Fourier Transforms and PDEs

date2026-07-15document_iddoc_02a2536de94f8a3da86698fa38543abedescriptionFourier 変換を、全空間上の PDE を周波数ごとの代数方程式または ODE へ変換する方法として整理する。prerequisites変数分離法と Fourier 級数 / フーリエ変換の基礎type講義content_typelecturestatusactive

mathpartial-differential-equationsfourier-transformlecture

1Introduction

This lecture explains how the Fourier transform converts spatial differentiation into multiplication by frequency and thereby decomposes a constant-coefficient PDE on the whole space into independent frequency problems.

2Strategy

For nonperiodic problems on the whole space, the Fourier transform is more natural than a Fourier series. Since differentiation becomes multiplication by the frequency variable, constant-coefficient heat and wave equations can be analyzed frequency by frequency.

3Why Differentiation Becomes Multiplication

Use the convention

\hat{f} (ξ) = \int_{R} f (x) e^{- i x ξ} d x .

Assume $f \in [PARSE ERROR: Undefined("Command(\"mathcal\")")] S (R)$ , the Schwartz class defined in the prerequisite lecture. Then $f$ and all its derivatives decay rapidly, so the integrals converge absolutely and integration by parts has no boundary contribution. Hence

\hat{f^{'}} (ξ) = \int_{R} f^{'} (x) e^{- i x ξ} d x = i ξ \hat{f} (ξ)

and

\hat{f^{'}'} (ξ) = - ξ^{2} \hat{f} (ξ) .

Thus a PDE containing spatial derivatives becomes an algebraic equation or an ODE at each frequency.

4Representative Example

Let $κ > 0$ . Transforming $u_{t} = κ u_{x x}$ in $x$ gives

\partial_{t} \hat{u} = - κ ξ^{2} \hat{u} .

For this calculation, assume that $t \mapsto u (\cdot, t)$ is a $C^{1}$ map with values in $[PARSE ERROR: Undefined("Command(\"mathcal\")")] S (R)$ . This condition permits interchange of the time derivative and Fourier transform. The transformed equation is an ODE in $t$ , and it shows that higher frequencies decay faster. If $f \in [PARSE ERROR: Undefined("Command(\"mathcal\")")] S (R)$ , the solution $u = G_{t} * f$ constructed below satisfies these conditions for $t > 0$ .

5Difference from Fourier Series

A Fourier series decomposes a periodic function or a function on a bounded interval into discrete frequencies. A Fourier transform decomposes a function on the whole space into continuous frequencies. Boundary-value problems naturally lead to Fourier series or other eigenfunction expansions, whereas translation-invariant whole-space problems naturally lead to Fourier transforms.

6Connection to the Heat Kernel

For $f \in [PARSE ERROR: Undefined("Command(\"mathcal\")")] S (R)$ and $u (x, 0) = f (x)$ ,

\hat{u} (ξ, t) = e^{- κ ξ^{2} t} \hat{f} (ξ) .

With the inverse-transform convention

f (x) = \frac{1}{2 π} \int_{R} \hat{f} (ξ) e^{i x ξ} d ξ,

the Gaussian Fourier-transform formula proved in the prerequisite lecture gives

\frac{1}{2 π} \int_{R} e^{- κ t ξ^{2}} e^{i x ξ} d ξ = \frac{1}{\sqrt{4 π κ t}} e^{- x^{2} / (4 κ t)} .

More explicitly, setting $η = \sqrt{κ t} ξ$ accounts for both the Jacobian and the scaled evaluation point:

\frac{1}{2 π \sqrt{κ t}} \int_{R} e^{- η^{2}} e^{i (x / \sqrt{κ t}) η} d η = \frac{1}{\sqrt{4 π κ t}} e^{- x^{2} / (4 κ t)} .

This is the scaled Gaussian Fourier-transform formula proved in the prerequisite lecture. Therefore the inverse transform of $e^{- κ t ξ^{2}}$ is the heat kernel, and convolution with it averages the initial distribution against a Gaussian.

7Solving the Whole-Space Heat Equation

Consider

u_{t} = κ u_{x x}, x \in R, t > 0, u (x, 0) = f (x),

where $κ > 0$ and $f \in [PARSE ERROR: Undefined("Command(\"mathcal\")")] S (R)$ . The unknown is $u (x, t)$ , the independent variables are space $x$ and time $t$ , and the spatial domain is the entire real line. Transforming in $x$ produces

\partial_{t} \hat{u} (ξ, t) = - κ ξ^{2} \hat{u} (ξ, t), \hat{u} (ξ, 0) = \hat{f} (ξ),

whose solution is

\hat{u} (ξ, t) = e^{- κ ξ^{2} t} \hat{f} (ξ) .

Inverting yields

u (x, t) = \frac{1}{\sqrt{4 π κ t}} \int_{R} e^{- (x - y)^{2} / (4 κ t)} f (y) d y = (G_{t} * f) (x),

where

G_{t} (x) = \frac{1}{\sqrt{4 π κ t}} e^{- x^{2} / (4 κ t)} .

For $t > 0$ , all derivatives of $G_{t}$ are integrable, and direct calculation gives

\partial_{t} G_{t} (x) = (- \frac{1}{2 t} + \frac{x^{2}}{4 κ t^{2}}) G_{t} (x) = κ \partial_{x x} G_{t} (x) .

Differentiation under the integral therefore verifies

u_{t} = (\partial_{t} G_{t}) * f = κ (\partial_{x x} G_{t}) * f = κ u_{x x} .

Moreover, $\int_{R} G_{t} (x) d x = 1$ , and for every $δ > 0$ ,

\int_{| x | > δ} G_{t} (x) d x ⟶ 0 (t ↓ 0) .

Thus $G_{t}$ is an approximate identity. For Schwartz $f$ , $G_{t} * f \to f$ pointwise and in $L^{2}$ , so the integral representation recovers the initial condition.

8Comparison: Problems with Boundaries

On $0 < x < L$ with Dirichlet conditions, the boundary destroys translation invariance. The appropriate frequencies are then the discrete modes $\sin (n π x / L)$ rather than the continuous variable $ξ$ . The domain and boundary conditions determine whether a Fourier transform, Fourier series, or another eigenfunction expansion is appropriate.

9Limitation

For a variable-coefficient equation

u_{t} = (a (x) u_{x})_{x},

the Fourier transform does not reduce the spatial operator to simple multiplication. Products in $x$ become convolutions in frequency, coupling distinct frequencies. The constant-coefficient procedure therefore does not close in the same way.

10Scope of Validity

The appropriate spectral representation depends strongly on the domain and boundary conditions. On bounded intervals, Fourier series and eigenfunction expansions are generally more natural.

11Related Lectures

data/lecture/math/analysis/introduction-to-fourier-transform.lecture.n.md data/lecture/math/partial-differential-equations/separation-of-variables-and-fourier-series.lecture.n.md

When an explicit solution is unavailable, maximum principles can still estimate its maximum and stability.

data/lecture/math/partial-differential-equations/maximum-principle-basics.lecture.n.md