**Sturm-Liouville Problems**

The homogeneous boundary conditions of 1D heat conduction problem are given by

\begin{align*}

-\kappa_1u_x(0,t)+h_1u(0,t)&=0,\ t>0\\

\kappa_2u_x(L,t)+h_2u(L,t)&=0,\ t>0

\end{align*}

(See here)

The homogeneous BCs for the second order linear differential equation \begin{equation}\label{eq:ho}X^{\prime\prime}=kX\end{equation} is then

\begin{equation}\label{eq:bc}\begin{aligned}

-k_1X’(0)+h_1X(0)&=0\\

k_2X’(L)+h_2X(L)&=0

\end{aligned}\end{equation}

Finding solutions of the second order linear differential equation \eqref{eq:ho} for $k=0$, $k=\lambda^2$, and $k=-\lambda^2$ that satisfy the BCs \eqref{eq:bc} is called a *Sturm-Liouville Problem*. Here, we study the Sturm-Liouville Theory with the following example.

*Remark*. In case of homogeneous heat BVPs, the eventual temperature would be 0 as there is no heat source. So, we see that $k=-\lambda^2<0$ is the only physically relevant case.

*Example*. [Fixed temperature at both ends]

Consider the heat BVP:

\begin{align*}

u_t&=\alpha^2 u_{xx}\ \mbox{PDE}\\

u(0,t)&=u(1,t)=0\ \mbox{(BCs)}

\end{align*}

From the above BCs, we obtain the BCs for $X(x)$:

$$X(0)=X(1)=0$$

For $k=0$ and $k=\lambda^2>0$ we have a trivial solution $X(x)=0$. For $k=-\lambda^2<0$ $X(x)=A\cos\lambda x+B\sin\lambda x$. With the BCS we find the eigenvalues

$$\lambda_n=n\pi,\ n=1,2,3,\cdots$$

and the corresponding eigenfunctions

$$X_n(x)=\sin n\pi x,\ n=1,2,3,\cdots$$

The $\{X_n: n=1,2,3,\cdots\}$ is a linearly independent set so they form a basis for the solution space which is infinite dimensional. The general solution to the heat BVP is given by

$$u(x,t)=\sum_{n=1}^\infty A_n e^{-n^2\pi^2\alpha^2t}\sin n\pi x$$

There are undetermined coefficients $A_n$ called Fourier coefficients. They can be determined by initial condition (initial temperature).

**Orthogonal Functions and Solution of a Homogeneous Heat IBVP**

Consider a heat distribution function $u(x,t)$ of the following form

$$u(x,t)=\sum_{n=0}^\infty A_ne^{-\lambda_n^2\alpha^2t}X_n(x)$$

where $X_n$’s are eigenfunctions corresponding to the eigenvalues $\lambda_n$’s respectively. The eigenfunctions $X_n$’s form a basis for the solution space (which is often infinite dimensional) of a given heat IBVP, furthermore they can form an orthonormal basis with respect to the inner product

\begin{equation}\label{eq:innerprod}\langle X_m,X_n\rangle=\int_0^LX_mX_ndx\end{equation}

We say that eigenfunctions $X_m$ and $X_n$ are orthogonal if $\langle X_m,X_n\rangle=0$.

*Example*. $X_n(x)=\sin n\pi x$, $n=1,2,3,\cdots$ form an orthogonal basis with respect to \eqref{eq:innerprod}, where $0<x<1$:

\begin{align*}

\langle X_m,X_n\rangle&=\int_0^1\sin m\pi x\sin n\pi xdx\\

&=\left\{\begin{aligned}

\frac{1}{2}\ &{\rm if}\ m=n\\

0\ &{\rm if}\ m\ne n.

\end{aligned}\right.

\end{align*}

*Remark*. [The Gram-Schmidt Orthogonalization Process]

If $\{X_n\}$ is not an orthogonal basis, one can construct an orthogonal basis from $\{X_n\}$ using the inner product (3). The standard process is called the Gram-Schmidt orthogonalization process. Details can be found in many standard linear algebra textbooks.

Now we assume that $\{X_n\}$ is an orthogonal basis for the solution space. Let $L_n:=\langle X_n,X_n\rangle=\int_0^LX_n^2dx$. Let the initial condition be given by

$u(x,0)=\phi(x)$. Then

$$\phi(x)=\sum_{n=0}^\infty A_nX_n$$

Multiply this by $X_m$ and then integrate:

$$\int_0^LX_m\phi(x)dx=\sum_{n=0}^\infty A_n\int_0^LX_nX_mdx$$

By orthogonality we obtain

$$L_mA_m=\int_0^LX_m\phi(x)dx$$

or

$$A_m=\frac{1}{L_m}\int_0^L\phi(x)X_mdx,\ m=0,1,2,\cdots$$

*Example*. Consider the heat BVP in the previous example with initial condition $\phi(x)=T$, a constant temperature. For $n=1,2,3,\cdots$, $X_n(x)=\sin n\pi x;\ 0<x<1$ so

$$L_n=\int_0^1\sin^2 n\pi x dx=\frac{1}{2}$$

The Fourier coefficients are then computed to be

\begin{align*}

A_n&=2\int_0^1\phi(x)\sin n\pi xdx\\

&=2T\int_0^1\sin n\pi xdx\\

&=\frac{2T}{n\pi}[1-\cos n\pi]\\

&=\frac{2T}{n\pi}[1-(-1)^n].

\end{align*}

$A_n=0$ for $n={\rm even}$ and $A_{2n-1}=\frac{4T}{(2n-1)\pi},\ n=1,2,3,\cdots$. Hence

$$u(x,t)=\sum_{n=1}^\infty\frac{4T}{(2n-1)\pi}e^{-(2n-1)^2\pi^2\alpha^2t}\sin(2n-1)\pi x.$$

**References:**

David Betounes, Partial Differential Equations for Computational Science with Maple and Vector Analysis, TELOS, Springer-Verlag

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