#### Abstract

A unified reproducing kernel method for solving linear differential equations with functional constraint is provided. We use a specified inner product to obtain a class of piecewise polynomial reproducing kernels which have a simple unified description. Arbitrary order linear differential operator is proved to be bounded about the special inner product. Based on space decomposition, we present the expressions of exact solution and approximate solution of linear differential equation by the polynomial reproducing kernel. Error estimation of approximate solution is investigated. Since the approximate solution can be described by polynomials, it is very suitable for numerical calculation.

#### 1. Introduction

The reproducing kernel method of solving differential equation is one of the important topics of reproducing kernel numerical methods. In recent years, reproducing kernel theories are widely applied to solve all kinds of the differential equations, which provide new description forms for exact solutions and approximate solutions of many classical equations. For example, Zhang and Cui [1] and Li [2] obtained the analytical solutions described as series for the steady state convecting equations and boundary value problems, separately. Wang et al. [3] not only got the exact solution described as series for a class of boundary value problems of ordinary differential equations, but also proposed iterative method for the approximate solutions. Gao et al. [4] and Li and Wu [5] investigated the reproducing kernel methods to solve singular second-order initial/boundary value problems and multipoint boundary value problems. Zhang [6] and Wu and Lin [7] specially introduced the reproducing kernel methods of solving linear differential equation based on reproducing kernel theory. Shi et al. [8] and Du and Zhang [9] obtained the analytical solutions and the corresponding approximate solutions of initial value problems of linear ordinary differential equations and error estimation was considered. Li and Wu [10] considered the error estimation of solving boundary value problem of linear equations. Castro et al. [11] introduced a new method to solve general initial value problems by means of reproducing kernel theory.

However, these studies were generally concentrated on constructing the corresponding reproducing kernels for solving specific equations. There was no unified description for these methods. Therefore, how to construct a common method of reproducing kernel for a class of differential equations is worthy of further study about theories and methods. In this paper, for a class of frequently used reproducing kernel space , a class of piecewise polynomial reproducing kernels is presented by a specific inner product. These reproducing kernels have a simple unified expression for all . They are suitable for numerical calculation and independent of the concrete form of the operator . According to the given inner product, the arbitrary th-order linear differential operator is proved to be bounded operator of   (if only ). Based on space decomposition, (defined as (1)) were proved to always be complete in a subspace. Finally, we give the error estimation of approximate solution. These results provide a unified reproducing kernel theory and method for solving arbitrary linear differential equation with arbitrary linear functional constraint.

#### 2. Preliminary

For the convenience of description, we first give a brief description of the problem concerned in this paper. Let , be reproducing kernel Hilbert spaces composed of functions defined on a certain . The inner products are , and reproducing kernels are , , separately. denotes a linear differential operator from to . For the given , under certain conditions, we need to find such that . The main idea of solving this kind of equation by reproducing kernel method is presented as follows:(i)For supposing to be conjugate operator of , prove to be a linear bounded operator from to .(ii)For the specified dense sequence , which is composed of distinct numbers, let and  then prove to be complete about .(iii)Based on (ii), orthonormalize to obtain and suppose ; then, from (1) and properties of reproducing kernel and conjugate operator, we can get the solution of as

From (2), we can obtain the exact expression of the solution and can do numerical calculation by values of the given .

According to the above process, we can see that if we want to solve differential equation, (i) and (ii) must be true. To verify these two conditions, the frequently used method is to firstly design an inner product and corresponding reproducing kernel according to the specified differential operator . Then, the boundedness of is considered and, for ensuring to be complete, has to be invertible. For each specific differential equation, we have the same process. Moreover, operator usually is not invertible. The forms of reproducing kernels obtained by solving high-order differential equation are complicated and have no unified structural properties. These also make it difficult to verify the boundedness of operator and the completeness of .

#### 3. Polynomial Reproducing Kernel in

For any positive integer , letwhere is a square integrable function space. According to [6], every can be written aswhereFor any , we have is a reproducing kernel Hilbert space with respect to inner product and has a reproducing kernel

For the convenience of numerical calculation, we explore more concrete form of reproducing kernel . Suppose that The key of calculating reproducing kernel is to calculate . Let then we can obtain where is as the definition in (5).

It follows that when , where is a Beta function. As a consequence, we have When , we can get by symmetry. By Beta function, we have

Then, from (8), we can getObviously, is a piecewise polynomial of two variables.

Lemma 1. For any dense sequence composed of distinct numbers in , function sequence is a linearly independent complete system in .

Proof. For any positive integer , if there are constants such that namely, , then, for any , it can follow that Take in the above equation as in turn; we can obtain . Therefore, are linearly independent.
For any , if , that is, , since are dense and is continuous in , then and are complete.
This completes the proof of Lemma 1.

Remark 2. The above proof is independent of the inner product of and the concrete form of reproducing kernel. Therefore, Lemma 1 is always true for arbitrary inner product of and the corresponding reproducing kernel.

#### 4. Reproducing Kernel Method for Solution of Linear Differential Equation

Suppose that th-order linear differential operator is where and integers and satisfy and . It is obvious that is a linear surjective map from to . The null space is a -dimensional subspace of .

When , then . Denote by ; that is, . The corresponding inner product is

Theorem 3. Suppose that exist and are bounded on ; then the differential operator is a bounded linear operator from to .

Proof. We only need to prove that is bounded. For any and , there is It follows thatAccording to the above inequality, for , we haveWhen , the norm of in is Sinceit is easily known that, for arbitrary numbers , we have and for . Because are bounded, there exists a constant (we might as well put ) such that Then, we can getLet ; we haveNote that is independent of , so, from (29) and (23), we can obtain Substituting the above two formulas into (24) and letting , we can get It shows that is a bounded operator from to .
When , then , and the norm of in is . According to the conditions, we can know that is bounded on . Therefore, there exists (we might as well put ) such that .
From (29) and (23), we can obtainThis shows that is a bounded operator from to .
This completes the proof of Theorem 3.

The null space of is -dimensional subspace of . Suppose that are continuous linear functional and linearly independent on and suppose that the values of are given. Now we consider the following definite solutions problem.

For a given , find such that it satisfiesDenoted by , is a vector-valued linear functional on ; namely, .

Lemma 4. has direct sum decomposition:

Proof. Because are linearly independent on , there is a basis of which is dual to ; namely, Since is a closed subspace of , every has a unique decomposition form: From (35), we can obtain ; then Let we have From (35) again, we can get and .
If , there is and . It follows that . Therefore, .
This completes the proof of Lemma 4.

According to the proof of Lemma 4, we have Since is a bounded linear operator, there exists a conjugate operator . For any dense sequence composed of distinct numbers in , when , suppose thatthen we have the following lemma.

Lemma 5. As in Lemma 4, suppose that ; then(i)for any , ;(ii)for any , .

Proof. Suppose that . If , then, from (41), we have Since and from Lemma 1, are complete in , so . As satisfies , therefore, ; that is to say, . By contrast, if , namely, , then . It follows that (i) is true. Meanwhile, according to the proof, we can also know that (ii) is true.
Thus, we finish the proof of Lemma 5.

Theorem 6. Suppose that ; then are linearly independent and complete in .

Proof. Clearly, . We first prove that are linearly independent. Suppose that there are a positive integer and constants such that ; namely, Since and is surjective, , we have From Lemma 1, are linearly independent; then . Therefore, are linearly independent.
Next, we prove that is a complete system of .
Suppose that satisfies . When , it is obvious that . From Lemma 5, it is easily obtained that when , and when , ; namely, . Then, and . As a result, . Therefore, is a complete system of .
This completes the proof of Theorem 6.

Theorem 7. Denote orthonormalized in by and suppose that then, the solution of (33) can be expressed as

Proof. From Lemma 4 and (39), can be expressed as Since , from Theorem 6, we suppose that ; according to the properties of conjugate operator and reproducing kernel, it follows thatSo we obtain (46).
Thus, the proof of Theorem 7 is completed.

From (48), we can obtain the approximation of as

Remark 8. Solving (33) by using (48) and (49) depends on the orthonormalization of . This can sometimes be troublesome. Therefore, we hope that can directly be written as a linear combination of .
Note that when , , so when , there isFrom (49) and Abel transformation of the partial sum of series, we have Let two sides of the above equation do inner product with ; by (49), we can obtain the system of equations where the coefficient matrix is symmetric and positive definite. Then, there exists only a set of solutions . By this solution, we can obtain the approximate solution .
In order to obtain matrix , we must calculate . For this purpose, we first deduce the concrete form of . According to the properties of conjugate operator and reproducing kernel, we haveUsing the reproducing kernel expression (14) and (15), we obtain the concrete form of aswhereIn order to calculate , from (54), we have When , it can be obtained thatWhen , by symmetry, there isThus, substituting (56)–(58) into (7), we can obtain .

#### 5. Convergence and Error Estimation

In this section, we will consider the convergence and error estimation of the approximate solution given by (49).

Lemma 9. is convergent uniformly to on .

Proof. Since is a complete orthonormal system of , it is obvious that From (14) and (15), it can be obtained that Since is continuous on , there exists a constant such that . Moreover,Hence, is convergent uniformly to on .
This completes the proof of Lemma 9.

Theorem 10. The exact solution and approximate solution of (33) satisfy

Proof. From (48) and (49), we have It follows thatSince , thenTherefore,Thus, we prove that (62) is true.
This completes the proof of Theorem 10.

Remark 11. When we use Theorem 10 to estimate , the value of has to be evaluated. It is usually difficult in practical problems. We will provide a new method of estimation in the following.

As in Lemma 4, suppose that is a basis of and is dual to . According to [6, 7], every has the decomposition form as where is Green function of differential operator and satisfies Since is continuous, there exists a constant such that .

Theorem 12. Suppose that ; then the exact solution and approximate solution of (33) satisfy The right-hand side of (69) converges to zero as .

Proof. From (39), (48), and (67), it follows that, for any , there is Then, we have From Lemmas 1 and 9, we know that the right-hand side of (71) converges to zero as . From (49), we can obtainSubstituting this equation into (71) and , we can get (69) and assert that the right-hand side of (69) converges to zero as .
This completes the proof of Theorem 12.

Remark 13. From (54), can be easily obtained as the piecewise polynomial.

#### 6. Conclusion

When we use the reproducing kernel method to solve differential equation, we always hope that the reproducing kernel is polynomial reproducing kernel. In general, to take the reproducing kernel of as polynomial reproducing kernel, the inner product taken must be as (7). But if we take this fixed inner product, it is difficult to prove the boundedness of operator and the completeness of . How do these two aspects integrate effectively with each other? For the linear differential operator equation with functional constraints, this paper investigates this topic and some meaningful results are presented. Moreover, the proofs of some results do not depend on whether operator is differential operator. Therefore, these processes can be extended for general linear bounded operator. In this paper, some conclusions are based on ; they need to be further explored for case .

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors thank the unknown reviewers for their careful reading and helpful comments. This work is supported by the Research Project of National University of Defense Technology (JC12-02-01).