# 1-Dimensional Heat Initial Boundary Value Problems 2: Sturm-Liouville Problems and Orthogonal Functions

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