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Sturm-Liouville Theory (Part 2)

\[ \frac{d}{dx} \left[ p(x) \frac{dy}{dx} \right] + q(x)y + \lambda w(x)y = 0 \]
\( a \leq x \leq b \)

Subject to the boundary conditions:

\[ \alpha_1 y(a) + \alpha_2 y'(a) = 0 \] \[ \beta_1 y(b) + \beta_2 y'(b) = 0 \]

Regular Sturm-Liouville Problems

A Sturm-Liouville (SL) problem is defined as regular on the interval \( a \leq x \leq b \) if the functions \( p(x) \), \( p'(x) \), \( q(x) \), and \( w(x) \) are continuous, and the following conditions hold:

\[ p(x) > 0, \quad w(x) > 0 \]

Key Properties of Regular SL Problems

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There exist infinitely many eigenvalues \( \lambda \) that form a strictly increasing sequence:

\[ \lambda_1 < \lambda_2 < \lambda_3 < \dots < \lambda_n < \dots \]

where \( \lambda_n \to \infty \) as \( n \to \infty \).

Note: While eigenvalues could be negative, if the coefficients \( \alpha_1, \alpha_2, \beta_1, \beta_2 \) are all non-negative, then the eigenvalues \( \lambda \) are also guaranteed to be non-negative.

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Each eigenvalue \( \lambda_n \) is associated with an eigenfunction \( y_n \). These eigenfunctions are mutually orthogonal with respect to the weight function \( w(x) \) over the interval \( [a, b] \):

\[ \int_{a}^{b} y_n y_m w(x) dx = 0 \quad \text{if } n \neq m \]
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Example Application

Problem Statement:

\[ y'' + \lambda y = 0 \] \[ y(0) = 0, \quad y(L) = 0 \] \[ 0 < x < L \]

This system is analogous to the spatial component of the heat equation:

\[ X'' + \lambda X = 0 \] \[ X(0) = 0, \quad X(L) = 0 \]

Identifying the Sturm-Liouville parameters:

\[ p = 1, \quad q = 0, \quad w = 1 \]
\[ \alpha_1 = 1, \quad \alpha_2 = 0, \quad \beta_1 = 1, \quad \beta_2 = 0 \]

Consequently, this is a regular Sturm-Liouville problem with non-negative eigenvalues.

From previous analysis, we know the eigenvalues are: \[ \lambda_n = \frac{n^2 \pi^2}{L^2} \] (These are non-negative, which agrees with SL theory)

The corresponding eigenfunctions are: \[ y_n = \sin\left( \frac{n\pi}{L} x \right) \]

Verifying orthogonality:

\[ \int_{0}^{L} y_n y_m w(x) dx = \int_{0}^{L} \sin\left( \frac{n\pi}{L} x \right) \sin\left( \frac{m\pi}{L} x \right) dx = 0 \]

As predicted by SL theory.

The existence of non-negative eigenvalues \( \lambda \) justifies why we originally assumed the separation constant to be \( -\lambda \) in the separation of variables method:

\[ \frac{X''}{X} = \frac{T'}{kT} = -\lambda \]
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The Fourier solution is simply a special case of the more general Sturm-Liouville problem.

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Example: Investigating Negative Eigenvalues

Problem Statement

Consider the following differential equation and boundary conditions:

\[ y'' + \lambda y = 0 \quad \text{for } 0 < x < L \]

\[ y(0) = 0 \]

\[ h y(L) - y'(L) = 0, \quad h > 0 \]

Parameters

\[ p = 1, \quad q = 0, \quad w = 1 \]

\[ \alpha_1 = 1, \quad \alpha_2 = 0 \]

\[ \beta_1 = h > 0, \quad \beta_2 = -1 \]

Since not all coefficients \( \alpha \) and \( \beta \) are non-negative, the eigenvalues \( \lambda \) are not guaranteed to be non-negative. Consequently, negative eigenvalues are a possibility in this system.

Investigation of Negative Eigenvalues

Let's explore the case where \( \lambda < 0 \). For mathematical convenience, we define:

\[ \lambda = -k^2 \quad (k > 0) \]
Note: Using \( k^2 \) is a convenience that avoids carrying square roots through the solution.

Substituting this into our original equation yields:

\[ y'' - k^2 y = 0 \]

The general solution can be expressed in two equivalent forms. We can choose whichever is more convenient for satisfying the given boundary conditions:

  • \[ y = C_1 e^{kx} + C_2 e^{-kx} \]
  • or \[ y = A \cosh(kx) + B \sinh(kx) \]
Choose the form that simplifies the application of boundary conditions.
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Applying the First Boundary Condition

First, let's test the exponential form with the boundary condition \( y(0) = 0 \):

\[ y = C_1 e^{kx} + C_2 e^{-kx} \implies 0 = C_1 + C_2 \]
This is valid, but it would be preferable if one of the constants vanished.

Now, let's apply \( y(0) = 0 \) to the hyperbolic form:

\[ y = A \cosh(kx) + B \sinh(kx) \implies 0 = A \]

This simplifies our solution significantly to:

\[ y = B \sinh(kx) \]
The eigenfunctions are therefore of the form \( y_n = \sinh(kx) \) for \( \lambda = -k^2 \).

Applying the Second Boundary Condition

We use the condition \( h y(L) - y'(L) = 0 \) to determine the allowed values for \( k \) (and thus \( \lambda \)).

\[ y = B \sinh(kx) \]

\[ y' = Bk \cosh(kx) \]

Substituting these into the boundary condition at \( x = L \):

\[ h B \sinh(kL) - Bk \cosh(kL) = 0 \]

Assuming non-trivial solutions where \( B \neq 0 \), and given \( L \neq 0 \) and \( h \neq 0 \), we rearrange to solve for \( k \):

\[ \tanh(kL) = \frac{k}{h} \]

To solve this transcendental equation, we can multiply both sides by \( L \) to normalize the argument:

\[ \tanh(kL) = \frac{kL}{hL} \]

Transcendental Equation:

Let \( z = kL \). We must solve for the intersection of:

\[ \tanh(z) = \frac{z}{hL} \]

This represents the intersection of the hyperbolic tangent function \( \tanh(z) \) and the linear function \( \frac{z}{hL} \).

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Analysis of Negative Eigenvalues

A coordinate plane showing the intersection of the curve y = tanh(z) and the line y = z / (hL). The intersection point is marked and projected down to the horizontal z-axis at a value labeled z_1.

There is at most one intersection between the curves.

If the product of the heat transfer coefficient and length, denoted as \( hL \), is small, the line becomes too steep to intersect the \( \tanh(z) \) curve.

The existence of at most one intersection point implies that there is at most one negative eigenvalue for the system.

\[ \lambda_1 < \lambda_2 < \lambda_3 < \dots < \lambda_n < \dots \]

\( \uparrow \)

negative

\( \uparrow \)

must be \( \ge 0 \)

This negative value \( \lambda \) represents the smallest eigenvalue. Note that there is no largest eigenvalue as the sequence continues indefinitely.

The first eigenvalue is given by: \[ \lambda_1 = -k^2 = -\left( \frac{z_1}{L} \right)^2 \]

The corresponding eigenfunction is: \[ y_1 = \sinh\left( \frac{z_1}{L} x \right) \]

Next Steps

  • Determine if \( \lambda = 0 \) is an eigenvalue.
  • Find the remaining eigenvalues where \( \lambda > 0 \).
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Revisiting the Heat Exchange Problem

Let's return to the heat exchange problem discussed in the previous session:

\[ u_t = k u_{xx} \quad \text{for } 0 < x < L \]

\[ u(0, t) = 0 \]

\[ u_x(L, t) = -h u(L, t) \quad \text{where } h > 0 \]

\[ u(x, 0) = 100 \quad \text{(initially heated to 100 units uniformly)} \]

Spatial Problem Solution

Previously, we solved the spatial component of the problem:

\[ y'' + \lambda y = 0 \]

\[ \lambda_n = \frac{z_n^2}{L^2} \]

Where \( z_n \) are the positive intersections of:

\[ \tan(z) \text{ and } -\frac{z}{hL} \]

General Solution

By solving the time-dependent portion of the problem as usual, we arrive at the general solution for the temperature distribution:

\[ u(x, t) = \sum_{n=1}^{\infty} C_n e^{-k \lambda_n t} \sin(\sqrt{\lambda_n} x) \]
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Solving for Coefficients using Orthogonality

Given the initial condition: \[ u(x, 0) = 100 \] We express this as a series expansion: \[ 100 = \sum_{n=1}^{\infty} C_n \sin(\sqrt{\lambda_n} x) \]

Note: Not a standard Sine Series

Since \( \lambda \) is not an integer multiple of \( \frac{\pi}{L} \), the coefficient \( C_n \) cannot be calculated using the standard formula: \[ C_n \neq \frac{2}{L} \int_{0}^{L} 100 \sin\left(\frac{n\pi}{L}x\right) dx \]

Applying Sturm-Liouville Theory

Although this is not a standard Fourier sine series, we can still utilize the orthogonality of eigenfunctions (as guaranteed by Sturm-Liouville theory) to determine the coefficients \( C_n \).

The orthogonality property states: \[ \int_{0}^{L} y_n y_m w(x) dx = 0 \quad \text{for } n \neq m \]

In this specific case, we have:

  • Weight function: \( w(x) = 1 \)
  • Eigenfunctions: \( y_n = \sin(\sqrt{\lambda_n} x) \)

Finding the Coefficient \( C_n \)

To isolate \( C_n \), we multiply both sides of the expansion by \( \sin(\sqrt{\lambda_m} x) \): \[ 100 \sin(\sqrt{\lambda_m} x) = \sum_{n=1}^{\infty} C_n \sin(\sqrt{\lambda_n} x) \sin(\sqrt{\lambda_m} x) \]

Next, we integrate both sides over the interval \( 0 < x < L \).

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Integrating the expression from the previous page over the domain: \[ \int_{0}^{L} 100 \sin(\sqrt{\lambda_m} x) dx = \sum_{n=1}^{\infty} \int_{0}^{L} C_n \sin(\sqrt{\lambda_n} x) \sin(\sqrt{\lambda_m} x) dx \]

Note: Due to orthogonality, the integral on the right side is zero for all terms except where \( n = m \).

This simplifies the equation to a single term: \[ \int_{0}^{L} 100 \sin(\sqrt{\lambda_n} x) dx = \int_{0}^{L} C_n \sin^2(\sqrt{\lambda_n} x) dx \]

Final Expression for \( C_n \)

\[ C_n = \frac{\int_{0}^{L} 100 \sin(\sqrt{\lambda_n} x) dx}{\int_{0}^{L} \sin^2(\sqrt{\lambda_n} x) dx} \]

Special Case: Standard Sine Series

If the eigenvalues are given by \( \lambda_n = \frac{n^2 \pi^2}{L^2} \), the expression reduces back to the standard Fourier sine coefficient formula: \[ \frac{2}{L} \int_{0}^{L} 100 \sin\left(\frac{n\pi}{L}x\right) dx \]