Applications to Differential Equations
Applications to Differential Equations
The Fourier transform converts differential equations into algebraic equations. For the heat equation $u_t=u_{xx}$, Fourier series diagonalizes the spatial operator: each mode $e^{inx}$ evolves independently as $e^{-n^2 t}$. For PDEs on $\mathbb{R}$, the Fourier transform converts $\partial_x$ to multiplication by $2\pi i\xi$, reducing PDEs to ODEs in time. Green's functions (fundamental solutions) are computed by inverting the transformed equation.
Fourier Methods for PDEs
The heat equation $u_t=\kappa u_{xx}$ on $[0,L]$ with $u(0,t)=u(L,t)=0$: expand $u(x,t)=\sum b_n(t)\sin(n\pi x/L)$. Substituting: $b_n'(t)=-\kappa(n\pi/L)^2 b_n(t)$, so $b_n(t)=b_n(0)e^{-\kappa n^2\pi^2 t/L^2}$. Initial data $u(x,0)=f(x)$ gives $b_n(0)$ via the sine series. High-frequency modes decay fastest — spatial structure smooths out exponentially.
Fundamental Solution of the Heat Equation
The heat equation $u_t=u_{xx}$ on $\mathbb{R}$ with initial data $u(x,0)=\delta(x)$ has solution $K(x,t)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}$ (heat kernel). Derivation via Fourier transform: $\hat{u}_t=-4\pi^2\xi^2\hat{u}$, so $\hat{u}(\xi,t)=e^{-4\pi^2\xi^2 t}$. Inverting: $u=\mathcal{F}^{-1}\{e^{-4\pi^2\xi^2 t}\}=K(x,t)$ (Gaussian). General solution: $u(x,t)=(K(\cdot,t)*f)(x)=\frac{1}{\sqrt{4\pi t}}\int f(y)e^{-(x-y)^2/(4t)}\,dy$.
Example 1
Solve $u_t=u_{xx}$, $u(x,0)=e^{-x^2}$ on $\mathbb{R}$.
Solution: $\hat{u}(\xi,0)=\hat{f}(\xi)=\sqrt{\pi}e^{-\pi^2\xi^2}$ (Fourier transform of Gaussian). After time $t$: $\hat{u}(\xi,t)=e^{-4\pi^2\xi^2 t}\hat{f}(\xi)=\sqrt{\pi}e^{-\pi^2\xi^2(1+4t)}$. Inverting: $u(x,t)=\frac{1}{\sqrt{1+4t}}e^{-x^2/(1+4t)}$. The Gaussian broadens: width $\sim\sqrt{1+4t}$, peak $\sim 1/\sqrt{1+4t}$.
Example 2
Use the wave equation $u_{tt}=c^2 u_{xx}$ with initial data $u(x,0)=f(x)$, $u_t(x,0)=0$ to derive d'Alembert's formula.
Solution: Fourier transform in $x$: $\hat{u}_{tt}=-4\pi^2\xi^2 c^2\hat{u}$. Solution: $\hat{u}(\xi,t)=\hat{f}(\xi)\cos(2\pi c\xi t)$. Invert: $u=\mathcal{F}^{-1}\{\hat{f}\cos(2\pi c\xi t)\}=\frac{1}{2}\mathcal{F}^{-1}\{\hat{f}(e^{2\pi ic\xi t}+e^{-2\pi ic\xi t})\}=\frac{1}{2}[f(x+ct)+f(x-ct)]$, d'Alembert's formula. Two waves travel at speed $\pm c$.
Practice
- Solve the heat equation $u_t=u_{xx}$, $u(0,t)=u(\pi,t)=0$, $u(x,0)=\sin x+\frac{1}{2}\sin 3x$.
- Derive the Poisson kernel for Laplace's equation in the disk: $\Delta u=0$, $u(1,\theta)=f(\theta)$.
- Show the fundamental solution of $u_t=u_{xx}+u$ is $e^t K(x,t)$ where $K$ is the heat kernel.
- Use Fourier transforms to solve $u_{xx}+u_{yy}=f(x)\delta(y)$ (2D Laplacian driven by a line source).
Show Answer Key
1. $u(x,t)=e^{-t}\sin x+\frac{1}{2}e^{-9t}\sin 3x$. Each mode $\sin(nx)$ decays as $e^{-n^2 t}$.
2. For $\Delta u=0$ in the unit disk with $u(1,\theta)=f(\theta)$: $u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}P_r(\theta-\phi)f(\phi)d\phi$ where the Poisson kernel is $P_r(\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2}$.
3. The heat equation $u_t=u_{xx}$ has fundamental solution $K(x,t)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}$. For $u_t=u_{xx}+u$, substitute $u=e^t v$: then $v_t=v_{xx}$, so $v$ has the standard heat kernel. Thus $u=e^t K(x,t)$.
4. Fourier transform in $x$: $-\xi^2\hat{u}+u_{yy}=\hat{f}(\xi)\delta(y)$. For $y\neq0$: $\hat{u}=A(\xi)e^{-|\xi||y|}$. Jump condition from $\delta(y)$: $A(\xi)=\hat{f}(\xi)/(2|\xi|)$. Invert to get $u(x,y)=\frac{1}{2\pi}\int\frac{\hat{f}(\xi)}{2|\xi|}e^{-|\xi||y|}e^{i\xi x}d\xi$.