Fourier Transforms and PDEs on Unbounded Domains, differential equations lesson

On an interval $[-L, L]$ a Fourier series builds a function from the discrete frequencies $n\pi/L$ . On the whole real line there is no length to fit, every real frequency $k$ can occur, and the sum over $n$ becomes an integral over $k$ . That is the situation for heat spreading along a rod so long that its ends never matter. This lesson fixes one Fourier transform convention, derives its rules for derivatives and Gaussians, solves the heat equation on the whole line with the heat kernel, and states the decay conditions that make each step valid.

Key terms

Fourier transform convention

Definition Let $f$ be absolutely integrable on $\mathbb{R}$ , which means $\int_{-\infty}^{\infty} \lvert f(x) \rvert\,dx < \infty$ . In the convention used in this course, the Fourier transform of $f$ is

\hat{f}(k) = \int_{-\infty}^{\infty} f(x)\,e^{-ikx}\,dx

for real $k$ , and the inverse Fourier transform rebuilds $f$ as

f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(k)\,e^{ikx}\,dk.

The forward integral converges for every real $k$ and defines a continuous function with $\lvert \hat{f}(k) \rvert \le \int \lvert f \rvert\,dx$ . The inverse formula needs conditions. If $f$ is also piecewise smooth on every bounded interval, then for every real $x$

\begin{aligned} &\lim_{R \to \infty} \frac{1}{2\pi}\int_{-R}^{R} \hat{f}(k)\,e^{ikx}\,dk \\[0.4em] &\qquad = \frac{f(x^+) + f(x^-)}{2}, \end{aligned}

which is $f(x)$ wherever $f$ is continuous. If $\hat{f}$ is absolutely integrable as well, the limit is an ordinary integral. Two piecewise continuous integrable functions with the same Fourier transform agree at every point where both are continuous. Other texts put the factor $1/(2\pi)$ on the forward integral, split it as $1/\sqrt{2\pi}$ on both integrals, or use $e^{-2\pi i\xi x}$ with no factor at all, and some reverse the signs in the two exponents. Every rule below carries constants that depend on this choice, so formulas from different sources have to be rewritten in one convention before they are combined.

In plain words The Fourier transform records how much of each frequency $k$ a function contains, and the inverse adds those frequencies back up, as Fourier coefficients do on an interval. Where the function jumps, the inverse returns the midpoint of the jump, just as a Fourier series does.

Example For $f(x) = e^{-\lvert x \rvert}$ , split the integral at $0$ ,

\begin{aligned} \hat{f}(k) &= \int_0^\infty e^{-(1 + ik)x}\,dx \\[0.4em] &\quad + \int_{-\infty}^0 e^{(1 - ik)x}\,dx \\[0.4em] &= \frac{1}{1 + ik} + \frac{1}{1 - ik} \\[0.4em] &= \frac{2}{1 + k^2}. \end{aligned}

This $\hat{f}$ is integrable and $f$ is continuous, so the inverse formula returns $e^{-\lvert x \rvert}$ at every $x$ . In the symmetric convention with $1/\sqrt{2\pi}$ in front, the same function has Fourier transform $\sqrt{2/\pi}/(1 + k^2)$ , so a formula without its convention is incomplete. The box function, equal to $1$ for $\lvert x \rvert < 1$ and $0$ for $\lvert x \rvert > 1$ , has Fourier transform $2\sin k/k$ , with the value $2$ at $k = 0$ . That function of $k$ is not absolutely integrable, but the symmetric limit above still converges, and at the jumps $x = \pm 1$ it gives $\tfrac{1}{2}$ .

Fourier transform of a derivative

Definition Let $f' = df/dx$ , and suppose that $f$ is continuous on $\mathbb{R}$ , continuously differentiable except possibly at finitely many points, and that $f$ and $f'$ are absolutely integrable. Then $f(x) \to 0$ as $x \to \pm\infty$ , because $f(x) = f(0) + \int_0^x f'(s)\,ds$ has limits there and an integrable function can only have the limit $0$ . Integration by parts gives

\widehat{f'}(k) = ik\,\hat{f}(k)

for every real $k$ . The boundary terms vanish at $\pm\infty$ because $f$ decays, and they cancel at the exceptional points because $f$ is continuous. If $f'$ meets the same conditions in turn, applying the rule twice gives

\widehat{f''}(k) = -k^2\,\hat{f}(k).

For a function $u(x, t)$ that satisfies these conditions in $x$ at each time $t$ , write $\hat{u}(k, t)$ for the Fourier transform of $u$ in $x$ . The rule then turns $u_{xx}$ into $-k^2\hat{u}(k, t)$ .

In plain words Differentiation in $x$ becomes multiplication by $ik$ at each frequency, and the factor grows with $\lvert k \rvert$ because derivatives magnify fast oscillations. A linear equation in $x$ with constant coefficients therefore turns into an algebraic relation at each frequency. The conditions matter, since a jump in $f$ adds a term that the rule does not show.

Example The function $f(x) = e^{-\lvert x \rvert}$ is continuous, and $f'(x) = -\operatorname{sgn}(x)\,e^{-\lvert x \rvert}$ for $x \ne 0$ . Integrating directly over each half-line gives

\begin{aligned} \widehat{f'}(k) &= -\frac{1}{1 + ik} + \frac{1}{1 - ik} \\[0.4em] &= \frac{2ik}{1 + k^2}, \end{aligned}

which is $ik$ times $\hat{f}(k) = 2/(1 + k^2)$ , as the rule predicts. The box function shows what goes wrong without continuity. Its derivative is $0$ at every point except $\pm 1$ , so the Fourier transform of that pointwise derivative is $0$ , while $ik \cdot 2\sin k/k = 2i\sin k$ . The missing term comes from the two jumps. As a distribution the derivative of the box is $\delta(x + 1) - \delta(x - 1)$ , and its Fourier transform $e^{ik} - e^{-ik}$ equals $2i\sin k$ .

Fourier transform of a Gaussian

Definition For every $a > 0$ and every real $k$ , the Fourier transform of $e^{-ax^2}$ is

F(k) = \sqrt{\frac{\pi}{a}}\;e^{-k^2/(4a)}.

One derivation solves a differential equation in $k$ , with $F' = dF/dk$ . Differentiating $F(k) = \int e^{-ax^2}e^{-ikx}\,dx$ under the integral sign is allowed, because the $k$ -derivative of the integrand has absolute value $\lvert x \rvert e^{-ax^2}$ , which is integrable and independent of $k$ . Since $-ix\,e^{-ax^2}$ equals $\tfrac{i}{2a}$ times the $x$ -derivative of $e^{-ax^2}$ , the derivative rule gives

\begin{aligned} F'(k) &= \frac{i}{2a}\,ik\,F(k) \\[0.4em] &= -\frac{k}{2a}\,F(k). \end{aligned}

The Gaussian integral $\int_{-\infty}^{\infty} e^{-z^2}\,dz = \sqrt{\pi}$ and the substitution $x = z/\sqrt{a}$ give the starting value $F(0) = \sqrt{\pi/a}$ . The solution of this linear first-order equation with that starting value is the formula above.

In plain words The Fourier transform of a Gaussian is again a Gaussian. The parameter $a$ sits in the numerator of one exponent and in the denominator of the other, so a narrow bump in $x$ has a wide Fourier transform in $k$ , and a wide bump has a narrow one.

Example With $a = 1$ the formula gives $\sqrt{\pi}\,e^{-k^2/4}$ , and with $a = \tfrac{1}{4}$ it gives $2\sqrt{\pi}\,e^{-k^2}$ . The case needed for diffusion is $a = 1/(4\kappa t)$ with $\kappa > 0$ and $t > 0$ ,

\begin{aligned} &\int_{-\infty}^{\infty} e^{-x^2/(4\kappa t)}\,e^{-ikx}\,dx \\[0.4em] &\qquad = \sqrt{4\pi\kappa t}\;e^{-\kappa k^2 t}. \end{aligned}

Dividing by $\sqrt{4\pi\kappa t}$ gives a Gaussian in $x$ whose Fourier transform is exactly $e^{-\kappa k^2 t}$ .

Heat kernel

Definition For a diffusivity $\kappa > 0$ , the heat kernel on the line is

K(x, t) = \frac{1}{\sqrt{4\pi\kappa t}}\,e^{-x^2/(4\kappa t)}

for real $x$ and $t > 0$ . It is positive, it satisfies $K_t = \kappa K_{xx}$ , its integral over $x$ equals $1$ for every $t > 0$ , and its Fourier transform in $x$ is $e^{-\kappa k^2 t}$ . It is not defined at $t = 0$ . As $t \to 0^+$ it concentrates at $x = 0$ and tends to the delta function in the sense of distributions. Consider the initial-value problem

\begin{gathered} u_t = \kappa u_{xx}, \quad x \in \mathbb{R},\ t > 0, \\[0.4em] u(x, 0) = f(x). \end{gathered}

For integrable data and a solution that decays in $x$ , the Fourier transform in $x$ gives the ordinary differential equation $d\hat{u}/dt = -\kappa k^2\hat{u}$ at each frequency $k$ , so $\hat{u}(k, t) = e^{-\kappa k^2 t}\hat{f}(k)$ . In this convention the Fourier transform of a convolution $(g * h)(x) = \int g(x - s)\,h(s)\,ds$ of integrable functions $g$ and $h$ is the product $\hat{g}\hat{h}$ , with no extra constant, and therefore

\begin{aligned} &u(x, t) \\[0.4em] &\quad = \int_{\mathbb{R}} K(x - s, t)\,f(s)\,ds. \end{aligned}

The formula holds well beyond the case used to derive it. If $f$ is bounded and continuous, it defines a smooth solution for $t > 0$ with $\lvert u \rvert \le \sup \lvert f \rvert$ , and $u(x, t) \to f(x)$ as $t \to 0^+$ for every $x$ . If $f$ is also integrable, the total amount $\int u(x, t)\,dx$ equals $\int f\,dx$ for every $t > 0$ .

In plain words Each frequency of the initial profile decays on its own at the rate $\kappa k^2$ , so fine detail fades first. Back in $x$ the same solution is a superposition of spreading bumps, one for each point $s$ of the initial profile, each carrying the amount $f(s)$ and keeping it. The width of the kernel grows in proportion to $\sqrt{t}$ while its height falls in proportion to $1/\sqrt{t}$ .

Example Take $\kappa = 1$ and $f(x) = e^{-x^2}$ . Then $\hat{f}(k) = \sqrt{\pi}\,e^{-k^2/4}$ , and each frequency decays to

\hat{u}(k, t) = \sqrt{\pi}\,e^{-(1/4 + t)k^2}.

The Gaussian formula with $1/(4a) = 1/4 + t$ , that is $a = 1/(1 + 4t)$ , inverts this to

u(x, t) = \frac{1}{\sqrt{1 + 4t}}\,e^{-x^2/(1 + 4t)}.

A direct check confirms it. Write $b = 1 + 4t$ , so $db/dt = 4$ . Then $u_x = -(2x/b)\,u$ and $u_{xx} = (4x^2/b^2 - 2/b)\,u$ , while differentiating $\ln u = -\tfrac{1}{2}\ln b - x^2/b$ in $t$ gives $u_t = (-2/b + 4x^2/b^2)\,u$ . So $u_t = u_{xx}$ , and $u(x, 0) = e^{-x^2}$ . The peak height $1/\sqrt{1 + 4t}$ falls to $\tfrac{1}{3}$ at $t = 2$ , yet the area under the curve stays $\sqrt{\pi}$ for every $t$ , because the substitution $x = \sqrt{b}\,z$ turns $\int u\,dx$ into $\int e^{-z^2}\,dz$ .

Unbounded domain

Definition A domain is unbounded when it extends to infinity, like the whole line $-\infty < x < \infty$ , a half-line or a half-plane. Where the domain has no end there is no point at which to impose a boundary condition, so a condition on the behavior as $\lvert x \rvert \to \infty$ takes its place, such as boundedness or decay. The discrete frequencies $n\pi/L$ of an interval give way to all real $k$ . For the heat equation on the whole line with bounded continuous initial data $f$ , the heat-kernel formula gives the only bounded solution. Without a condition of this kind uniqueness fails, since there are nonzero solutions with zero initial data that grow extremely fast as $\lvert x \rvert \to \infty$ . Two features of whole-line diffusion follow from the formula.

Infinite propagation speed. If $f \ge 0$ is bounded, continuous and not identically zero, then $u(x, t) > 0$ for every $x$ and every $t > 0$ , because $K > 0$ everywhere. The wave equation, by contrast, has a finite propagation speed.
Instability backward in time. Recovering data from the values a time $t$ later multiplies each frequency by $e^{+\kappa k^2 t}$ , so tiny high-frequency errors become enormous, and the backward heat problem is not well posed.

In plain words With no ends to hold the solution in place, a condition at infinity does the job that boundary conditions do on an interval. On the whole line heat reaches every point at once, though only in tiny amounts far away. That is a feature of the idealized diffusion equation, not a claim that heat carriers move infinitely fast.

Example Let $\kappa = 1$ and let $f = 1$ on $[-1, 1]$ and $f = 0$ elsewhere. This $f$ has jumps at $\pm 1$ , but the formula still gives a smooth solution for $t > 0$ . With $\operatorname{erf}(z) = \frac{2}{\sqrt{\pi}}\int_0^z e^{-s^2}\,ds$ it reads

\begin{aligned} u(x, t) &= \tfrac{1}{2}\operatorname{erf}\left(\frac{x + 1}{2\sqrt{t}}\right) \\[0.4em] &\quad - \tfrac{1}{2}\operatorname{erf}\left(\frac{x - 1}{2\sqrt{t}}\right). \end{aligned}

At $x = 3$ the initial temperature was $0$ , yet at $t = 0.1$ the formula gives $u \approx 3.9 \times 10^{-6}$ , small but positive. Going backward in time behaves very differently. The mode $\sin 10x$ decays by the factor $e^{-100}$ between $t = 0$ and $t = 1$ , so an error of $10^{-6}\sin 10x$ in data measured at $t = 1$ corresponds to an error with amplitude about $2.7 \times 10^{37}$ in the reconstructed data at $t = 0$ .

Decay conditions

Definition Decay conditions are assumptions on how fast a function and its derivatives vanish as $\lvert x \rvert \to \infty$ , made so that the Fourier transform and its rules apply. Absolute integrability of $f$ makes $\hat{f}(k)$ defined and continuous for every real $k$ , with $\hat{f}(k) \to 0$ as $\lvert k \rvert \to \infty$ . The derivative rule needs $f$ and $f'$ integrable, so that the boundary terms vanish. Taking the Fourier transform of the heat equation also needs $u$ , $u_x$ and $u_{xx}$ integrable in $x$ at each time, which is a condition on the unknown solution, so a formula found this way should be confirmed afterward by substituting it back. A convenient class is the Schwartz functions, the smooth functions that decay, together with all their derivatives, faster than every power of $1/\lvert x \rvert$ . Gaussians belong to it, the Fourier transform maps it onto itself, and every rule on this page holds there without exceptions. Square-integrable functions are handled by extending the Fourier transform, with Plancherel’s identity

\begin{gathered} \int_{-\infty}^{\infty} \lvert f(x) \rvert^2\,dx \\[0.4em] = \frac{1}{2\pi}\int_{-\infty}^{\infty} \lvert \hat{f}(k) \rvert^2\,dk \end{gathered}

in this convention, and functions such as constants and sines need the Fourier transform of distributions.

In plain words The Fourier transform method assumes that the functions involved die off far away. Each rule needs a definite amount of decay, and when the data or the solution lack it, the integrals can diverge or the rules can fail. Saying which class of functions is in use is part of stating a result.

Example The function $1/(1 + x^2)$ is integrable, and its Fourier transform is $\pi e^{-\lvert k \rvert}$ . If $x/(1 + x^2)$ were integrable, this Fourier transform would be differentiable, but $x/(1 + x^2)$ decays only like $1/x$ , and indeed $\pi e^{-\lvert k \rvert}$ has a corner at $k = 0$ . Slow decay in $x$ shows up as missing smoothness in $k$ . The constant function $1$ has no classical Fourier transform at all, since $\int e^{-ikx}\,dx$ diverges for every $k$ , and as a distribution its Fourier transform is $2\pi\delta(k)$ . The heat-kernel formula still handles this data directly, because $\int K(x - s, t)\,ds = 1$ gives the constant solution $u = 1$ for every $t > 0$ .

Check yourself

In the convention of this page, find the function whose Fourier transform is $e^{-3k^2}$ .

Show the answer

The Gaussian formula with $a = \tfrac{1}{12}$ gives $\sqrt{12\pi}\,e^{-3k^2}$ as the Fourier transform of $e^{-x^2/12}$ . Dividing by $\sqrt{12\pi}$ gives

f(x) = \frac{1}{\sqrt{12\pi}}\,e^{-x^2/12}.

This is the heat kernel with $\kappa t = 3$ , and its peak value is $1/\sqrt{12\pi} \approx 0.163$ . Leaving out the factor $1/(2\pi)$ of the inverse, or using a formula from a text with another convention, would give the wrong constant in front.

Solve $u_t = 2u_{xx}$ on the whole line with $u(x, 0) = e^{-x^2}$ . When has the peak fallen to half its initial height, and what is $\int u(x, t)\,dx$ at that time?

Show the answer

The Fourier transform in $x$ gives $\hat{u}(k, t) = \sqrt{\pi}\,e^{-k^2/4}e^{-2k^2t} = \sqrt{\pi}\,e^{-(1 + 8t)k^2/4}$ . This is $\sqrt{\pi}\,e^{-k^2/(4a)}$ with $a = 1/(1 + 8t)$ , and the Gaussian formula inverts it to

u(x, t) = \frac{1}{\sqrt{1 + 8t}}\,e^{-x^2/(1 + 8t)}.

To check, write $b = 1 + 8t$ , so $db/dt = 8$ . Then $u_{xx} = (4x^2/b^2 - 2/b)\,u$ and $u_t = (-4/b + 8x^2/b^2)\,u$ , which is $2u_{xx}$ , and $u(x, 0) = e^{-x^2}$ . The peak $1/\sqrt{1 + 8t}$ equals $\tfrac{1}{2}$ when $1 + 8t = 4$ , that is at $t = \tfrac{3}{8}$ . The area is $\sqrt{\pi}$ then, as at every time, because the initial data are integrable and the heat kernel has total mass $1$ .

The function $f(x) = e^{-\lvert x \rvert}$ satisfies $f''(x) = f(x)$ at every $x \ne 0$ . A student takes the Fourier transform of $f'' = f$ , uses the derivative rule twice, and gets $-k^2\hat{f} = \hat{f}$ , which would force $\hat{f} = 0$ . Where is the mistake, and what is the correct relation?

Show the answer

The first use of the rule is valid, because $f$ is continuous, and it gives $\widehat{f'} = ik\hat{f}$ . The second use needs $f'$ to be continuous, but $f'(x) = -\operatorname{sgn}(x)\,e^{-\lvert x \rvert}$ jumps from $f'(0^-) = 1$ to $f'(0^+) = -1$ . Integrating by parts separately on $x < 0$ and $x > 0$ keeps the boundary terms at $0$ , and the Fourier transform of the pointwise second derivative is

\begin{aligned} &f'(0^-) - f'(0^+) - k^2\hat{f}(k) \\[0.4em] &\qquad = 2 - k^2\hat{f}(k). \end{aligned}

So the correct relation is $\hat{f} = 2 - k^2\hat{f}$ , which gives $\hat{f}(k) = 2/(1 + k^2)$ , the Fourier transform found directly. In the language of distributions, $f'' = f - 2\delta$ .

Fourier Transforms and PDEs on Unbounded Domains

Key terms

Fourier transform convention

Fourier transform of a derivative

Fourier transform of a Gaussian

Heat kernel

Unbounded domain

Decay conditions

Check yourself

How this lesson connects

Builds on