TITLE: How does changing metrics help to find solutions to a partial differential equation?
QUESTION [11 upvotes]: I am taking a course in functional analysis and while reviewing the definition of a metric and various examples, my professor mentioned that one of the reason we care about swapping (possibly equivalent?) metrics is in pde, since a solution to a pde may be very difficult to find in one metric space, but easier in another.
I was hoping the mathstack exchange community might be able to expand on this and maybe provide me with some examples hopefully that I can understand. 
Thanks!

REPLY [12 votes]: This is really embedded into one of the central questions in PDE, namely well-posedness.  The well-posedness of a PDE requires three things: existence of a solution, uniqueness of solutions, and continuous dependence of the solution on the data in the problem.  In each of these statements there is actually a crucial missing ingredient: existence in what space and in what sense, uniqueness in what space or collection of functions, and continuity with respect to which metric or topology?
It is in the qualification of the spaces / metrics in these three issues that we encounter your professor's point.  Here's a very important example in the theory of elliptic equations.
Let $\Omega \subseteq \mathbb{R}^n$ be bounded and open with a smooth boundary and consider the Poisson problem of finding $u$ such that
$$
\begin{cases}
-\Delta u =f & \text{in }\Omega \\
u =0 & \text{on }\partial \Omega
\end{cases}
$$
for a given function $f$ (the data in the problem).  Remarkably, it's possible to find explicit examples of functions $f \in C^0(\bar{\Omega})$ such that the above problem does not admit a solution $u \in C^2(\Omega)$, where here the notion of solution is pointwise, i.e. $-\Delta u(x) = f(x)$ for all $x \in \Omega$.  Framed in terms of the above discussion, the problem is not well-posed in the space $C^2(\Omega)$ for data in $C^0(\Omega)$.  
However, if we switch to a different function space and a different notion of solution, then we can produce a solution for the same $f$ given above.  If we work with weak solutions, then we require $u \in H^1_0(\Omega)$ (the $L^2$ Sobolev space of order $1$ with functions vanishing on the boundary), and a solution must satisfy 
$$
\int_\Omega \nabla u \cdot \nabla v = \int_\Omega f v \text{ for all }v \in H^1_0(\Omega).
$$
Within this context it's possible to prove that there exists a unique $u \in H^1_0(\Omega)$ solving the PDE in the above weak sense, and that $u$ depends continuously on $f$ in the $H^1_0$ metric in the sense that 
$$
\Vert u \Vert_{H^1_0} \le C_1 \Vert f \Vert_{L^2} \le C_2 \Vert f \Vert_{C^0}
$$
for some constants $C_1,C_2>0$ that don't depend on $f$.  This means that the problem is well-posed in this weak sense, and the key difference is changing our notion of solution and the metric that we use to define the function space for solutions and to "measure" the dependence of solutions on the data.
This is, of course, just one small example.  There are many, many more...