Boundary-Value Problems

Characteristic Values and Characteristic Functions

Introduction

In many problems the solution of an ordinary differential equation must satisfy certain conditions which are specified for two or more values of the independent variable. Such problems are called boundary-value problems, in distinction with initial-value problems, wherein all conditions are specified at one point.

In illustration, we may require a solution of a homogeneous linear equation of second order of the following form

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+a_1(x)\frac{\mathrm{d}y}{\mathrm{d}x}+a_2(x)y=0.$$

which vanishes at the two points $x=a$ and $x=b$

$$y(a)=0,y(b)=0.$$

The general solution of the differential equation is of the form

$$y=c_1{u}_1(x)+c_2{u}_2(x).$$

where $u_1$ and $u_2$ are linearly independent solutions and $c_1$ and $c_2$ are constants, the boundary conditions constitute the requirements

$$\begin{array}{c}{c}_1{u}_1(a)+c_2{u}_2(a)=0\\ c_1{u}_1(b)+c_2{u}_2(b)=0\end{array}\}$$

In order that nontrivial solutions exist (no meaning in physics), it is necessary that the determinant of coefficients vanish

$$\left|\begin{array}{cc}{u}_1(a)& u_2(a)\\ u_1(b)& u_2(b)\end{array}\right|=0$$

OR $c_1=c_2\equiv 0$, has no nontrivial solution

If this condition exists, then we can easily derive (by solving the linear equations about $c_1,c_2$ )

$$y=C[u_2(a)u_1(x)-u_1(a)u_2(x)]$$

Characteristic Values and Characteristic Functions

In many cases, one or both coefficients $a_1(x)$ and $a_2(x)$ and hence the solutions $u_1(x)$ and $u_2(x)$, depend upon a constant parameter $\lambda$ which may take on various constant values in a discussion.

In such cases, the determinant may vanish for certain definite values of $\lambda$, say $\lambda ={\lambda }_1,{\lambda }_2,\cdots$

the values of $\lambda$ for which nontrivial solutions exist are called the characteristic values (or eigenvalues) of $\lambda$, and the corresponding solutions are called the characteristic functions (or eigenfunctions) of the problem.

Example: boundary value problem

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+\lambda y=0,y(0)=y(L)=0$$

How to solve this BVP? Assume $y=e^{rx}$

$$\begin{array}{c}{r}^2{e}^{rx}+\lambda e^{rx}=0\\ r^2=-\lambda \end{array}$$

1. Suppose $\lambda <0$, let $\lambda =-\beta$

$$\begin{array}{rl}& r=\pm \sqrt{\beta }\\ \Rightarrow & y=c_1{e}^{\sqrt{\beta }x}+c_2{e}^{-\sqrt{\beta }x}\\ & y(0)=0\Rightarrow c_1+c_2=0;\\ & y(L)=0\Rightarrow c_1{e}^{\sqrt{\beta }L}+c_2{e}^{-\sqrt{\beta }L}=0\\ & \left|\begin{array}{cc}1& 1\\ e^{\sqrt{\beta }L}& e^{-\sqrt{\beta }L}\end{array}\right|\ne 0\\ \Rightarrow & y=0\text{ (trivial solution) }\end{array}$$

2. Suppose $\lambda >0$

$$\begin{array}{rl}& r^2+\lambda =0\\ & r=\pm \sqrt{\lambda }i\\ \Rightarrow & y=c_1{e}^{\sqrt{\lambda }}xi+c_2{e}^{-\sqrt{\lambda }xi}\\ \text{Since }& \sin x=\frac{e^{ix}-e^{-ix}}{2i},\cos x=\frac{e^{ix}+e^{-ix}}{2}\\ \Rightarrow & y=A_1\sin (\sqrt{\lambda }x)+A_2\cos (\sqrt{\lambda }x)\\ & y(0)=0\Rightarrow A_2=0;\\ & y(L)=0\Rightarrow A_1\sin (\sqrt{\lambda }L)+A_2\cos (\sqrt{\lambda }L)=0\\ \Rightarrow & \{\begin{array}{rl}& A_1=0y=0,\text{ trivial solution }\\ & \sin (\sqrt{\lambda }L)=0\Rightarrow \sqrt{\lambda }L=n\pi ,n=1,2,\dots \end{array}\\ \Rightarrow & y=A_1\sin \left(\frac{n\pi }{L}x\right)\text{ nontrivial solution }\end{array}$$

Orthogonality of Functions

An orthogonal basis can expand an indefinite dimensional function space, and eigenfunctions in S-L equations with specific boundary values can construct an orthogonal basis

The eigenfunctions of second-order linear differential equations have orthogonality

Two functions ${\phi }_m(x)$ and ${\phi }_n(x)$ are said to be orthogonal over an interval $(a,b)$ if the integral of the product ${\phi }_m(x),{\phi }_n(x)$ over that interval vanishes

$${\int }_a^b{\phi }_m(x){\phi }_n(x)\mathrm{d}x=0.$$

More generally, the functions ${\phi }_m(x)$ and ${\phi }_n(x)$ are said to be orthogonal with respect to a weighting function $r(x)$, over an interval $(a,b)$, if

$${\int }_a^{b}r(x){\phi }_m(x){\phi }_n(x)\mathrm{d}x=0.$$

Sturm-Liouville Problem

The linear homogeneous second-order differential equation with suitably prescribed homogeneous boundary conditions at the ends of an interval $(a,b)$

$$\frac{\mathrm{d}}{\mathrm{d}x}[p(x)\frac{\mathrm{d}y}{\mathrm{d}x}]+[q(x)+\lambda r(x)]y=0$$

The functions $p,q$, and $r$ are assumed to be real. In terms of the operator

$$\begin{array}{c}L=\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}}{\mathrm{d}x}\right)+q=p\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}+\frac{\mathrm{d}p}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x}+q\\ Ly+\lambda r(x)y=0\end{array}$$

Note that any equation of the following form

$$a_0(x)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+a_1(x)\frac{\mathrm{d}y}{\mathrm{d}x}+[a_2(x)+\lambda a_3(x)]y=0$$

can be written in the Sturm-Liouville form by setting

$$p=e^{\int \frac{a_1}{a_0}\mathrm{d}x},q=\frac{a_2}{a_0}p,r=\frac{a_3}{a_0}p$$

Orthogonality of Characteristic Functions or Eigenfunctions

Suppose now that ${\lambda }_1$ and ${\lambda }_2$ are any two different characteristic values of the problem considered and that the corresponding characteristic functions are $y={\phi }_1(x)$ and $y={\phi }_2(x)$, respectively. There then follows

$$\begin{array}{l}\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)+(q+{\lambda }_{1}r){\phi }_1=0\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)+(q+{\lambda }_{2}r){\phi }_2=0\end{array}\}$$

If the first of these equations is multiplied by ${\phi }_2(x)$ and the second by ${\phi }_1(x)$, and the resultant equations are subtracted from each other, there follows

$${\phi }_2\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)-{\phi }_1\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)+({\lambda }_1-{\lambda }_2)r{\phi }_1{\phi }_2=0,$$

and hence

$$({\lambda }_2-{\lambda }_1){\int }_a^{b}r{\phi }_1{\phi }_2\mathrm{d}x={\int }_a^b[{\phi }_2\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)-{\phi }_1\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)]\mathrm{d}x$$

Integrating the right member by parts, we obtain

$$\mathrm{RHS}={\int }_a^b[{\phi }_2\mathrm{d}\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)-{\phi }_1\mathrm{d}\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)]$$

$$\begin{array}{rl}({\lambda }_2-{\lambda }_1){\int }_a^{b}r{\phi }_1{\phi }_2\mathrm{d}x=& {[{\phi }_2\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)-{\phi }_1\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)]}_a^b\\ & -{\int }_a^b[\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\right)-\frac{\mathrm{d}{\phi }_1}{\mathrm{d}x}\left(p\frac{\mathrm{d}{\phi }_2}{\mathrm{d}x}\right)]\mathrm{d}x\end{array}$$

Since the last integrand vanishes identically, there follows finally

$$({\lambda }_2-{\lambda }_1){\int }_a^{b}r{\phi }_1{\phi }_2\mathrm{d}x={[p(x)\{{\phi }_2(x)\frac{\mathrm{d}{\phi }_1(x)}{\mathrm{d}x}-{\phi }_1(x)\frac{\mathrm{d}{\phi }_2(x)}{\mathrm{d}x}\}]}_a^b.$$

if at each end point a prescribed condition is of one of the following forms:

$$\begin{array}{rl}y& =0\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =0\\ y+\alpha \frac{\mathrm{d}y}{\mathrm{d}x}& =0\end{array}\}(\text{ when }x=a\text{ or }x=b)$$

$${\phi }_2{\phi }_1^{\prime }-{\phi }_1{\phi }_2^{\prime }\equiv ({\phi }_2+\alpha {\phi }_2^{\prime }){\phi }_1^{\prime }-({\phi }_1+\alpha {\phi }_1^{\prime }){\phi }_2^{\prime }$$

$$⟹{\int }_a^{b}r(x){\phi }_1(x){\phi }_2(x)\mathrm{d}x=0\text{ if }{\lambda }_2\ne {\lambda }_1$$

that is, if the characteristic functions correspond to different characteristic numbers, then they are orthogonal with respect to the function $r(x)$.

As an example, the problem

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+\lambda y=0,y(0)=y(L)=0$$

has the characteristic values ${\lambda }_n=n^2{\pi }^2/L^2$ with corresponding characteristic functions proportional to the functions $y_n=\sin (n\pi x/L)$. Since in this case $r(x)=1$,

$${\int }_0^L{y}_m(x)y_n(x)\mathrm{d}x={\int }_0^L\sin \frac{m\pi x}{L}\sin \frac{n\pi x}{L}\mathrm{d}x=0(m\ne n)$$

when $m$ and $n$ are positive integers. This fact is readily verified independently by direct integration.

Expansion of arbitrary functions in series of orthogonal functions

Suppose that we have a set of functions $\{{\phi }_n(x)\}$, orthogonal in a given interval $(a,b)$ with respect to a certain known weighting function $r(x)$, and desire to expand a given function $f(x)$ in terms of a series of these functions, of the form

$$\begin{array}{rl}f(x)& =A_0{\phi }_0(x)+A_1{\phi }_1(x)+A_2{\phi }_2(x)+\cdots \\ & =\sum _{n=0}^{\mathrm{\infty }}{A}_n{\phi }_n(x)\end{array}$$

If we assume that such an expansion exists, and multiply both sides by $r(x){\phi }_k(x)$, where ${\phi }_k(x)$ is the $k$ th function in the set, we have

$$r(x)f(x){\phi }_k(x)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n}r(x){\phi }_n(x){\phi }_k(x)$$

Next, if we integrate both sides of this last equation over the interval $(a,b)$ and assume that the integral of the infinite sum is equivalent to the sum of the integrals, there follows formally

The assumption is justified if $\sum _{n=0}^{\mathrm{\infty }}{A}_n{\phi }_n(x)$ is uniformly convergent in $(a,b)$.

$${\int }_a^{b}r(x)f(x){\phi }_k(x)\mathrm{d}x=\sum _{n=0}^{\mathrm{\infty }}{A}_n{\int }_a^{b}r(x){\phi }_n(x){\phi }_k(x)\mathrm{d}x$$

But by virtue of the orthogonality of the set $\{{\phi }_n(x)\}$, all terms in the sum on the right are zero except that one for which $n=k$

$$A_n{\int }_a^{b}r(x){[{\phi }_n(x)]}^2\mathrm{d}x={\int }_a^{b}r(x)f(x){\phi }_n(x)\mathrm{d}x$$

With these values of the constants a formal series $\sum A_n{\phi }_n(x)$ is determined. It should be emphasized, however, that we have not established the fact that this series actually does represent the function $f(x)$ in the interval $(a,b)$. In fact, we have not even shown that the series converges in $(a,b)$ and hence represents any function in that interval. We may, however, speak of the series so obtained as the formal representation of the function $f(x)$.

BVP involving inhomogeneous differential equations

Boundary-value problems involving inhomogeneous differential equations

$$\begin{array}{rl}& [\frac{\mathrm{d}}{\mathrm{d}x}\left(p\frac{\mathrm{d}y}{\mathrm{d}x}\right)+qy]+\lambda ry=F(x)\\ & \text{ or }Ly+\lambda ry=F(x)\end{array}$$

if at each end point a prescribed condition is of one of

$$\begin{array}{r}y=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}=0\\ y+\alpha \frac{\mathrm{d}y}{\mathrm{d}x}=0\end{array}\}(\text{ when }x=a\text{ or }x=b)$$

If the required solution exists, we express it as a series of the form (method of series solution)

$$y=\sum a_n{\phi }_n(x)$$

in which:

$$L{\phi }_n(x)+{\lambda }_{n}r(x){\phi }_n(x)=0,n=1,2,\cdots$$

Substituting the above expression of $y$ into the inhomogeneous equation

$$r(x)\sum (\lambda -{\lambda }_n)a_n{\phi }_n(x)=F(x)$$

Taking

$$\begin{array}{rl}& f(x)=\frac{F(x)}{r(x)}=\sum A_n{\phi }_n(x)\text{ where }\\ & A_n{\int }_a^{b}r(x){[{\phi }_n(x)]}^2\mathrm{d}x={\int }_a^{b}r(x)f(x){\phi }_n(x)\mathrm{d}x\\ ⟹& \sum (\lambda -{\lambda }_n)a_n{\phi }_n(x)=\sum A_n{\phi }_n(x)\\ & y=\sum \frac{A_n}{\lambda -{\lambda }_n}{\phi }_n(x)=\frac{A_0}{\lambda -{\lambda }_0}{\phi }_0(x)+\frac{A_1}{\lambda -{\lambda }_1}{\phi }_1(x)+\cdots \end{array}$$

Example

$$\begin{array}{c}\text{ Solve }\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+\lambda y=F(x),y(0)=y(L)=0\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+\lambda y=0⟹{\lambda }_n={\left(\frac{n\pi }{L}\right)}^2,y=\sum A_n\sin \left(\frac{n\pi }{L}x\right)\end{array}$$

What happens to $\lambda$ if $F(\mathrm{x})$ exists?

Substitute the homogeneous solution into question

$$\sum _{n=0}^{\mathrm{\infty }}[-A_n{\left(\frac{n\pi }{L}\right)}^2\sin \left(\frac{n\pi }{L}x\right)]+\lambda \sum _{n=0}^{\mathrm{\infty }}{A}_n\sin \left(\frac{n\pi }{L}x\right)=F(x)$$

Orthogonal expand $F(x):F(x)=\sum _{n=0}^{\mathrm{\infty }}{F}_n\sin \left(\frac{n\pi }{L}x\right)$

$$\begin{array}{c}{A}_n[\lambda -{\left(\frac{n\pi }{L}\right)}^2]=F_n\Rightarrow A_n=F_n/[\lambda -{\left(\frac{n\pi }{L}\right)}^2]\\ y=\sum _{n=0}^{\mathrm{\infty }}\frac{F_n}{\lambda -{\left(\frac{n\pi }{L}\right)}^2}\sin \left(\frac{n\pi }{L}x\right)\end{array}$$

BV Problems in Engineering and Science

• Quantum mechanics
• Semi-conductor and electronics
• Continuum mechanics
• Elastic waves
• Thermal hydraulics