In the functional analysis books that I have read, they do not explain how the ideas and theorems of functional analysis (in the sense of operators on Banach spaces) help to deal with differential equations, such as proving existence or uniqueness of solutions.
Could someone give me an example of using the ideas and theorems of functional analysis to actually say something about a (partial) differential equation?
Best Answer
I am a little surprised you haven't gotten more responses here; perhaps it is because once you see the functional analytic perspective on PDEs, it becomes hard to think about PDEs without functional analysis.
Anyway here is a sketch of an example. Depending on how much functional analysis you've seen, it might not totally make sense; but I wanted to keep the example short to preserve the high level point. Fix a domain $\Omega$ and consider the Laplace's equation $\Delta u = 0$ on $\Omega$ with Dirichlet boundary conditions. Multiplying by a test function $v$ and integrating by parts, we see that $$-\int_{\Omega}v\Delta u=\int_{\Omega}\nabla u\cdot\nabla v-\int_{\partial\Omega}\frac{\partial u}{\partial \nu}vd\mathcal{H}^{d-1}.$$ If we stipulate that the test function $v$ is also zero on the boundary of $\Omega$, this implies that $$\int_\Omega \nabla u \cdot \nabla v = 0.$$ We say that this is the "weak" form of Laplace's equation.
Here is where the functional analysis perspective comes in handy. The "natural" function space here is the Sobolev space $H_0^1 (\Omega)$. On this space, it turns out[1] that $\int_\Omega \nabla u \cdot \nabla v$ is an equivalent inner product. Therefore, we can rewrite the weak form of Laplace's equation as $\langle u, v \rangle_{H_0^1 (\Omega)} = 0$. Now, we allow $v$ to range over the entire space $H_0^1 (\Omega)$. Defining the continuous linear functional $L$ on $H_0^1(\Omega)$ by $L(v)=0$, we can further rewrite the weak form of Laplace's equation as $$\forall v \in H_0^1 (\Omega) \quad \langle u, v \rangle_{H_0^1 (\Omega)} = L(v).$$ In other words, there is a solution to the weak form of Laplace's equation if there is a $u$ which solves the problem above. But because $H_0^1(\Omega)$ is a Hilbert space, this is automatically true thanks to the Riesz representation theorem (and moreover, the $u$ is unique!). That is, there is a unique $u\in H_0^1 (\Omega)$ such that $\forall v \in H_0^1 (\Omega)$, $\int_\Omega \nabla u \cdot \nabla v = 0.$
In particular, since any "bona fide" solution $u$ to $\Delta u = 0$ automatically satisfies the weak Laplace equation, this implies that there is at most one solution to $\Delta u = 0$.
In summary: we have taken a fairly standard PDE, transformed it slightly into a "weak" analogue in the "right" space, and then shown that solving the weak analogue is just a direct application of some functional analysis result.
[1] See here: Norm and scalar product of $H_0^1(\Omega)$