Visualizing differential equations and the Fourier transform.
Convolution.
Recall that the convolution of two functions $f, g:\R \to \C$ is the function
$$(f \ast g)(x) = \int_{\R} f(x-y) g(y)\,dy.$$
One reason for interest in convolution is because the Fourier transform of the convolution $f*g$ of two functions $f(x)$ and $g(x)$ is the product of the Fourier transforms of $f$ and $g$:
$$\cF(f \ast g) = (\cF f)(\cF g).$$
Convolution may look asymmetric in how it treats $f$ and $g$ (we put $x-y$ inside $f$ rather than $g$),
but it works the same way with the roles of $f$ and $g$ swapped: in other words,
$(g \ast f)(x) = (f \ast g)(x)$.
What is the meaning of convolution? An instructive example is $f \ast u_L$ where $$u_L(x) = \begin{cases} L, & -1/(2L) \le x \le 1/(2L)\\ 0, & |x| > 1/(2L)\end{cases}$$
is a square bump concentrated over a short interval $[-1/(2L), 1/(2L)]$ of length $1/L$ centered at the origin
and having height $L$ (so total area 1). Then
$$(f \ast u_L)(x) = \int_{\R} f(x-y) u_L(y)\,dy = L \int_{-1/(2L)}^{1/(2L)} f(x-y)\,dy = \frac{1}{1/L} \int_{x-1/(2L)}^{x+1/(2L)} f(w)\,dw$$
(with the substitution $w = x-y$). This value at $x$ is the average of $f$ over the interval centered
at $x$ with length $1/L$. The effect of such convolution is to “smoothen” $f$ by
replacing its value at each $x$ with the average of its values nearby (for large $L$).
So for large $L$, $f \ast u_L$ is a good approximation to $f$ that is smoother.
Below we give two examples of $f \ast u_L$. Observe how $f \ast u_L$ better approximates $f$ as $L$ get larger.
The heat equation via Fourier transform.
The heat equation governing the time evolution of heat distribution along a single spatial dimension is the PDE
$$\pd{u}{t}=\alpha \frac{\partial^2 u}{\partial x^2}$$
for a constant $\alpha > 0$ (thermal diffusivity, with dimension (distance)$^2$/time), where $u=u(t,x)$ is defined for $t\geq 0$ and $x\in\R$. In earlier chapters we made a change of variable so that $\alpha=1$, but here we keep the equation in its general form. As usual, we also impose an initial condition:
$$
u(0,x)=f(x)
$$
where $f(x)$ is the initial heat distribution on the line. Here is the solution:
For $t > 0$, the solution $u(t,x)$ of the heat equation with initial condition $u(0,x) = f(x)$ is
the convolution of $f(x)$ with a scaled Gaussian that depends on $t>0$:
$$
u(t,x)=\int_{\R} f(y) \frac{1}{\sqrt{4\pi \alpha t}} e^{-(x-y)^2/(4\alpha t)} \,dy=\frac{1}{\sqrt{4\pi \alpha t}}\int_{\R}f(y) e^{-(x-y)^2/(4\alpha t)} \,dy.
$$
In the textbook, there is a detailed discussion about how to obtain this solution. First, we apply Fourier transform and find for every
$t \ge 0$ and every $\lambda \in \R$:
$$\widehat u(t,\lambda) = \widehat{f}(\lambda) e^{-\alpha t \lambda^2}.$$
We then apply the inverse Fourier transform (with respect to $\lambda$) to $\widehat{u}(t,\lambda)$ and obtain
$$u(t,x)= \cF^{-1}(\widehat{u}(t,\lambda)) =
\frac{1}{2\pi}\int_{\R} \widehat{f}(\lambda) e^{-\alpha t \lambda^2} e^{i\lambda x}\,d\lambda.$$
Lastly we use convolution to rewrite it into the desired form.
Below we visualize $\widehat{u}(t,\lambda)$ and $u(t,x)$ for the heat equation
with Gaussian initial condition $f(x) = e^{-x^2}$. Adjust the parameter $\alpha$,
and hit the “Run the animation!” button to see the animation. Observe
that the animation begins at time $t = 0$ with the graphs of $u(0, x) = f(x)$ in
blue and $\widehat u(t,\lambda) =\sqrt{\pi}e^{-\lambda^2/4}$ in red that are both
independent of $\alpha$ (but as $t$ grows, the new red and blue curves for
$t > 0$ very much depend on $\alpha$: this is because the PDE depends on $\alpha$,
and can also be seen explicitly in the formulas for each).