“ … partial differential equations are the basis of all physical theorems. In the theory of sound in gases, liquid and solids, in the investigations of elasticity, in optics, everywhere partial differential equations formulate basic laws of nature which can be checked against experiments.”
Bernhard Riemann

And since “who said it” sometimes carries more importance than “what is said”, let me emphasize that this quote is attributed to Bernhard Riemann! The same Riemann of geometry, complex analysis, topology, number theory; the “pure” mathematician Riemann. This post is for those “pure” mathematicians who believe that applications of a mathematical result mar its aesthetic beauty. Once not very long ago, I too was of this opinion. “PDE is not mathematics”, a friend quipped and I agreed wholeheartedly. I now realize how shallow this notion was.

In what follows, I will describe a smooth PDE result that involves interesting real analysis results. The Cauchy problem asks for a solution to a PDE with initial data defined on a hypersurface. In the case we are discussing, we have a linear PDE of first order with initial data. The “transport equation” is the simplest PDE in the sense that any simplification will result in it being an ordinary differential equation (ODE). (If I were to rechristen the subject, I’d rather call PDE as EDE, extra-ordinary differential equations, not because they are NOT ODE but because when learnt properly, they can be extraordinarily amazing).

\displaystyle c \displaystyle\frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = 0, \qquad c>0.

\displaystyle u(x,0) = \phi(x); \qquad \phi \in \mathcal C^1(\mathbb R).

We shall take Fourier transform on both sides of both equations, convert the PDE into an ODE, use the initial data {\phi} and finally prove that the solution is

\displaystyle u(x,t) = \phi ( x-ct).

The justification of taking Fourier transforms, doing some algebra with them and equating two functions whose Fourier transforms are themselves equal, was discussed in this post. There we  saw that two square-integrable functions with the same FT are equal almost everywhere.

We fix the variable {t} and take Fourier transforms on the given PDEwith respect to the variable {x} which gives,

\displaystyle 2 \pi i \,\xi c \:\widehat{u}(\xi, t) + \widehat{u}_t (\xi, t) = 0.

The Fourier transform converts an {n}-th order derivative into a polynomial of degree {n}, so here the PDE in {x} and {t} has been converted into an easy ODE in {t}, which we know how to solve. (Say by Clairut’s method to solve the ODE – {y' + p(t)y = q(t)}). Its solution is given by

\displaystyle \widehat{u}(\xi, t) = \widehat{\tau_{-ct} \phi}(\xi).

Here, {\tau_{-ct}} denotes shifting {\phi} by {-ct}. (See the Wikipedia link for basic properties of Fourier transforms).

By taking inverse-Fourier transform, we see that

\displaystyle \boxed{u(x,t) = \tau_{-ct}(\phi)(x) = \phi(x-ct).}

An assertion on uniqueness of the solution {u} (which we shall not prove) shows that this solution is unique. Observe that the solution is just a shift (transport) of the initial wave {\phi} with speed {c}. If we started with a sine wave at {t=0}, we would end up with a sine wave propagating in space-time. Isn’t it remarkable?