Recipe

Solve a linear PDE by Fourier transform

Use when: the PDE holds for all $x$ on the whole line and its coefficients don't depend on $x$

Goal: $u(x,t)$ , recovered by inverting a transformed equation that was solved as an ordinary differential equation in $t$

Taught in Solving the heat equation by Fourier transform

Quick recipe

Transform the PDE in $x$
Solve the resulting equation as an ODE in $t$
Substitute the transformed initial condition
Invert the transform to return to $x$ -space

1
Apply $\mathcal F$ to both sides. Use the differentiation property to replace every $x$ -derivative with a factor of $i\xi$ . A second derivative becomes $(i\xi)^2=-\xi^2$ , not $+\xi^2$ .

Example

Transforming $u_t=ku_{xx}$ turns $u_{xx}$ into $-\xi^2\hat u$ , so the PDE becomes $\hat u_t=-k\xi^2\hat u$ .

Why: the differentiation property trades derivatives in $x$ for multiplication by $i\xi$ . That collapses the space variable entirely into a fixed parameter.
2
Treat $\xi$ as a constant. Solve the now-ordinary differential equation for $\hat u(\xi,t)$ . It is always the familiar first-order linear form.

Example

$\hat u_t=-k\xi^2\hat u$ solves to $\hat u(\xi,t)=\hat u(\xi,0)e^{-k\xi^2t}$ .

Why: every $x$ -dependence was already absorbed into $\xi$ in Step 1. So nothing about solving in $t$ is genuinely new calculus.
3
Transform the initial data $g(x)$ to get $\hat g(\xi)=\mathcal F[g](\xi)$ . Use a known transform pair when one applies. Substitute it for $\hat u(\xi,0)$ .

Example

For $g(x)=e^{-x^2}$ , the known Gaussian pair gives $\hat g(\xi)=\sqrt\pi\,e^{-\xi^2/4}$ , so $\hat u(\xi,t)=\sqrt\pi\,e^{-\xi^2(1/4+t)}$ .

Why: $\hat u(\xi,0)$ is exactly the constant of integration Step 2's ODE needs. It comes from the problem's own initial data, nowhere else.
4
Match $\hat u(\xi,t)$ against a known transform pair -- often another Gaussian, widened over time. Invert it to recover $u(x,t)$ .

Example

Inverting $\sqrt\pi\,e^{-\xi^2(1/4+t)}$ gives $u(x,t)=\sqrt{1/(1+4t)}\,e^{-x^2/(1+4t)}$ , so at $x=0,t=2$ : $u(0,2)=\sqrt{1/9}=\tfrac13$ .

Why: solving in frequency space is only useful once translated back, since $\hat u(\xi,t)$ alone doesn't answer the original question about $u(x,t)$ .

Common mistakes

Dropping the $i$ before squaring: writing $\mathcal F[u_{xx}]=+\xi^2\hat u$ instead of $-\xi^2\hat u$ . This flips a decaying solution into one that grows without bound.
Stopping at $\hat u(\xi,t)$ without inverting back to $x$ -space.
Applying the differentiation property once instead of twice for a second derivative.
Trusting a growing-in- $t$ solution, instead of treating it as a red flag. The heat equation always smooths things out.