# Approximate Solutions to Three-Point Boundary Value Problems with Two-Space Integral Condition for Parabolic Equations

Authors: Jing Niu; Yingzhen Lin; Minggen Cui
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We construct a novel reproducing kernel space and give the
expression of reproducing kernel skillfully. Based on the orthogonal basis of
the reproducing kernel space, an efficient algorithm is provided firstly to solve
a three-point boundary value problem of parabolic equations with two-space
integral condition. The exact solution of this problem can be expressed by
the series form. The numerical method is supported by strong theories. The
numerical experiment shows that the algorithm is simple and easy to implement
by the common computer and software.


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## Body

## 1. Introduction

Nonclassical boundary value problems with nonlocal boundary conditions arise naturally in various engineering models and physical phenomena, for example, chemical engineering, thermoelasticity, underground water flow, and population dynamics [1–4]. The importance of boundary value problems with integral boundary conditions has been pointed out by Samarskiĭ [5].Boundary value problems for parabolic equations with an integral boundary condition are investigated in the literature for the development, analysis, and implementation of accurate methods [6–11]. Integral boundary conditions of models emerged in previous literatures can be summed up as(1.1)∫abu(x,t)dx=h(t),u(x,t)∈[a,b]×[0,T].
However, Marhoune [12] studied the parabolic equation with a generalized integral boundary condition (1.2)–(1.4). This model is more universal, and it extended usual integral boundary conditions. The form is as follows:(1.2)∂u(x,t)∂t-a(t)∂2u(x,t)∂x2=f(x,t),(x,t)∈(0,1)×(0,T),
subject to the initial-boundary value conditions(1.3)u(x,0)=φ(x),x∈(0,1),u(0,t)=u(1,t),t∈(0,T),
and the integral condition(1.4)∫0αu(x,t)dx+∫β1u(x,t)dx=0,0<α<β<1,α+β=1,
with the function a(t),  and its derivatives are bounded on the interval [0,T]: 0<a0<a(t)<a1, 0<a2<a′(t)<a3. In the following, we may assume φ(x)=0  because it can be got from homogeneous boundary conditions. The existence and uniqueness of the solution for (1.2)–(1.4) have been proved in [12].A practical model is typed by various definite conditions under different environment. Investigation about the definite conditions is the key problem to the model. Due to condition (1.4), it is difficult to construct reproducing kernel space, so nobody gives the algorithm for the above problem by applying reproducing kernel theory. In this paper, the author successfully constructs a novel reproducing kernel space which includes boundary conditions (1.3)-(1.4) and gains the expression of the reproducing kernel skillfully. Meanwhile, we provide a simple algorithm for solving (1.2)–(1.4). Based on the orthogonal basis in the reproducing kernel space, the exact solution is given by the form of series. Meanwhile, using a similar process, it is possible to solve other linear ordinary differential equations, partial differential equations with the same boundary value conditions.


## 2. Constructive Method for the Reproducing Kernel SpaceH0(Ω)

H(Ω)  and H0(Ω)(Ω=[0,1]×[0,T])  are inner product spaces and H0(Ω)⊂H(Ω), and  they are defined in the following.Definition 2.1.
The inner product spaceH(Ω) is defined by(2.1)H(Ω)={u(x,t)∣∂x2t3u(x,t)isabsolutelycontinuous,u(x,0)=0,u(0,t)=u(1,t),∂x3t25u(x,t)∈L2(Ω)}.
The inner product ofH(Ω) is defined by
(2.2)〈u(x,t),v(x,t)〉H=∑j=01[∑i=01∂tjxij+iu(0,0)∂tjxij+iv(0,0)+∫01∂tjx3j+3u(x,0)∂tjx3j+3v(x,0)dx]+∑i=01∫0T∂t2xi2+iu(0,t)∂t2xi2+iv(0,t)dt+∫0T∫01∂t2x35u(x,t)∂t2x35v(x,t)dxdt.
And it possesses associated norm ∥·∥.Lemma 2.2.
Inner spaceH(Ω) is a Hilbert reproducing kernel space. Its reproducing kernel function
(2.3)K(x,y,t,s)=R(x,y)G(t,s),
and for any u(x,t)∈H(Ω),
(2.4)u(x,t)=(u(y,s),K(x,y,t,s))H,
where
(2.5)R(x,y)={ry(x),x≤y,rx(y),y<x,G(t,s)={st(2+t)2-t36,t≤s,ts(2+s)2-s36,t>s,
where
(2.6)ry(x)=118720(18720+x(y-1)(((5-x)x3(2+y)-120)y(18+(y-6)y)-156x4-10xy(3+x)(y(6+(y-4)y)-120))).This proof can be found in [13–17].Clearly, one has(2.7)(1)G(t,s)=G(s,t),R(x,y)=R(y,x),(2)∀s∈[0,T],G(0,s)=0,(3)∀y∈[0,1],R(0,y)=R(1,y).Lemma 2.3.
Fix ans∈[0,T], G(t,s)(∫0αR(x,y)dy+∫β1R(x,y)dy)∈H(Ω).Definition 2.4.
The subspaceH0(Ω) is defined by
(2.8)H0(Ω)={u(x,t)∈H(Ω),∫0αu(x,t)dx+∫β1u(x,t)dx=0}.
Obviously,H0(Ω) is a closed subspace of the reproducing kernel H(Ω).  It is very important to obtain the representation of reproducing kernel in H0(Ω),  which is the base of our algorithm. Therefore, our work begins with some lemmas to provide constructive method for reproducing kernel in H0(Ω).Lemma 2.5.
Fix ans∈[0,T],  G(t,s)(∫0αR(x,y)dy+∫β1R(x,y)dy)≢0.Proof.
Otherwise, for allu(x,t)∈H(Ω), by (2.4),
(2.9)∫0αu(x,t)dx+∫β1u(x,t)dx=∫0α〈u(y,s),R(x,y)G(t,s)〉Hdx+∫β1〈u(y,s),R(x,y)G(t,s)〉Hdx=〈u(y,s),G(t,s)(∫0αR(x,y)dx+∫β1R(x,y)dx)〉H≡0,
then u(x,t)∈H0(Ω), which is contradictory.Lemma 2.6.
Fix ans∈[0,T], G(t,s)(∫0αR(x,y)dy+∫β1R(x,y)dy)∉H0(Ω), namely, ∫0α[∫0αRdy+∫β1Rdy]dx+∫β1[∫0αRdy+∫β1Rdy]dx≠0.Proof.
Otherwise,for allu(x,t)∈H0(Ω)⊂H(Ω),  then
(2.10)0=∫0αu(x,t)dx+∫β1u(x,t)dx=∫0α〈u(y,s),R(x,y)G(t,s)〉Hdx+∫β1〈u(y,s),R(x,y)G(t,s)〉Hdx=〈u(y,s),G(t,s)(∫0αR(x,y)dx+∫β1R(x,y)dx)〉H,
which implies that G(t,s)(∫0αR(x,y)+∫β1R(x,y))≡0, and this contradicts with Lemma 2.5.Consider a function(2.11)K0(x,y,t,s)=K(x,y,t,s)+G(t,s)(∫0αRdy+∫β1Rdy)(∫β1Rdx-∫0αRdx)∫0α[∫0αRdy+∫β1Rdy]dx+∫β1[∫0αRdy+∫β1Rdy]dx,
and one can check carefully that K0(x,y,t,s)∈H0(Ω)  is the reproducing kernel of H0(Ω).


## 3. An Orthogonal Basis ofH0(Ω)

Let an operator𝕃:H0(Ω)→L2(Ω),  putting(3.1)(Lu)(x,t)=∂tu(x,t)-a(t)∂x22u(x,t),
rewriting (1.2)–(1.4) as(3.2)(Lu)(x,t)=f(x,t),u(x,0)=0,u(0,t)=u(1,t),∫0αu(x,t)dx+∫β1u(x,t)dx=0.Lemma 3.1.
𝕃 is a bounded linear operator.Proof.
Noting that(3.3)|∂tu(x,t)|=|〈u(y,s),∂tK0(x,y,t,s)〉H0|≤M1‖u‖H0,|∂x22u(x,t)|=|〈u(y,s),∂x22K0(x,y,t,s)〉H0|≤M2‖u‖H0,
where M1,M2 are positive real numbers,
(3.4)‖Lu‖L22=∫0T∫01((Lu)(x,t))2dxdt=∫0T∫01(∂tu(x,t)-a(t)∂x22u(x,t))2dxdt≤∫0T∫01(|∂tu|2+a2(t)|∂x22u|2+2|a(t)||∂tu||∂x22u|)dxdt.
Combining with the bounded function a(t), it holds that 𝕃 is a bounded linear operator.We will choose and fix a countable dense subsetS={(x1,t1),(x2,t2),…}⊂Ω   and define ψi(x,t) by(3.5)ψi(x,t)=def(L(y,s)K0(x,y,t,s))(xi,ti).Lemma 3.2.
Consider the following:ψi(x,t)∈H0(Ω),i=1,2,…,  (see [11]).Lemma 3.3.
The function system{ψi(x,t)}i=1∞ is a complete system of the space H0(Ω).Proof.
For everyi, we have
(3.6)0=〈u(x,t),ψi(x,t)〉H0=〈u(x,t),(L(y,s)K0(x,y,t,s))(xi,ti)〉H0=L(y,s)(〈u(x,t),K0(x,y,t,s)〉H0)(xi,ti)=L(y,s)(u(y,s))(xi,ti)=(Lu)(xi,ti),
which shows that (𝕃u)(x,t)=0  due to the denseness of S. It follows that u(x,t)≡0  from the existence of 𝕃-1.Applying Gram-Schmidt process, we obtain an orthogonal basis{ψ̃i(x,t)}i=1∞  of H0(Ω), such that(3.7)ψ̃i(x,t)=∑k=1iβikψk(x,t),
where βik are orthogonal coefficients.


## 4. Numerical Algorithm

In this section, it is explained how to deduce the exact solution from the orthogonal basis{ψ̃i(x,t)}i=1∞  of H0(Ω).Theorem 4.1.
The exact solution of (3.2) can be expressed by
(4.1)u(x,t)=∑i=1∞∑k=1iβiku(xk,tk)ψ̃i(x,t).Proof.
The exact solutionu(x,t)  can be expanded to a Fourier series in terms of normal orthogonal basis ψ̃i(x,t)  in H0(Ω),
(4.2)u(x,t)=∑i=1∞〈u(x,t),ψ̃i(x,t)〉H0ψ̃i(x,t)=∑i=1∞∑k=1iβik〈u(x,t),ψk(x,t)〉H0ψ̃i(x,t)=∑i=1∞∑k=1iβik〈u(x,t),(L(y,s)K0(x,y,t,s))(xk,tk)〉H0ψ̃i(x,t)=∑i=1∞∑k=1iβikL(y,s)(〈u(x,t),K0(x,y,t,s)〉H0)(xk,tk)ψ̃i(x,t)=∑i=1∞∑k=1iβik(Lu)(xk,tk)ψ̃i(x,t)=∑i=1∞∑k=1iβikf(xk,tk)ψ̃i(x,t).We obtain then-truncation approximate solution of (3.2),(4.3)un(x,t)=∑i=1n∑k=1iβikf(xk,tk)ψ̃i(x,t),
which is n-truncation Fourier series of the exact solution u(x,t) in (3.2), so un(x,t)→∥·∥H0u(x,t),  as n→∞.Theorem 4.2.
Consider the following:|un(x,t)-u(x,t)|→0,  as  n→∞.Proof.
SinceK0(x,x,t,t)  is continuous on Ω, it follows that K0(x,x,t,t)≤C, for all (x,t)∈Ω,  where C is a constant. When ∥un-u∥H0→0,
(4.4)|un(x,t)-u(x,t)|=|〈un(x,t)-u(x,t),K0(x,y,t,s)〉H0|≤‖un-u‖H0‖K0(x,y,t,s)‖H0=K0(x,x,t,t)‖un-u‖H0⟶0.


## 5. Numerical Example

In this section, a numerical example is studied to demonstrate the accuracy of the present algorithm. The example is computed by Mathematica 5.0. Results obtained by the algorithm are compared with the analytical solution and are found to be in good agreement.Example 5.1.
Consider a three-point boundary value problem of parabolic equations with two-space integral condition(5.1)∂u(x,t)∂t-sin⁡t∂2u(x,t)∂x2=et(7-36x+36x3)-216(et-1)xsin⁡(t),u(x,0)=0,x∈(0,1),u(0,t)=u(1,t),∫01/3u(x,t)dx+∫2/31u(x,t)dx=0,t∈(0,1).
The exact solution is u(x,t)=(et-1)(7-36x+36x3). The numerical results are collected in Table 1.Table 1
Numerical results.
(x,t)u(x,t)|(u-u25)/u||(u-u100)/u||(u-u400)/u|(0.00,0.0001)0.0007000350.0000000002.45579E-101.12923E-9(0.95,0.05)0.1879340.0002304420.00001121121.94207E-6(0.85,0.15)−0.2413760.00002246856.90183E-61.36152E-6(0.75,0.25)−1.366870.000863020.00007153747.60769E-8(0.65,0.35)−2.72960.003225830.0002506075.17148E-7(0.55,0.45)−3.870490.002871170.0005717661.33238E-6(0.45,0.55)−4.340490.005173340.0008940532.81685E-6(0.35,0.65)−3.713890.001426320.0009135715.62867E-6(0.25,0.75)−1.605690.002135250.0004808411.37775E-5(0.15,0.85)2.30620.004508650.0005934522.81487E-6(0.05,0.95)8.252830.00209220.000486785.00647E-6


## 6. Conclusion

In this paper, we construct a reproducing kernel space by new method, in which each function satisfies boundary value conditions of considered problems. In this space, a numerical algorithm is presented for solving a class of parabolic equations with two-space integral boundary condition. Exact solution with series form is given. Approximate solution obtained by present algorithm converges to exact solution uniformly.



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