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In this post, we shall see the maximum principles for subharmonic ({\Longleftrightarrow} the Laplacian is nonnegative) {{\mathcal C}^2} functions({\Longleftrightarrow} twice differentiable with continuous second derivative). As an application, we shall see the uniqueness of a solution to the Dirichlet problem. The Dirichlet problem is:

Suppose {\Omega} is an open bounded subset of {{\mathbb R}^n}. Let {\phi} be a continuous function defined on {\partial \Omega}, the boundary of {\Omega}. Does there exist a function {f\in {\mathcal C}^2(\Omega) \cap {\mathcal C}(\overline\Omega)} that solves

\displaystyle \begin{gathered} \triangle f = 0 \qquad \text{ in } \Omega, \\ f = \phi \qquad \text{ on } \partial \Omega. \end{gathered} \ \ \ \ \ (1)

Historically, the Dirichlet problem has been important to solve PDEs that arise naturally in Physics and engineering. But as mathematicians, let us not worry about that. For us, it is important to know when a solution to the DP exists and when its unique. Curiously, a sufficient condition on the existence of a solution is how {\Omega} looks! For example, if {\Omega = B(0,r)} then {f} exists! In the following part, we shall show using the “Maximum Principle” that if an {f} exists, then it must be unique.

The Maximum Principle

Theorem 1 (Strong form) Suppose {\Omega } is an open subset of {{\mathbb R}^n} and {f \in {\mathcal C}^2(\Omega), \triangle f >0} in {\Omega}. Then {f} cannot attain a maximum in {\Omega}.

Idea of proof:

By contradiction: Suppose {f} attains its maximum at {x_0\in \Omega}. Find an {r>0} such that {B(x_0, r)\subseteq \Omega} and let {\phi_i : (-r,r) \rightarrow {\mathbb R}} be {\phi_i(t) = f(x_0 + te_i)}. Show that {\phi_i \in {\mathcal C}^2} and {\phi_i''(0) \leq 0}. Add over all {i = 1, \cdots, n} to get {\triangle f(x_0) \leq 0}, a contradiction. \square

In the weak form, we assume the function to be subharmonic, i.e., {\triangle f \geq 0}.

Theorem 2 ( Weak form) Suppose {\Omega} is an open bounded subset of {{\mathbb R}^n}. (So now, both {\partial \Omega} and {\overline\Omega} are compact sets). If {\triangle f \geq 0} in {\Omega}, then

\displaystyle \max_{\overline \Omega} f = \max_{\partial \Omega} f.

Idea of proof: We use the {\varepsilon} trick.


\displaystyle g_\varepsilon(x) = f(x) + \varepsilon \frac{|x|^2}{2n}, \qquad x \in \overline \Omega.
Show that {\triangle g_\varepsilon\geq \varepsilon > 0}, and use the strong form of the maximum principle. \square

Proof of uniqueness in the DP:

Suppose that {f_1} and {f_2} both solve the DP equation (1). Define {f = f_1 - f_2}. Then, {\triangle f = 0} in {\Omega} and {f=0} on {\partial \Omega}. Hence by the weak form of the maximum principle, {f\leq 0} in {\Omega}. Since we could have taken {f} to be {f_2-f_1} instead, it follows that {f = 0}, by which {f_1 = f_2} in {\Omega} and uniqueness has been established. \square

“ … 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?

About me

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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