Download original notes
PAGE 1

Laplace Transform for PDEs

we know

\[ \mathcal{L} \{ y(t) \} = \int_{0}^{\infty} y(t) e^{-st} dt = Y(s) \] \[ \mathcal{L} \{ y'(t) \} = s Y(s) - y(0) \] \[ \mathcal{L} \{ y''(t) \} = s^2 Y(s) - s y(0) - y'(0) \]

\( t \) is integrated away

replaced w/ \( s \)

now let's do the same for \( u(x, t) \)

\[ \mathcal{L} \{ u(x, t) \} = \int_{0}^{\infty} u(x, t) e^{-st} dt = U(x, s) \]
\( x \) is "constant"
\[ \mathcal{L} \{ u_t(x, t) \} = s U(x, s) - u(x, 0) \] \[ \mathcal{L} \{ u_{tt}(x, t) \} = s^2 U(x, s) - s u(x, 0) - u_t(x, 0) \]

how about \( u_x(x, t) \)?

\[ \mathcal{L} \{ u_x(x, t) \} = \int_{0}^{\infty} u_x(x, t) e^{-st} dt \]
\( x \) does not get involved here
\[ = \int_{0}^{\infty} \frac{\partial}{\partial x} u(x, t) e^{-st} dt \]
PAGE 2
\[ = \frac{\partial}{\partial x} \left[ \int_{0}^{\infty} u(x, t) e^{-st} dt \right] \] \[ = \frac{\partial}{\partial x} U(x, s) \]
\[ \mathcal{L} \{ u_{xx}(x, t) \} = \frac{\partial^2}{\partial x^2} U(x, s) \]

let's do a simple heat eq.

example

\( u_t = u_{xx} \) \( 0 < x < \pi \) , \( t > 0 \)
  • \( u(0, t) = 0 \)
  • \( u(\pi, t) = 0 \)
  • \( u(x, 0) = \sin x \)
A diagram showing a horizontal line segment representing a spatial domain from x=0 to x=pi. A semi-circular arc is drawn above the segment labeled u=0 at the endpoints and u=sin(x) for the initial state.
\[ \mathcal{L} \{ u_t \} = \mathcal{L} \{ u_{xx} \} \]
\[ s U(x, s) - u(x, 0) = \frac{\partial^2}{\partial x^2} U(x, s) \]
\( u(x, 0) = \sin x \)
\[ s U - \sin(x) = U'' \]
(prime \( \rightarrow \) deriv. w/ \( x \))
PAGE 3
\[ U'' - sU = -\sin(x) \]
think of it as \( y'' - ay = -\sin(x) \)

find complementary solution: \( U'' - sU = 0 \) \( s \) is "constant"

\[ U = C_1 e^{\sqrt{s}x} + C_2 e^{-\sqrt{s}x} \]

to find \( C_1, C_2 \), use the BCs

  • \( u(0, t) = 0 \rightarrow \mathcal{L}\{u(0, t)\} = \mathcal{L}\{0\} \rightarrow U(x=0) = 0 \)
  • \( u(\pi, t) = 0 \rightarrow U(x=\pi) = 0 \)

w/ those we get

\[ 0 = C_1 + C_2 \] \[ 0 = C_1 e^{\sqrt{s}\pi} + C_2 e^{-\sqrt{s}\pi} \]
\[ \vdots \]
\[ C_1 = C_2 = 0 \]

now the particular solution

using undetermined coeff: \( U_p = A \cos x + B \sin x \)

plug into \( U'' - sU = -\sin(x) \)

PAGE 4
\[ \vdots \]
\[ A = 0, \quad B = \frac{1}{s+1} \]
\[ U(x, s) = \frac{1}{s+1} \sin x \]

solution is \( s \)-domain

we often want the solution in \( t \)-domain

\[ u(x, t) = \mathcal{L}^{-1} \left\{ \frac{1}{s+1} \sin x \right\} \]
\( s \) is variable
\( x \) is "constant"
\[ = \sin x \, \mathcal{L}^{-1} \left\{ \frac{1}{s+1} \right\} \]
\[ u(x, t) = e^{-t} \sin x \]

Laplace transform is particularly effective w/ infinite domains and time-vary boundary conditions

\( \rightarrow \) separation of variables can't handle these easily

PAGE 5

Example: Heat Equation on a Semi-Infinite Rod

\[ u_t = u_{xx} \quad 0 < x < \infty \]

\( u(0, t) = \sin(t) \) (left end temp = \( \sin(t) \))

\( u(x, 0) = 0 \) initially frozen

A horizontal cylinder representing a rod, with arrows pointing to the right indicating an infinite domain.

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

\[ sU - u(x, 0) = U'' \rightarrow U'' - sU = 0 \]

\[ U = C_1 e^{\sqrt{s}x} + C_2 e^{-\sqrt{s}x} \]

"hidden" BC at \( \infty \): temp must be bounded at \( \infty \)

\( \rightarrow C_1 = 0 \)

\[ U = C_2 e^{-\sqrt{s}x} \]

BC at \( x = 0 \) : \( u(0, t) = \sin(t) \)

\[ U(x=0) = \frac{1}{s^2 + 1} \]

so, \( C_2 = \frac{1}{s^2 + 1} \)

PAGE 6

\[ U(x, s) = \frac{1}{s^2 + 1} e^{-\sqrt{s}x} \]

Solution in S-domain

back to t: convolution

\[ \mathcal{L} \left\{ \int_0^t f(t-\tau) g(\tau) d\tau \right\} = FG \]

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s^2 + 1} \right\} = f(t) = \sin(t) \]

\[ \mathcal{L}^{-1} \left\{ e^{-\sqrt{s}x} \right\} = \frac{x}{2\sqrt{\pi t^3}} e^{-x^2/4t} = g(t) \]

\[ u(x, t) = \int_0^t \sin(t-\tau) \frac{x}{2\sqrt{\pi \tau^3}} e^{-x^2/4\tau} d\tau \]

\( \sin(t) \): the heat we put in

\[ \frac{x}{2\sqrt{\pi t^3}} e^{-x^2/4t} \]

: how heat moves in this rod

(heat kernel)

PAGE 7

Heat Kernel as Impulse Response

Heat kernel is also the impulse response in reaction to an impulse of heat.

A coordinate graph showing a decaying curve starting from a point on the vertical u-axis and tapering off along the horizontal x-axis.
+
A coordinate graph showing a similar decaying curve, but shifted further along the x-axis compared to the first graph.
+
A coordinate graph showing a small impulse or curve near the origin, followed by a series of dots indicating a continuation of the sequence.

What is measured is the sum (integral) of all those.

PAGE 8

Thermal Wave Propagation into the Rod

A 3D surface plot showing thermal wave propagation. The vertical axis represents temperature u(x,t), ranging from -0.5 to 0.5. The horizontal axes represent Distance x (0 to 5) and Time t (0 to 10). The surface shows two distinct peaks of heat propagating and decaying over time and distance.

The visualization above illustrates how heat waves propagate through a rod over time, showing the spatial and temporal distribution of temperature \( u(x,t) \).

PAGE 9

Oscillatory Boundary Condition: \( u(0, t) = \sin(t) \)

The following graph illustrates the temperature distribution \( u(x, t) \) as a function of distance \( x \) for various time steps \( t \). The boundary condition at \( x = 0 \) is driven by a sinusoidal oscillation, causing waves of temperature to propagate into the medium and decay as distance increases.

A line graph showing temperature u(x, t) versus distance x for five different time values: t = 0.00, 1.57, 3.14, 4.71, and 6.28. At x = 0, the temperature values follow a sine wave. As x increases, the oscillations dampen significantly, converging toward zero by x = 6.

Key Observations

  • At \( x = 0 \), the temperature matches the boundary condition \( \sin(t) \).
  • The amplitude of the temperature oscillations decreases exponentially with distance \( x \).
  • Phase shifts are visible as the peaks and troughs of the waves move further into the material over time.