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PAGE 1

Sturm-Liouville

\[ y'' + \lambda y = 0 \]
\( a < x < b \)

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

BCs:

\( a = 0, \quad b = 3 \)
\( \alpha_1 = 1, \quad \alpha_2 = 0 \)
\( \beta_1 = 5, \quad \beta_2 = 10 \)

goal: find \( \lambda_n \) and \( y_n \)

express some function using the \( y_n \)

\( y = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \) (\(\lambda > 0\))

\( y(0) = 0 \rightarrow A = 0 \)

\( y = B \sin(\sqrt{\lambda} x) \)

\( y' = \sqrt{\lambda} B \cos(\sqrt{\lambda} x) \)

\( 5 y(3) + 10 y'(3) = 0 \)
\( y(3) + 2 y'(3) = 0 \)

\( B \sin(3\sqrt{\lambda}) + 2\sqrt{\lambda} B \cos(3\sqrt{\lambda}) = 0 \) \( B \neq 0, \lambda \neq 0 \)

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\( \sin(3\sqrt{\lambda}) + 2\sqrt{\lambda} \cos(3\sqrt{\lambda}) = 0 \)

\( \sin(3\sqrt{\lambda}) = -2\sqrt{\lambda} \cos(3\sqrt{\lambda}) \)

\( \tan(3\sqrt{\lambda}) = -2\sqrt{\lambda} \)

let \( z = 3\sqrt{\lambda} \)

\( \tan(z) = -\frac{2}{3} z \) intersections (positive) of \( \tan(z) \) and \( -\frac{2}{3} z \)

A graph showing the intersection of the tangent function tan(z) with the linear function -2/3 z. The tangent curves have vertical asymptotes, and the line has a negative slope passing through the origin. Intersection points are labeled z1 and z2 on the positive z-axis.

\( z = 3\sqrt{\lambda} \)

\[ \lambda_n = \frac{z_n^2}{9} \]

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

\[ y_n = \sin\left(\frac{z_n}{3} x\right) \]

can \( \lambda \) be 0?

\( y'' + \lambda y = 0 \rightarrow y = Ax + B \)

\( y(0) = 0 \rightarrow B = 0 \)

\( y(3) + 2y'(3) = 0 \rightarrow 3A + 2A = 0 \rightarrow A = 0 \) trivial solution, so \( \lambda \neq 0 \)

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eigenfunction \( y_n = \sin\left(\frac{n\pi}{3}x\right) = \sin(\sqrt{\lambda}x) \)

mutually orthogonal: \( \int_{0}^{3} y_n y_m dx = 0 \quad n \neq m \)

we can express any function using these eigenfunctions

\[ f(x) = \sum_{n=1}^{\infty} c_n y_n \]

find \( c_n \):

multiply by \( y_m \)

\[ f(x) y_m = \sum_{n=1}^{\infty} c_n y_n y_m \]
\[ \int_{0}^{3} f(x) y_m dx = \underbrace{\int_{0}^{3} \sum_{n=1}^{\infty} c_n y_n y_m dx}_{\text{all 0 except } n=m} \]
\[ \int_{0}^{3} f(x) y_n dx = \int_{0}^{3} c_n (y_n)^2 dx \]
\(\rightarrow\)
\[ c_n = \frac{\int_{0}^{3} f(x) y_n dx}{\int_{0}^{3} (y_n)^2 dx} \]
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Heat eq

\( u_t = u_{xx} \quad 0 < x < 2 \)

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

\( u(2, t) = 0 \)

\( u(x, 0) = f(x) \)

\[ \frac{\bar{X}''}{\bar{X}} = + \frac{T'}{T} = -\lambda \]

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

\( \bar{X}(0) = \bar{X}(2) = 0 \)

\( \lambda_n = \frac{n^2 \pi^2}{4} \)

\( \bar{X}_n = \sin\left(\frac{n\pi}{2} x\right) \)

\( T' + \lambda T = 0 \)

\( T' + \frac{n^2 \pi^2}{4} T = 0 \)

\( T = A e^{-\frac{n^2 \pi^2}{4} t} \)

\( T_n = e^{-\frac{n^2 \pi^2}{4} t} \)

\[ u(x, t) = \sum_{n=1}^{\infty} c_n e^{-n^2 \pi^2 / 4 t} \sin\left(\frac{n\pi}{2} x\right) \]

\[ f(x) = u(x, 0) = \sum_{n=1}^{\infty} c_n \sin\left(\frac{n\pi}{2} x\right) \]

Sine series

\[ c_n = \frac{2}{2} \int_0^2 f(x) \sin\left(\frac{n\pi}{2} x\right) dx \]

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Insulated / steady-state

\( u_t = u_{xx} \)

\( u_x(0, t) = 0 \)

\( u_x(1, t) = 0 \)

\( u(x, 0) = f(x) \)

\( 0 < x < 1 \)
\[ \frac{X''}{X} = \frac{T'}{T} = -\lambda \]
\( X'' + \lambda X = 0 \) \( X'(0) = X'(1) = 0 \)

\( X = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \) \( (\lambda > 0) \)

\( X' = -A \sqrt{\lambda} \sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x) \)

\( 0 = B \sqrt{\lambda} \rightarrow B = 0 \)

\( 0 = -A \sqrt{\lambda} \sin(\sqrt{\lambda}) \)

\( \sin(\sqrt{\lambda}) = 0 \rightarrow \sqrt{\lambda} = n\pi \) \( n = 1, 2, 3, \dots \)

\( \lambda_n = n^2 \pi^2 \)

\( X_n = \cos(n\pi x) \)

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if \( \lambda = 0 \),

\( X'' = 0 \)

\( X = Ax + B \)

\( X' = A \)

\( \underbrace{X'(0) = X'(1) = 0}_{\text{implies } A = 0} \)

\( \lambda = 0 \)

\( X = 1 \)

\[ u(x, t) = \frac{1}{2} c_0 \cdot 1 + \sum_{n=1}^{\infty} c_n e^{-n^2 \pi^2 t} \cos(n\pi x) \] \[ f(x) = u(x, 0) = \frac{1}{2} c_0 + \sum_{n=1}^{\infty} c_n \cos(n\pi x) \] \[ c_n = \frac{2}{1} \int_{0}^{1} f(x) \cos(n\pi x) dx \]

steady-state: \( t \to \infty \)

\( e^{-n^2 \pi^2 t} \to 0 \)

\( u \to \frac{1}{2} c_0 \)

average value of initial heat distribution

Two coordinate graphs showing the evolution of heat distribution. The left graph shows an irregular initial curve f(x) over x. The right graph shows the steady-state result where the distribution has flattened into a horizontal line at the average value, with an arrow indicating the transition and a note 'same area'.
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Laplace transform

\[ \mathcal{L} \{u(x,t)\} = \int_{0}^{\infty} u(x,t) e^{-st} dt \]
\(\leftarrow\) transform away \(t\)

\[ \mathcal{L} \{u\} = U(x,s) \]

\[ \mathcal{L} \{u_t(x,t)\} = s U(x,s) - u(x,0) \]
(just like \( \mathcal{L} \{y'\} = sY - y(0) \))

\[ \mathcal{L} \{u_{tt}\} = s^2 U - s u(x,0) - u_t(x,0) \]

\[ \mathcal{L} \{u_x\} = \frac{d}{dx} U \]

\[ \mathcal{L} \{u_{xx}\} = \frac{d^2}{dx^2} U \]

example heat eq:

\[ u_t = 3 u_{xx} \]

\[ u(x,0) = f(x) = 1 \]

\[ u(0,t) = 10 \]

\( 0 < x < \infty \)

temp is finite at right "end"

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\[ \mathcal{L} \{u_t\} = \mathcal{L} \{3 u_{xx}\} \]

\[ s U - u(x,0) = 3 U'' \quad \rightarrow \quad 3 U'' - s U = -1 \]

solve \[ 3 U'' - s U = 0 \]

\[ U'' - \frac{s}{3} U = 0 \]

\[ U = A e^{\sqrt{\frac{s}{3}}x} + B e^{-\sqrt{\frac{s}{3}}x} \]

Left BC:

\[ u(0,t) = 10 \]

\[ U(x=0) = \frac{10}{s} \]

\[ \rightarrow A + B = \frac{10}{s} \]

Right BC:

bounded so \( A = 0 \) and \( B = \frac{10}{s} \)

\[ U = \frac{10}{s} e^{-\sqrt{\frac{s}{3}}x} \]
solution in s-domain

then nonhomogeneous part

then inverse transform