10.1 Sturm-Liouville Problem

Introduction to Separation of Variables Problems

In heat, wave, and Laplace's equations, we repeatedly encountered the ordinary differential equation (ODE):

$$X'' + \lambda X = 0$$

Case 1: Fixed Boundary Conditions (BC's)

Boundary Conditions: $X(0) = X(L) = 0$ (Ends at fixed temperature)

• Eigenvalues: $$\lambda_n = \frac{n^2 \pi^2}{L^2}$$
• Eigenfunctions: $$X_n = \sin\left(\frac{n\pi x}{L}\right)$$ where $n = 1, 2, 3, \dots$

Case 2: Insulated Boundary Conditions (BC's)

Boundary Conditions: $X'(0) = X'(L) = 0$ (Ends are insulated)

• Eigenvalues: $$\lambda_n = \frac{n^2 \pi^2}{L^2}$$
• Eigenfunctions: $$X_n = \cos\left(\frac{n\pi x}{L}\right)$$ where $n = 0, 1, 2, 3, \dots$

Questions on Changing the Problem

What if we change the BC's, for example:

• $X(0) = X'(L) = 0$ (Left fixed, right insulated)
• $X(0) - X'(0) = 0$ (Left temperature = heat flux at left!)
• $X(L) + X'(L) = 0$

Or instead of $u_t = k u_{xx}$ (constant conductivity), we use a heat equation where conductivity is a **function of position**?

• Will the eigenvalues still be nonnegative and form an increasing sequence?
• Will the eigenfunctions remain mutually orthogonal?

The Sturm-Liouville Problem (SL)

The general form of a Sturm-Liouville ODE is:

$$\frac{d}{dx}\left[p(x)\frac{dy}{dx}\right] - q(x)y + \lambda r(x)y = 0$$

The Boundary Conditions (BC's) are of the general homogeneous form for $a < x < b$:

At $x=a$: $$\alpha_1 y(a) - \alpha_2 y'(a) = 0$$

At $x=b$: $$\beta_1 y(b) + \beta_2 y'(b) = 0$$

**Constraint:** Not both $\alpha_1$ and $\alpha_2$ can be zero; not both $\beta_1$ and $\beta_2$ can be zero.

Regular Sturm-Liouville Problem

The SL problem is said to be **regular** if:

• $p'(x)$, $p(x)$, $q(x)$, and $r(x)$ are continuous on the interval $[a, b]$.
• $p(x) > 0$ and $r(x) > 0$ on $[a, b]$.

Key Properties of a Regular SL Problem

If $q(x)$, $\alpha_1$, $\alpha_2$, $\beta_1$, and $\beta_2$ are all **nonnegative**, then:

• The eigenvalues $(\lambda)$ are also **nonnegative** (and real).
• The eigenvalues form an **increasing sequence**: $$\lambda_1 < \lambda_2 < \lambda_3 < \dots < \lambda_n < \dots$$
• The eigenfunctions $y_n$ are **mutually orthogonal** with respect to the **weight function** $r(x)$: $$\int_a^b y_i y_j r(x) dx = 0 \quad \text{if } i \neq j$$

Examples and Non-Regular Cases

Example 1: Fixed Ends (Standard Heat Eq.)

ODE: $$y'' + \lambda y = 0 \quad \text{for } 0 < x < L$$

BC's: $y(0) = 0$ and $y(L) = 0$

This ODE can be mapped to the general SL form with: $p(x)=1$, $q(x)=0$, $r(x)=1$.

The BC's map to: $\alpha_1=1, \alpha_2=0$ (at $x=0$) and $\beta_1=1, \beta_2=0$ (at $x=L$).

Since $p, q, r$ are continuous, $p>0$, $r>0$, and all coefficients ($\alpha_1, \alpha_2, \beta_1, \beta_2, q$) are nonnegative, this is a **regular SL problem**.

It agrees with the known heat equation solution: $$\lambda_n = \frac{n^2 \pi^2}{L^2}, \quad y_n = \sin\left(\frac{n\pi x}{L}\right)$$

Properties: $\lambda_1 < \lambda_2 < \dots$ and $\int_0^L y_i y_j dx = 0$ (since $r(x)=1$).

Example 2: Non-Regular BC's (Mixed Condition)

ODE: $$y'' + \lambda y = 0 \quad \text{for } 0 < x < L$$

BC's: $y(0) + y'(0) = 0$ and $y(L) = 0$

At $x=0$: $y(0) + y'(0) = 0 \implies \alpha_1=1, \alpha_2=-1$. Since $\alpha_2$ is negative, this problem is **not guaranteed to have nonnegative eigenvalues**.

This is **not a regular** SL problem because not all $\alpha_1, \alpha_2, \beta_1, \beta_2$ are nonnegative.

Example 3: Finding Negative Eigenvalues in a Non-Regular Case

ODE: $$y'' + \lambda y = 0$$

BC's: $y(0) = 0$ ($\alpha_1=1, \alpha_2=0$) and $h y(L) - y'(L) = 0$ (where $h>0$).

At $x=L$: $h y(L) - y'(L) = 0 \implies \beta_1=h, \beta_2=-1$. Since $\beta_2 < 0$, this is **not a regular** SL problem, and a negative eigenvalue could exist.

Solving for the Smallest Eigenvalue ($\lambda < 0$)

Since the problem is not regular, we start by looking for negative eigenvalues. Let $\lambda = -k^2$, where $k$ is a positive constant ($k>0$).

The ODE becomes: $$y'' - k^2 y = 0$$

General Solution: $y = d_1 \cosh(kx) + d_2 \sinh(kx)$ (more convenient for BC's).

1. Apply $y(0) = 0$:

Solution simplifies to: $y = d_2 \sinh(kx)$, and its derivative is $y' = k d_2 \cosh(kx)$.

2. Apply $h y(L) - y'(L) = 0$:

Since the eigenfunction cannot be trivial (i.e., $d_2 \neq 0$), we can divide by $d_2$:

Divide by $\cosh(kL)$ (since $\cosh(x)$ is never zero):

Let $X = kL$. The equation becomes a transcendental equation:

We solve $\tanh(X) = \frac{1}{hL} X$. This is the intersection of the $\tanh(X)$ curve and a line with positive slope $\frac{1}{hL}$ through the origin.

Let $\beta_0$ be the **smallest positive intersection** of the two graphs.

The **smallest eigenvalue** (which is negative) is:

If $\frac{1}{hL}$ is close to $1$ (the maximum slope $\tanh(X)$ can have at the origin), then $\beta_0 \approx hL$, so $k \approx h$. This gives an approximation for the smallest eigenvalue:

The associated eigenfunction is $y = d_2 \sinh(kx)$. Setting $d_2=1$ for a simplified form gives:

This process is the first step. Next, one would find the remaining eigenvalues ($\lambda > 0$), collect all eigenfunctions, and form a solution that will resemble a Fourier-like series.

End of Lesson 34 notes.