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PAGE 1

Laplace Transform for PDEs

we know

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

we can do the same for \( u(x, t) \)

\( \rightarrow \) we transform the variable \( t \) away, leave \( x \) alone

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

now

\( u_x(x, t) \)

\[ \mathcal{L} \{ u_x(x, t) \} = \int_{0}^{\infty} u_x(x, t) e^{-st} dt = \int_{0}^{\infty} \frac{\partial}{\partial x} u(x, t) e^{-st} dt \]
\( x \) does NOT get involved
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\[ = \frac{\partial}{\partial x} \left[ \int_{0}^{\infty} u(x, t) e^{-st} dt \right] \]
Swap order of \( \int_{0}^{\infty} \dots dt \) and \( \frac{\partial}{\partial x} \)
\[ \mathcal{L} \{ u_x \} = \frac{\partial}{\partial x} U(x, s) \] \[ \mathcal{L} \{ u_{xx} \} = \frac{\partial^2 U(x, s)}{\partial x^2} \]

now let's solve some PDEs that we have seen before

Example

\[ u_t = u_{xx} \quad 0 < x < \pi \quad t > 0 \]

\[ u(0, t) = u(\pi, t) = 0 \quad \text{ends fixed at } 0 \]

\[ u(x, 0) = \sin x \quad \text{initial heat profile is } \sin(x) \]

A diagram showing a horizontal line segment on an x-axis from x=0 to x=pi. Above the segment is a curved arc labeled sin x, representing the initial temperature distribution along the rod.
\[ \mathcal{L} \{ u_t \} = \mathcal{L} \{ u_{xx} \} \] \[ s U(x, s) - u(x, 0) = \frac{\partial^2 U(x, s)}{\partial x^2} \]

for convenience, let's drop \( (x, s) \) from now on

\[ s U - \sin x = U'' \]
prime is deriv. with respect to \( x \)
PAGE 3

Solving Differential Equations with Laplace Transforms - Page 3

rewrite: \[ U'' - sU = \sin x \]

treat \( s \) as constant for now

\( \hookrightarrow \) treat this equation like \[ y'' - ay = \sin x \]

first, find complementary solution

\[ U'' - sU = 0 \]

(\( s > 0 \) "constant")

\[ U = c_1 e^{\sqrt{s}x} + c_2 e^{-\sqrt{s}x} \]

BCs:

  • \( u(0, t) = 0 \quad \rightarrow \quad U(0) = 0 \)
  • \( u(\pi, t) = 0 \quad \rightarrow \quad U(\pi) = 0 \)

(LT of the BCs, too)

\( U(0) = 0 \rightarrow 0 = c_1 + c_2 \)

\( U(\pi) = 0 \rightarrow 0 = c_1 e^{\sqrt{s}\pi} + c_2 e^{-\sqrt{s}\pi} \)

}
\( c_1 = c_2 = 0 \)

now, the particular solution due to \( \sin x \) (right side)

using undetermined coefficients, \( U_p = A \cos x + B \sin x \)

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

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Solving Differential Equations with Laplace Transforms - Page 4

\( \vdots \)

\[ A = 0, \quad B = -\frac{1}{s+1} \]

general solution:

\[ U(x, s) = -\frac{1}{s+1} \sin x \]

in s-domain

back to t-domain:

\[ u(x, t) = \mathcal{L}^{-1} \{ U(x, s) \} = \mathcal{L}^{-1} \left\{ -\frac{1}{s+1} \sin x \right\} \]

\[ = -\sin x \cdot \mathcal{L}^{-1} \left\{ \frac{1}{s+1} \right\} \]

\( x \) doesn't take part

so, \( \sin x = \) "constant"

\[ u(x, t) = -e^{-t} \sin x \]

Laplace transform is particularly effective w/ infinite domains or when BCs are time-varying

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Example: Heat Equation on a Semi-Infinite Domain

PDE: \[ u_t = u_{xx} \quad 0 < x < \infty \]
  • 1 BC: \( u(0, t) = \sin(t) \)
  • 1 IC: \( u(x, 0) = 0 \)
A horizontal bar starting at a boundary marked x=0 on the left, with an arrow pointing to the right towards infinity.

Left end varying w/ time; initially whole bar is frozen

Taking the Laplace Transform of the PDE:

\[ \mathcal{L} \{ u_t \} = \mathcal{L} \{ u_{xx} \} \] \[ sU - u(x, 0) = U'' \rightarrow U'' - sU = 0 \]

The general solution for the ODE in \( x \) is:

\[ U = c_1 e^{\sqrt{s}x} + c_2 e^{-\sqrt{s}x} \]

“BC at infinity” \( \rightarrow \) solution must be bounded

\( \rightarrow c_1 = 0 \)

\[ U = c_2 e^{-\sqrt{s}x} \]

Left BC:

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

\[ \mathcal{L} \{ u(0, t) \} = \mathcal{L} \{ \sin(t) \} \] \[ U(0) = \frac{1}{s^2 + 1} \]
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\[ U = c_2 e^{-\sqrt{s}x} \] \[ U(x=0) = \frac{1}{s^2 + 1} \rightarrow c_2 = \frac{1}{s^2 + 1} \]
\[ U(x, s) = \frac{1}{s^2 + 1} e^{-\sqrt{s}x} \]

Now back to t-domain

Convolution: \[ \mathcal{L} \{ f * g \} = F \cdot G \]
\[ \underbrace{\frac{1}{s^2 + 1}}_{F} \underbrace{e^{-\sqrt{s}x}}_{G} \]

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

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

\[ u(x, t) = \int_{0}^{t} \sin(t - \tau) \frac{x}{2\sqrt{\pi \tau^3}} e^{-\frac{x^2}{4\tau}} d\tau \]
PAGE 7

Convolution and the Heat Kernel

Convolution: treating input function \( (\sin t) \) as a bunch of impulses.

Each impulse of heat propagates down according to

\[ \frac{x}{2\sqrt{\pi t^3}} e^{-x^2/4t} \]
("heat kernel")
A coordinate graph showing a single impulse response curve starting at the origin on the u-axis and decaying as it moves along the x-axis.
+
A coordinate graph showing a second impulse response curve, shifted further along the time/distance axis, illustrating the additive nature of convolution.
+ ...
PAGE 8

Thermal Wave Propagation into the Rod

A 3D surface plot showing thermal wave propagation. The vertical axis represents temperature u(x,t), while the horizontal axes represent Distance x and Time t. The surface shows periodic peaks and valleys that dampen as distance x increases.
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Oscillatory Boundary Condition: \( u(0, t) = \sin(t) \)

The following graph illustrates the spatial distribution of temperature \( u(x, t) \) over distance \( x \) at various time intervals \( t \). The boundary condition at \( x = 0 \) is driven by a sinusoidal oscillation, causing thermal waves that 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. The curves oscillate at the origin (x=0) and rapidly dampen toward zero as x increases beyond 6. At t=1.57, the temperature starts at 1.00, while at t=4.71, it starts at -1.00.

Key Observations

  • At \( x = 0 \), the temperature follows the boundary condition \( \sin(t) \).
  • As \( x \) increases, the amplitude of the temperature oscillations decreases exponentially, demonstrating the diffusive nature of the heat equation.
  • By approximately \( x = 6 \), the temperature fluctuations have effectively dampened to zero for all shown time steps.