How does distribution theory plays role in solving differential equations?
This question might seem to be very general. I will try to explain, please bear with me.
I understand, distributions make it possible to differentiate functions whose derivatives do not exist in the classical sense and any locally integrable function has a distributional derivative. In terms of differential equations, if coefficients of a differential operator are piece-wise continuous then we make use of distributions (how and why it works?).
I am more interested in their relation with Green's function. Please help me understand, how can I use distribution theory for solving differential equations.
Best Answer
I'll try to present the role that distributions play when we're looking for solutions of differential equations. For the moment, let's work in $\mathbb{R}^n$ and with smooth functions only. Suppose that we would like to solve the Poisson equation $\Delta u = f$, given the function $f \in C^{\infty}$. If such a function $u \in C^{\infty}$ exists, then for any $v \in C^{\infty}_0$, we have \begin{equation} \int_{\mathbb{R}^n} v \Delta u = \int_{\mathbb{R}^n} fv. \tag{1} \end{equation} Using integration by parts, \begin{equation} -\int_{\mathbb{R^n}} \nabla u \cdot \nabla v = \int_{\mathbb{R}^n} fv. \tag{2} \end{equation} Let us now reverse the logic: suppose we can find a function $u$ that satisfies $(2)$ for every $v \in C^{\infty}_0$, then does it necessarily satisfy $\Delta u =f$? Whether or not it does, we'll call such a function $u$ a $\textbf{weak solution}$ of the differential equation $\Delta u = f$. It is in general much easier to show that a particular differential equation has a weak solution than a solution in the usual sense.
The idea of distributions takes this idea one step further: we can view the map $v \mapsto -\int_{\mathbb{R}^n} \nabla u \cdot \nabla v$ as a linear functional $C^{\infty}_0 \to \mathbb{R}$. We'll define a $\textbf{distribution}$ as a linear functional $C^{\infty}_0 \to \mathbb{R}$; as before, we can ask if we have a distribution $\nu$ that agrees with the linear functional $v \mapsto \int_{\mathbb{R}^n} fv$, can we build from it a solution of $\Delta u =f$? (Such a distribution will often also be called a weak solution.) While it is not clear whether or not this will always occur, this is the setting in with the machinery of functional analysis is best suited to tackle the problem.
In many nice cases (such as the Poisson equation for suitable $f$), if we have a distrubution that solves our PDE in the weak sense, then it will turn out that this is in fact a function, and its regularity will depend on that of the given function $f$ (for example, if $f \in C^{\infty}$ and $\Delta u =f$, then $u \in C^{\infty}$ as well; this is an example of elliptic regularity.)
The takeaway from the above is the following: we would like to find $u$ satisfying $\Delta u = f$ in some function space, so enlarge our function space to the space of distributions. There, show we can find a solution, and show that this solution in fact was a member of the original function space to start with (of course, this last sentence will not always work out, but this is the philosophy and the hope!).