We consider again the heat equation:
\[
u_t = \alpha^2 u_{xx}
\]
Our PDE here looks the same, however, we are going to consider when our rod is semi-infinite and infinite in length. That means we will have
1 and 0 boundary conditions respectively. Until now, our techniques have only worked on homogenous boundary conditions. We now introduce a method
that does not rely on homogenous boundary conditions. This is due to the fact that these transforms essentially transform our derivatives into
multiplication. In other words, we can sometimes turn differential equations into algebraic ones. Even when this does not happen, we essentially
get an ODE in the variable that we did not transform, and this ODE is one degree lower than our PDE. This makes solving problems much easier. Once
we solve the transformed problem, we can just invert the transformation to get our solution to the original problem. Now that we know the motivation
behind these transforms, we can start to consider some very common and useful ones.
The Fourier Transform
The Fourier Transform is defined as:
\[
\mathcal{F}[f](\xi) = F(\xi) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-i\xi x}\, dx
\]
With the Inverse Fourier Transform defined as:
\[
\mathcal{F}^{-1}[F](x) = f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(\xi) e^{i\xi x}\, d\xi
\]
Example transform:
\[
\mathcal{F}[e^{-a^2x^2}] = \frac{1}{a\sqrt{2}} e^{-\omega^2/(4a^2)}
\]
Useful derivative properties:
\[
\mathcal{F}[u_x] = i\xi \mathcal{F}[u], \quad
\mathcal{F}[u_{xx}] = -\xi^2 \mathcal{F}[u]
\]
\[
\mathcal{F}[u_t] = \frac{\partial}{\partial t}\mathcal{F}[u], \quad
\mathcal{F}[u_{tt}] = \frac{\partial^2}{\partial t^2}\mathcal{F}[u]
\]
Convolution
\[
(f * g)(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x-\xi)g(\xi)\, d\xi
\]
Key property:
\[
\mathcal{F}^{-1}[\mathcal{F}[f]\mathcal{F}[g]] = f * g
\]
Solving the Heat Equation on an Infinite Rod
PDE with initial condition:
\[
u_t = \alpha^2 u_{xx}, \quad u(x,0) = \phi(x)
\]
After transform:
\[
\frac{\partial}{\partial t} U(\xi,t) = -\alpha^2 \xi^2 U(\xi,t), \quad U(\xi,0) = \mathcal{F}[\phi(x)]
\]
Solution:
\[
U(\xi,t) = \Phi(\xi) e^{-(\alpha \xi)^2 t}
\]
Inverse transform:
\[
u(x,t) = \mathcal{F}^{-1}\!\left[\Phi(\xi) e^{-(\alpha \xi)^2 t}\right]
\]
By convolution:
\[
u(x,t) = \phi(x) * \frac{1}{\alpha \sqrt{2t}} e^{-x^2/(4\alpha^2 t)}
\]
\[
u(x,t) = \frac{1}{2\alpha\sqrt{\pi t}} \int_{-\infty}^{\infty} \phi(\xi) e^{-(x-\xi)^2/(4\alpha^2 t)}\, d\xi
\]
The Laplace Transform
The Laplace Transform is defined as:
\[
\mathcal{L}[f](s) = F(s) = \int_0^\infty f(t)e^{-st}\, dt
\]
With inverse:
\[
\mathcal{L}^{-1}[F](t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} F(s)e^{st}\, ds
\]
Useful derivative properties:
\[
\mathcal{L}[u_t] = sU(x,s) - u(x,0), \quad
\mathcal{L}[u_{tt}] = s^2U(x,s) - s u(x,0) - u_t(x,0)
\]
\[
\mathcal{L}[u_x] = \frac{\partial U}{\partial x}(x,s), \quad
\mathcal{L}[u_{xx}] = \frac{\partial^2 U}{\partial x^2}(x,s)
\]
Convolution property:
\[
(f*g)(t) = \int_0^t f(\tau)g(t-\tau)\, d\tau, \quad
(f*g)(t) = \mathcal{L}^{-1}[\mathcal{L}[f]\mathcal{L}[g]]
\]
Heat Equation on a Semi-Infinite Rod
PDE:
\[
u_t = \alpha^2 u_{xx}
\]
After Laplace transform:
\[
sU(x,s) - u(x,0) = \alpha^2 U''(x,s)
\]
Which simplifies to:
\[
\alpha^2 U''(x) - sU(x) + u(x,0) = 0
\]
This is now an ODE in \(x\). Solve using transformed boundary conditions, then invert with the Laplace inverse to get \(u(x,t)\).