  10. This is to show the nontriviality of the function .

By the equations satisfied by and , respectively, we see that, for each ,

Because , it follows from (105) that, for each ,

By (86), we can make use of the very similar arguments as those in the proof of (99) to verify that

which, together with (90), (91), and (106), implies that, for each ,

This shows that the function is not trivial.

Step  11. This is to prove (12).

Indeed, by (23) in Lemma 7, (12) holds.

Thus, we complete the proof of Theorem 2 in the case, where with and .

We can use similar arguments in the proof of Theorem 2 to prove Theorem 3 (see [9] for the detailed proofs).

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is Supported by the National Natural Science Foundation of China (no. 11071036) and the National Basis Research Program of China (973 Program no. 2011CB808002).

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