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2D Case

A rectangular domain in the xy-plane with vertices at (0,0), (a,0), (a,b), and (0,b). The boundary conditions are labeled as 0 on all four sides of the rectangle.

\[ u_t = k(u_{xx} + u_{yy}) \quad 0 < x < a, \quad 0 < y < b \]

  • \( u(x, 0) = 0 \)
  • \( u(x, b) = 0 \)
  • \( u(0, y) = 0 \)
  • \( u(a, y) = 0 \)

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

\[ u = X Y T \]

\[ X Y T' = k(X'' Y T + X Y'' T) \]

\[ \frac{T'}{k T} = \frac{X''}{X} + \frac{Y''}{Y} \]
goal: want to see \( X'' + \lambda X = 0 \) or the \( Y \) version
\[ \frac{X''}{X} = \frac{T'}{k T} - \frac{Y''}{Y} = -\lambda \]

\( \rightarrow X'' + \lambda X = 0 \quad X(0) = X(a) = 0 \)

\[ \lambda_n = \frac{n^2 \pi^2}{a^2} \quad X_n = \sin\left(\frac{n \pi}{a} x\right) \]

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\[ \frac{Y''}{Y} = \frac{T'}{k T} + \lambda = -\mu \]

\( \rightarrow Y'' + \mu Y = 0 \quad Y(0) = Y(b) = 0 \)

\[ \mu_m = \frac{m^2 \pi^2}{b^2} \quad Y_m = \sin\left(\frac{m \pi}{b} y\right) \]

\[ \frac{T'}{k T} = -(\lambda + \mu) \]

\[ T' + k(\lambda + \mu) T = 0 \]

\[ T_{nm} = e^{-k \left( \frac{n^2 \pi^2}{a^2} + \frac{m^2 \pi^2}{b^2} \right) t} \]

\[ u(x, y, t) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} C_{nm} e^{-k \left( \frac{n^2 \pi^2}{a^2} + \frac{m^2 \pi^2}{b^2} \right) t} \sin\left(\frac{n \pi}{a} x\right) \sin\left(\frac{m \pi}{b} y\right) \]

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

\[ f(x, y) = \sum_{n=1}^{\infty} \left[ \underbrace{\sum_{m=1}^{\infty} C_{nm} \sin\left(\frac{m \pi}{b} y\right)}_{\text{Constant if } y \text{ is fixed}} \right] \sin\left(\frac{n \pi}{a} x\right) \]

sine series w/ \( L = a \)

w/ constants \( \sum_{m=1}^{\infty} C_{nm} \sin\left(\frac{m \pi}{b} y\right) \)

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\[ D = \sum_{m=1}^{\infty} C_{nm} \sin\left(\frac{n\pi}{b}y\right) = \underbrace{\frac{2}{a} \int_{0}^{a} f(x,y) \sin\left(\frac{n\pi x}{a}\right) dx}_{\text{This term is substituted for } D \text{ below}} \]

\[ C_{nm} = \frac{2}{b} \int_{0}^{b} D \sin\left(\frac{n\pi}{b}y\right) dy = \frac{4}{ab} \int_{0}^{a} \int_{0}^{b} f(x,y) \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{b}\right) dy dx \]

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#6 practice problems

\[ 2L = 4 \]

BCs:

\[ u(-2, t) = u(2, t) \]

\[ u_x(-2, t) = u_x(2, t) \]

A horizontal line segment labeled with length 2L, with endpoints at x equals negative 2 and x equals 2. An arrow points from the segment to a circle, indicating periodic boundary conditions.

BCs must match

\[ u_t = k u_{xx} \quad \text{for } -2 < x < 2 \]

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

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

\[ T' + k\lambda T = 0 \]

\[ u(-2, t) = u(2, t) \]

\( \rightarrow \)

\[ X(-2) = X(2) \]

\[ u_x(-2, t) = u_x(2, t) \]

\( \rightarrow \)

\[ X'(-2) = X'(2) \]

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

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

\( X(-2) = A \cos(-2\sqrt{\lambda}) + B \sin(-2\sqrt{\lambda}) \)

\( X(2) = A \cos(2\sqrt{\lambda}) + B \sin(2\sqrt{\lambda}) \)

\( \} \) same

Note: From the equality above,

\( -B \sin(2\sqrt{\lambda}) = B \sin(2\sqrt{\lambda}) \)

\( 2B \sin(2\sqrt{\lambda}) = 0 \)

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

\( X'(-2) = -\sqrt{\lambda} A \sin(-2\sqrt{\lambda}) + \sqrt{\lambda} B \cos(-2\sqrt{\lambda}) \)

\( X'(2) = -\sqrt{\lambda} A \sin(2\sqrt{\lambda}) + \sqrt{\lambda} B \cos(2\sqrt{\lambda}) \)

\( \} \) same

Note: From the derivative equality,

\( \sqrt{\lambda} A \sin(2\sqrt{\lambda}) = -\sqrt{\lambda} A \sin(2\sqrt{\lambda}) \)

\( 2A \sqrt{\lambda} \sin(2\sqrt{\lambda}) = 0 \)

\( \sin(2\sqrt{\lambda}) = 0 \)

\( 2\sqrt{\lambda} = n\pi \)

find \( \lambda \) the usual way

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

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

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\( u_t = u_{xx} \)

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

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

\( 0 < x < 1 \)

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

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

\( X'' + \lambda X = 0 \quad X'(0) = X'(1) = 0 \)

\( \lambda_n = \frac{n^2 \pi^2}{1^2} = n^2 \pi^2 \)

\( X_n = \cos(n\pi x) \quad n = 0, 1, 2, 3, \dots \)

\( T' + n^2 \pi^2 T = 0 \)

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

\[ u(x, t) = \sum_{n=0}^{\infty} C_n e^{-n^2 \pi^2 t} \cos(n\pi x) \]

at \( t = 0 \)

\[ f(x) = \sum_{n=0}^{\infty} C_n \cos(n\pi x) = \underbrace{\frac{1}{2} C_0}_{\text{artificially introduced}} + \sum_{n=1}^{\infty} C_n \cos(n\pi x) \]

Note: \( \frac{1}{2} C_0 \) is artificially introduced so we have cosine series

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Problem 12: Separation of Variables on a Rectangular Domain

A coordinate graph showing a rectangle in the first quadrant. The horizontal axis is x and the vertical axis is y. The rectangle has width 'a' and height 'b'. The boundary conditions are written on each side: the left side at x=0 is u=0, the right side at x=a is u=g(y), the bottom side at y=0 is u_y=0, and the top side at y=b is u_y=0.

Using separation of variables, we assume a solution of the form \( u(x,y) = X(x)Y(y) \). This leads to:

\[ \frac{X''}{X} = -\frac{Y''}{Y} = \lambda \]

Solving for \( Y(y) \)

\[ Y'' + \lambda Y = 0 \quad \text{with boundary conditions} \quad Y'(0) = Y'(b) = 0 \]

The eigenvalues and eigenfunctions are:

\[ \lambda_n = \frac{n^2 \pi^2}{b^2}, \quad Y_n = \cos\left(\frac{n \pi}{b} y\right) \]

for \( n = 0, 1, 2, \dots \)

Solving for \( X(x) \)

Case 1: \( \lambda \neq 0 \)

\[ X'' - \lambda X = 0, \quad X(0) = 0 \] \[ X = A \cosh\left(\frac{n \pi}{b} x\right) + B \sinh\left(\frac{n \pi}{b} x\right) \]

Applying the boundary condition \( X(0) = 0 \):

\[ 0 = A \implies X_n = \sinh\left(\frac{n \pi}{b} x\right) \]

Case 2: \( \lambda = 0 \)

\[ X'' = 0, \quad X(0) = 0 \] \[ X = Ax + B \]

Applying the boundary condition \( X(0) = 0 \):

\[ X(0) = 0 \implies B = 0 \implies X_0 = x \]