When can one expect a classical solution of a PDE?

When solving a PDE, there may be a classical solution or a weak solution (or distribution solution). But I am wondering that when people talk about "finding a solution" to some PDE, what do they refer to?

Consider for example, in Folland's Introduction to Partial Differential Equations, Chapter 3, which is devoted to the solution of the Dirichlet and Neumann problems for the Laplacian by the method of layer potentials.

OK, one may say that it quite depends on the context. It may be too vague to ask such questions here. Let me "rephrase" the question a little bit:

• When can one expect a classical solution of a PDE?
• Is there a rule of thumb for when a PDE may fail to have a classical solution, such that one has to look for a weak one?

I think it may be convenient to restrict the attention to the three classical types of PDEs; namely, the Poisson equations, heat equations and wave equations.

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What a solution is depends on the context, in the context of physics problems (which are usually the problems in undergraduate PDEs) they usually mean the "strong" solutions. Green's functions usually yield us strong solutions. –  Jonas Teuwen Jun 3 '11 at 23:02
I was taught there are three steps to solving a PDE: existence of a weak solution, regularity of the solution (so that not only is it strong, it is even "nice"), and uniqueness of the solution subject to certain sanity constraints (so it is the right solution for the physics guys). Undergraduate texts are unlikely to cover PDE where the 3 steps cannot be completed, so the solutions given must be strong. –  Jack Schmidt Jun 3 '11 at 23:34
As I've recently learned your first question depends strongly on which type of equation you're studying, for example in a typical elliptic problem Jack Schmidt's plan is the usual, but for example in case of certain hyperbolic equations (eg. conservation laws) weak solutions are seeked but regularity is less of an issue (they shouldn't be too bad though), and uniqueness is not guaranteed without some additional assumptions (hence the so called entropy conditions). In this last case you're not really interested (eventhough the problem is ususally locally well posed) in classical solutions. –  Jose27 Jun 4 '11 at 1:45
@J.M.: Thanks for your edit.:-) –  Jack Jul 30 '11 at 1:13
@Jac: What does "regularity" mean in your comment? –  Jack Aug 3 '11 at 23:14

Well, since no one has given a proper answer I will describe the regularity for the two equations I am familiar with. Maybe someone else knows how to answer in the case of the wave equations. For the sake of simplicity, lets assume Dirichlet boundary conditions in both cases, although this does not really affect the answer.

Poisson's Equation $\Delta u = f$:

This is an elliptic PDE. The usual way to attack the regularity of these is to use the Elliptic Regularity Theorem. A very basic form of this says, for example:

If $\Omega$ is a "nice" open set in $R^n$, $f \in H^k(\Omega)$ and $u$ satisfies $\Delta u = f$, then $u \in H^{k+2}(\Omega)$.

This implies a variety of generalizations to more general equations. For example, if $\Psi \in L^{\infty}(\Omega)$, $f \in L^2(\Omega)$ and $u \in L^2(\Omega)$ satisfies $\Delta u + \Psi u = f$, then we know that $\Delta u = f - \Psi u$. But the RHS is in $L^2(\Omega)$ by the assumptions on $u$, $f$ and $\Psi$. Thus $u \in H^2(\Omega)$. Also notice that the need to know a priori that $u \in L^2(\Omega)$ is not really a restriction since it is trivial by Lax-Milgram that $u \in H^1(\Omega)$. Moreover, if $f \in C^{\infty}(\Omega)$, then $f \in H^k(\Omega)$ for every $k > 0$. Thus if $\Delta u = f$ we know that $u \in H^{k+2}$ for every $k > 0$. Then the Sobolev Embedding Theorem implies that $u$ is also smooth. In general, you should expect solutions to be classical if $f \in H^k(\Omega)$ with $k + 2$ large enough so that the Sobolev Embedding implies that $u \in C^2_b(\Omega)$ These are the type of things to keep in mind when dealing with elliptic PDE.

Heat equation $u_t = \Delta u$, $u(0, x) = f(x)$:

This is a prime example of an evolution problem (it is also parabolic). These tend to have the property that the operator on the RHS ($\Delta$ in this case) generates a continuous (sometimes even analytic) semigroup (I recommend Roger & Renardy's book for an accessible introduction). It turns out that $\Delta$ generates an analytic semigroup $e^{t\Delta}$ on $L^2(\Omega)$ so the time regularity comes pretty much for free. Moreover, if $T(t)$ is an analytic semigroup with generator $A$ then $T(t)$ maps into the domain of $A^k$ for every $k > 0$. In the case of $A = \Delta$, this means that $e^{t\Delta}$ maps into $H^k(\Omega)$ for every even $k > 0$. Thus by applying the Sobolev Embedding Theorem, $e^{t\Delta}$ maps into $C^{\infty}(\Omega)$. Therefore, for every $f\in L^2(\Omega)$, the solution $u(t) = e^{t\Delta}f$ is a strong solution on $(0, \infty) \times \Omega$ and is immediately smoothed. So for this type of equation you get both existence and regularity if you can show that the operator on the RHS generates an analytic semigroup. A similar approach works for the Schrodinger equation, which is hyperbolic, so the main thing to notice here is the "evolution form" $u_t = Au$. So for these the solution will almost always be classical

I have never studied the wave equation, so in this case I have no idea.

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For the wave equation it has persistence of regularity in the $L^2$ Sobolev classes (if initial data is $W^{k,2}$, the solution will also be $W^{k,2}$). But the wave equation can, in principle, lose regularity in the classical function spaces. That is, if the data is $C^{k}$, you only expect the solution to be $C^{\lfloor k-(n-1)/2\rfloor}$ where $n$ is the spatial dimension. You can even see this at the level of the Green's function. –  Willie Wong Jul 21 '11 at 23:55
@Willie Wong how can it be done with Green's functions when $\Omega \neq R^n$? –  user12014 Jul 23 '11 at 2:15
For sufficiently short time, finite speed of propagation guarantees that the Green's function is the same as that of $\mathbb{R}^n$. So you already see the $(n-1)/2$ derivative loss there. In general, intersecting characteristic curves only make regularity worse, not better. –  Willie Wong Jul 23 '11 at 2:38
Ah, just in case I wasn't clear in the first comment, the sentence "You can even see this at the level of the Green's function" applies only to the part about regularity loss of classical solutions, not to the persistence of regularity in Sobolev classes (which you get from Energy methods). –  Willie Wong Jul 23 '11 at 2:40

Instead of considering the three types of PDE in the OP, one can restrict the attention to the constant-coefficient operators. Then one has the following nice result, which I learned from Folland's Introduction to Partial Differential Equations(Chapter 1 (1.54)):

If $L$ is a differential operator with constant coefficients on ${\mathbb R}^n$ and $f\in C^{\infty}_c({\mathbb R}^n)$, there exists $u\in C^{\infty}({\mathbb R}^n)$ such that $Lu=f$.

This is an immediate corollary of the existence of a fundamental solution:

The Malgrange-Ehrenpreis Theorem
Every differential operator $L$ with constant coefficients has a fundamental solution.

Note:
With a fundamental solution $K$ in hand, we can solve the equation $Lu=f$ not only when $f\in C^{\infty}_c$ but when $f$ is any distribution with compact support; the solution $u$ will then be a distribution.

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