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The Fourier transform and the heat kernel

Derivatives become multiplication, the heat equation solved on the line, the Gaussian kernel and Plancherel.

On the whole line there are no boundary conditions to pick out a discrete set of modes, and the Fourier series becomes the Fourier transform \[ \hat f(\xi) = \int_{-\infty}^\infty f(x)\,e^{-ix\xi}\,dx, \qquad f(x) = \frac1{2\pi}\int_{-\infty}^\infty\hat f(\xi)\,e^{ix\xi}\,d\xi. \] Its key property: integration by parts gives \( \widehat{f'}(\xi) = i\xi\,\hat f(\xi) \). Differentiation becomes multiplication, so constant-coefficient PDEs become ODEs in \( t \), one for each frequency \( \xi \).

Transform \( u_t = u_{xx} \) in \( x \): \( \hat u_t = -\xi^2\hat u \), so \( \hat u(\xi, t) = e^{-\xi^2t}\hat f(\xi) \). A product of transforms is the transform of a convolution, and \( e^{-\xi^2t} \) is the transform of a Gaussian, so \[ u(x,t) = \int_{-\infty}^\infty K_t(x - y)f(y)\,dy, \qquad K_t(x) = \frac{1}{\sqrt{4\pi t}}\,e^{-x^2/(4t)}. \] The heat kernel \( K_t \) is positive, has total mass 1 for every \( t \) (the example checks \( t = 1 \)), has variance \( \int x^2K_t\,dx = 2t \), and concentrates at the origin as \( t \to 0^+ \). In \( \mathbb R^n \) it is \( (4\pi t)^{-n/2}e^{-|x|^2/(4t)} \).

Plancherel's theorem says \( \int|\hat f|^2\,d\xi = 2\pi\int|f|^2\,dx \): the transform preserves energy up to a constant. Combined with the formula for \( \hat u \) it shows at once that \( \int|u(x,t)|^2\,dx \) decreases, and that high frequencies are damped by \( e^{-\xi^2t} \), which is the smoothing effect in its cleanest form.

Picture it: put a unit of heat at one point. At time \( t \) it has become a bell curve of width about \( \sqrt{2t} \) and height \( 1/\sqrt{4\pi t} \): half as tall and twice as wide after four times as long, with the area underneath fixed at 1.

Think it: the kernel obeys the scaling \( K_t(x) = t^{-1/2}K_1(x/\sqrt t) \), the same parabolic scaling as the equation. The Navier-Stokes equations can be rewritten as \( u(t) = e^{t\nu\Delta}u_0 - \int_0^te^{(t-s)\nu\Delta}\mathbb P\nabla\cdot(u\otimes u)(s)\,ds \) using this kernel (with \( \mathbb P \) the projection onto divergence-free fields), and local existence theory for Navier-Stokes is a fixed-point argument for that integral equation.

Радни пример · integrate exp(-x^2/4)/sqrt(4*pi) dx from -oo to oo

Integrate e^(-x^2/4)/(2·sqrt(pi)) from -oo to oo

\int_{-\infty}^{\infty} \frac{e^{- \frac{x^{2}}{4}}}{2 \sqrt{\pi}}\, dx

Корак по корак

  1. \int_{-\infty}^{\infty} \frac{e^{- \frac{x^{2}}{4}}}{2 \sqrt{\pi}}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{e^{- \frac{x^{2}}{4}}}{2 \sqrt{\pi}}\, dx = \frac{1}{2 \sqrt{\pi}} \int e^{- \frac{x^{2}}{4}}\, dx

    Pull the constant \frac{1}{2 \sqrt{\pi}} out of the integral.

  3. \int e^{- \frac{x^{2}}{4}}\, dx = \sqrt{\pi} \operatorname{erf}{\left(\frac{x}{2} \right)}

    Erf rule.

  4. F(\infty) - F(-\infty) = \left(\frac{1}{2}\right) - \left(- \frac{1}{2}\right)

    Fundamental theorem of calculus: plug in the limits.

  5. = 1

    Simplify.

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Како да: The Fourier transform and the heat kernel

  1. Transform the equation in the space variable, replacing each ∂_x by iξ.
  2. Solve the resulting ODE in t for each fixed ξ.
  3. Recognise the multiplier (e^{-ξ²t} for heat) as the transform of a known kernel.
  4. Write the solution as a convolution of the kernel with the initial data.
  5. Check mass, positivity and the limit t → 0⁺.

Питања људи постављају

Why do different books have different constants in the Fourier transform?

The 2π can sit in the exponent, in front of the transform, or in front of the inverse. Each convention is consistent; what changes is where the 2π appears in Plancherel and in the convolution theorem.

If heat spreads at infinite speed, why does a hot spot not warm the far side of the world instantly?

It does, by an amount like e^{-x²/(4t)}, which for a kilometre and a second is far smaller than anything measurable.

What should I know before starting?

Partial derivatives, the divergence theorem and multiple integrals from multivariable calculus; linear second-order ODEs; eigenvalues from linear algebra; and enough analysis to be comfortable with uniform convergence and integrals over infinite intervals.

Why are there so few formulas for solutions?

Explicit solutions exist for linear equations with constant coefficients on simple domains. Almost everything else (curved domains, variable coefficients, nonlinear terms) has none, so the modern subject proves that a solution exists and estimates its size and smoothness without ever writing it down.

How does this course lead to the Navier-Stokes problem?

The Navier-Stokes equations are a heat equation for the velocity with a transport term and a pressure that is found from a Poisson equation. Reading the Millennium problem needs energy estimates, weak solutions, Sobolev spaces, local existence with a blow-up criterion, and the scaling argument that explains why three dimensions is hard. The last six lessons build those.

What do elliptic, parabolic and hyperbolic mean?

Elliptic equations (Laplace) describe equilibrium and smooth everything; parabolic equations (heat) describe diffusion forward in time; hyperbolic equations (waves) carry signals at finite speed and keep their sharp edges. The type decides which data make a sensible problem.

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