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Sturm-Liouville Problem (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:

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

The Sturm-Liouville (SL) problem is considered regular if on the interval \( a \leq x \leq b \), the functions \( p(x) \), \( p'(x) \), \( q(x) \), and \( w(x) \) are continuous, and furthermore, \( p(x) > 0 \) and \( w(x) > 0 \).

If the coefficients \( \alpha_1, \alpha_2, \beta_1, \beta_2 \) are all non-negative, then the resulting eigenvalues \( \lambda \) are also non-negative.

If the SL problem is regular, then the eigenvalues are all real and form an increasing sequence:

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

where \( \lambda_n \to \infty \) as \( n \to \infty \). Each eigenvalue is paired up with an eigenfunction \( y_n \) that are mutually orthogonal with respect to the weight function \( w(x) \):

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

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

\( 0 < x < L \)

\( y(0) = 0 \)

\( y(L) = 0 \)

\[ \underbrace{p=1, \; q=0, \; w=1, \; \alpha_1=1, \; \alpha_2=0, \; \beta_1=1, \; \beta_2=0}_{\text{regular SL problem with non-negative } \lambda\text{'s}} \]

This is equivalent to:

\[ \bar{X}'' + \lambda \bar{X} = 0 \] \[ \bar{X}(0) = \bar{X}(L) = 0 \]

(Reason for separation constant \( = -\lambda \) because this is a regular SL problem)

Solutions

\[ \bar{X}_n = \sin\left(\frac{n\pi}{L}x\right) \]
\[ \lambda_n = \frac{n^2\pi^2}{L^2} \]

\( n = 1, 2, 3, \dots \)

\( \lambda_1 < \lambda_2 < \lambda_3 < \dots < \lambda_n < \dots \)
\[ \int_0^L \bar{X}_n \bar{X}_m w(x) dx = \int_0^L \sin\left(\frac{n\pi}{L}x\right) \sin\left(\frac{m\pi}{L}x\right) dx = 0 \]

Fourier series is just a special case of outcome of SL problem.

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Example: Regular Sturm-Liouville Problem with Potential Negative Eigenvalues

Consider the differential equation:

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

With boundary conditions:

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

Identifying the parameters for the Sturm-Liouville form:

\[ p = 1, \quad q = 0, \quad w = 1 \] \[ \alpha_1 = 1, \quad \alpha_2 = 0 \] \[ \beta_1 = h > 0, \quad \underbrace{\beta_2 = -1}_{\text{not nonnegative}} \]

This is still a regular Sturm-Liouville problem, but since \(\alpha\) and \(\beta\) are not all nonnegative, the eigenvalues \(\lambda\) are not guaranteed to be nonnegative. This means \(\lambda\) could be negative.

We need to check if a negative \(\lambda\) could satisfy the differential equation.

Checking for Negative Eigenvalues

Starting with the equation:

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

Assume a negative \(\lambda\). Let \(\lambda = -k^2\) where \(k > 0\). The choice of \(k^2\) is for convenience only, to avoid \(\sqrt{\lambda}\) in the solution.

Substituting this into the differential equation gives:

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

The general solution can be written in two forms:

\[ y = C_1 e^{kx} + C_2 e^{-kx} \]

or

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

Choose whichever form is easier to work with given the boundary conditions.

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Applying Boundary Conditions

First Boundary Condition: \(y(0) = 0\)

If we use the exponential form, applying \(y(0) = 0\) gives:

\[ 0 = C_1 + C_2 \]

This is not the best choice since we usually want one of the constants to be zero.

However, if we use the hyperbolic form:

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

Applying \(y(0) = 0\) gives:

\[ 0 = A \]

Therefore, the solution simplifies to:

\[ y = B \sinh(kx) \quad \text{where } B \neq 0 \]

Second Boundary Condition: \(h y(L) - y'(L) = 0\)

We use the other boundary condition to find out more about \(k\) (recall \(\lambda = -k^2\)).

The boundary condition is:

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

Using our simplified solution and its derivative:

\[ y = B \sinh(kx) \] \[ y' = k B \cosh(kx) \]

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

\[ h B \sinh(kL) = k B \cosh(kL) \]

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

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

To solve this, we can rewrite the right side:

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

Letting \(z = kL\), we need to solve:

\[ \underbrace{\tanh(z) = \frac{z}{hL}}_{\text{intersections (positive) of } \tanh(z) \text{ and } \frac{z}{hL}} \]
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Eigenvalue Analysis

Graph showing the intersection of the curves y = tanh(z) and y = z/(hL) in the first quadrant. The intersection point is labeled z_1.

Assume \( hL \) such that there is an intersection.

There is only one positive intersection: \( z_1 \).

Given \( z = kL \) and \( \lambda = -k^2 \), having only one \( z \) implies one \( k \), which in turn implies one negative \( \lambda \).

\[ \lambda_1 < \lambda_2 < \dots < \lambda_n < \dots \]
\( \uparrow \) negative
\( \uparrow \) must be \( \ge 0 \)

This negative \( \lambda \) is the smallest eigenvalue.

\[ \lambda_1 = -\left(\frac{z_1}{L}\right)^2 \]
Corresponding eigenfunction: \( y_1 = \sinh\left(\frac{z_1}{L}x\right) \)

Next: see if \( \lambda = 0 \) is an eigenvalue, then \( \lambda > 0 \).

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Revisiting the Heat Exchange Problem

Let's revisit the heat exchange problem from last time:

\[ u_t = k u_{xx} \quad 0 < x < L \] \[ u(0,t) = 0 \] \[ u_x(L,t) = -h u(L,t) \quad (h > 0) \] \[ u(x,0) = 100 \quad \text{initially heated to } u=100 \text{ uniformly} \]

Spatial Problem

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

\( \vdots \)

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

where \( z_n \) are positive intersections of \( \tan(z) \) and \( -\frac{z}{hL} \).

Time Part and General Solution

The time part is solved as usual. This led to the general solution:

\[ u(x,t) = \sum_{n=1}^{\infty} c_n e^{-k \lambda_n t} \sin(\sqrt{\lambda_n} x) \]
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Initial Condition and Orthogonality

Given the initial condition:

\[ u(x,0) = 100 \]
\[ 100 = \sum_{n=1}^{\infty} C_n \sin(\sqrt{\lambda_n} x) \]

Note: This is NOT a standard sine series since \( \sqrt{\lambda_n} \neq \frac{n\pi}{L} \).

Therefore, we cannot use the standard formula: \( C_n = \frac{2}{L} \int_0^L 100 \sin\left(\frac{n\pi}{L}x\right) dx \).

Orthogonality of Eigenfunctions

However, since this is a regular Sturm-Liouville (SL) problem, the theory guarantees the orthogonality of eigenfunctions:

\[ \int_0^L y_n y_m w(x) dx = 0 \quad \text{if } n \neq m \]

In this specific case, the weight function and eigenfunctions are:

\[ w(x) = 1 \] \[ y_n = \sin(\sqrt{\lambda_n} x) \]

Finding the Coefficients \( C_n \)

To find \( C_n \), we start with our series equation:

\[ 100 = \sum_{n=1}^{\infty} C_n \sin(\sqrt{\lambda_n} x) \]

Multiply both sides by \( \sin(\sqrt{\lambda_m} x) \):

\[ 100 \sin(\sqrt{\lambda_m} x) = \sum_{n=1}^{\infty} C_n \sin(\sqrt{\lambda_n} x) \sin(\sqrt{\lambda_m} x) \]

Then, integrate over the domain \( 0 < x < L \).

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Evaluating the Integral for \( C_n \)

Integrating both sides from \( 0 \) to \( L \) yields:

\[ \int_0^L 100 \sin(\sqrt{\lambda_n} x) dx = \int_0^L C_n [\sin(\sqrt{\lambda_n} x)]^2 dx \]

Because of orthogonality, all other terms in the sum are 0:

\[ \int_0^L y_n y_m dx = 0 \]

Solving for \( C_n \), we obtain the final expression:

\[ C_n = \frac{\int_0^L 100 \sin(\sqrt{\lambda_n} x) dx}{\int_0^L \sin^2(\sqrt{\lambda_n} x) dx} \]

Note: If \( \lambda_n = \frac{n^2\pi^2}{L^2} \), then this expression will reduce to the standard Fourier sine coefficient formula:

\[ \frac{2}{L} \int_0^L 100 \sin\left(\frac{n\pi}{L}x\right) dx \]