Fourier Transform Methods for PDEs

The Fourier transform method for PDEs is the direct ancestor of the general theory of pseudodifferential operators, where a variable-coefficient differential operator is analyzed by tracking how it acts approximately as multiplication in frequency space, generalizing the exact diagonalization seen here for constant coefficients. In probability theory, the heat kernel $G(x,t)$ derived above is precisely the transition density of Brownian motion, and the Fourier-transform computation of $\hat G(\xi,t) = e^{-k\xi^2 t}$ is the same calculation that produces the characteristic function of a Gaussian random variable, linking PDE theory directly to the study of stochastic processes. In signal processing and engineering, the same differentiation rule $\widehat{f'}=i\xi \hat f$ underlies the design of frequency-domain filters, since a linear time-invariant system acts on signals exactly as multiplication by a fixed "transfer function" in frequency space, mirroring how the heat semigroup acts as multiplication by $e^{-k\xi^2 t}$. In numerical analysis, spectral methods for solving PDEs on the whole line or with periodic boundary conditions use the fast Fourier transform to implement this same algebraic-diagonalization idea on a computer, achieving high accuracy for smooth solutions. In quantum mechanics, the free particle Schrödinger equation $i\hbar\psi_t = -\frac{\hbar^2}{2m}\psi_{xx}$ is solved by an essentially identical Fourier computation, producing a complex analogue of the heat kernel that describes the spreading of a wave packet. Finally, this method is the natural bridge to Laplace Transform Methods for PDEs, which handle initial-boundary value problems on half-lines by transforming in $t$ instead of $x$, and to the broader question of Well-Posedness of PDE Problems, since the explicit solution formula derived here makes it possible to directly check continuous dependence on initial data and hence well-posedness in the sense of Hadamard.

:::faq[What goes wrong on a bounded domain, like $x \in [0, L]$?] The two-sided Fourier transform integrates over all of $\mathbb{R}$, so it implicitly assumes the function is defined (or extendable) on the whole line; a function only given on $[0,L]$ has no natural such extension without imposing extra structure. This is exactly the gap that Convergence of Fourier Series Solutions fills for bounded intervals, using discrete Fourier series built from the boundary conditions (Dirichlet, Neumann, or periodic) instead of the continuous transform. The transform method of this article is the $L \to \infty$ limit of that series method, valid precisely when there is no boundary to dictate discrete allowed frequencies.

Ask the tutor: How does the same three-step Fourier transform algorithm change when applied to the wave equation instead of the heat equation?

Fourier series express a function on a bounded interval as a superposition of countably many pure oscillations $e^{in\pi x/L}$ with frequencies spaced $\pi/L$ apart, using the periodic boundary structure of the interval to select discrete allowed frequencies. Send $L \to \infty$ and the spacing between allowed frequencies shrinks to zero: the discrete sum becomes a continuous integral, and the Fourier coefficients become the Fourier transform $\hat f(\xi)$, a continuous density of how much of each frequency $\xi$ is present in $f$. The payoff for PDEs is algebraic. If $u(x,t) = e^{i\xi x}$ formally, then $u_x = i\xi u$ and $u_{xx} = -\xi^2 u$: each spatial derivative simply multiplies by a factor of $i\xi$. A linear constant-coefficient PDE in $x$ and $t$, applied to a superposition of such exponentials, becomes — after transforming — an ODE in $t$ with $\xi$ as a frozen parameter, because the transform diagonalizes constant-coefficient differential operators in $x$ exactly the way an eigenbasis diagonalizes a matrix. Solving that ODE for every $\xi$ and reassembling via the inverse transform is the entire method. For the heat equation this reassembly has a beautifully concrete form: the solution at time $t$ is the initial data convolved with a spreading Gaussian bump, the heat kernel. The kernel starts (as $t \to 0^+$) as a spike concentrated at a point and flattens into an ever-wider, ever-shorter bell curve as $t$ increases, always keeping total area $1$ — a mathematical picture of a hot point cooling and diffusing its heat outward.

Example 1: An asymmetric bump on an infinite rod. Consider a rod occupying all of $\mathbb{R}$, with diffusivity $k = 1/4$ and an initial temperature distribution

Note the interval is deliberately asymmetric about $x=0$ (width from $-1$ to $2$, not $-a$ to $a$), so that no symmetry can be used to shortcut the computation. By the theorem, with $4kt = t$,

Substituting $\eta = \dfrac{y - x}{\sqrt{t}}$, so $dy = \sqrt t\, d\eta$, transforms the limits to $\eta = \dfrac{-1-x}{\sqrt t}$ and $\eta = \dfrac{2-x}{\sqrt t}$:

using the standard error function $\operatorname{erf}(z) = \frac{2}{\sqrt\pi}\int_0^z e^{-\eta^2}\,d\eta$ and $\int_a^b e^{-\eta^2}d\eta = \frac{\sqrt\pi}{2}\left[\operatorname{erf}(b)-\operatorname{erf}(a)\right]$. Check the value at $x = 0$, $t = 1$: $u(0,1) = \frac{3}{2}\left[\operatorname{erf}(2) - \operatorname{erf}(-1)\right] = \frac{3}{2}\left[\operatorname{erf}(2) + \operatorname{erf}(1)\right]$. Numerically $\operatorname{erf}(2) \approx 0.99532$ and $\operatorname{erf}(1) \approx 0.84270$, giving $u(0,1) \approx \frac{3}{2}(1.83802) \approx 2.757$. This is sensible: at $t=1$ diffusion has not yet fully erased the initial bump of height $3$ over $x \in [-1,2]$, and $x=0$ lies inside that interval, so the temperature there has decreased only modestly from $3$. As $t \to 0^+$, since $x=0$ is an interior continuity point of $f$ (with $f \equiv 3$ near $0$), the theorem guarantees $u(0,t) \to 3$; indeed as $t\to0^+$, $\operatorname{erf}(2/\sqrt t) \to 1$ and $\operatorname{erf}(1/\sqrt t)\to 1$, giving $u(0,t)\to \frac32(1+1)=3$, confirming consistency.

Example 2: Verifying the transform algebra directly for a decaying exponential-type initial condition. To illustrate Step 1–2 of the proof concretely (rather than just the final formula), take $k=1$ and $f(x) = e^{-|x|}$, which is continuous, bounded, and in $L^1(\mathbb{R})$ (it fails to be $C^1$ at $x=0$, but this does not affect the validity of the solution formula, only the direct differentiation-rule justification at that single point — the integral formula for $u$ remains valid by the convolution argument). First compute $\hat f(\xi)$ directly:

(Each piece is a convergent exponential integral: $\int_{-\infty}^0 e^{(1-i\xi)x}dx = \frac{1}{1-i\xi}$ since $\operatorname{Re}(1-i\xi)=1>0$, and similarly for the other piece.) By Step 2 of the theorem's proof, $\hat u(\xi,t) = \dfrac{2}{1+\xi^2} e^{-\xi^2 t}$. The solution is then, by Step 3,

which for general $x,t$ has no elementary closed form in terms of standard functions, but can be evaluated at a specific point to check consistency with known asymptotics. As $t \to \infty$, the Gaussian kernel becomes very flat and wide compared to the decay scale of $f$, so $u(x,t)$ should approach $0$ uniformly (total heat $\int f = 2$ spreads over an infinite line, diluting to zero temperature everywhere) — consistent with $\hat u(\xi,t) = \frac{2}{1+\xi^2}e^{-\xi^2 t} \to 0$ pointwise in $\xi$ as $t\to\infty$, matching the physical expectation that a fixed finite quantity of heat, once spread over all of $\mathbb{R}$, dilutes to zero temperature at any fixed point. At $x=0$, $t=1/4$ ($4kt=1$): $u(0,1/4) = \int_{-\infty}^\infty e^{-|y|} \frac{1}{\sqrt\pi} e^{-y^2}\,dy = \frac{2}{\sqrt\pi}\int_0^\infty e^{-y - y^2}\,dy$ by evenness of the integrand. Completing the square, $-y^2-y = -(y+\tfrac12)^2+\tfrac14$, so this equals $\frac{2}{\sqrt\pi}e^{1/4}\int_0^\infty e^{-(y+1/2)^2}dy = \frac{2e^{1/4}}{\sqrt\pi}\int_{1/2}^\infty e^{-w^2}dw$. Using $\int_{1/2}^\infty e^{-w^2}dw = \frac{\sqrt\pi}{2}(1-\operatorname{erf}(1/2))$ with $\operatorname{erf}(0.5)\approx 0.5205$, this gives $u(0,1/4) \approx e^{0.25}(1-0.5205) \approx 1.2840 \times 0.4795 \approx 0.6157$, a modest smoothing-down from the initial peak value $f(0)=1$, as expected for early but nonzero diffusion time.