Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have noticed that two omissions are frequent in PDE theory, to the point that it has even cropped up in textbooks and very much discouraged my study of the field. This is my 5th attempt at learning the subject seriously, and I would like some advice, say in the form of answers to the following specific questions, thanks:

  1. Why is the solution space rarely specified? Even in ODE theory, we need to say "We are looking for solutions in the space of continuous functions." (Or more usually $C^k$ where $k$ is the order of the DEQ, although vacuously no function with not enough smoothness could solve an ODE in the traditional sense.)
  2. Speaking of the traditional sense as opposed to distributional or whatever else exists in PDE theory (I've heard terms like "weak solution"), how come none of this stuff showed up in ODE theory? In $d=1$ are all distributions functions? I am aware, without proof, of a result that says that all linear systems have a distributional solution if and only if it is, in some appropriate sense, a function, for $d=1$. What about other systems? Couldn't richer collections of solutions be obtained if we allowed distributions?
  3. In both PDE and ODE theory, I have noticed only half the problem is usually solved, in the sense that the authors usually take a given equation, and perform operations on it and end up at some "solution." But this solution need not actually work, especially if noninvertible operations are used, such as the Fourier transform as it's defined on $L^1$. In ODE theory, without allowing for the generalities in 2., it has usually been easy to verify a guessed solution indeed works. Although it may not be as trivial, is this what authors are assuming in 3.? Or is there some agreement among those in PDE theory that not all guessed solutions may actually work?
share|cite|improve this question
Very, very reasonable question! If there's not a good answer by tomorrow I will make one... if only from the viewpoint of "an outsider" who has studied some history and cares considerably about the issues raised in the question, in non-physical situations where ... therefore... "physical reasoning" is interesting but just a heuristic. But, sure, indeed, heuristics are good... :) – paul garrett Feb 10 '14 at 2:33
Hans Lewy shook the world of analysis when he showed that there are first order PDEs with polynomial coefficients which have no solutions. See, for example, . By enlarging the space of objects considered as valid "solutions", this problem went away, but was replaced with a new problem of finding conditions so that weak solutions are represented by actual functions (smoothness is even harder.) This is much different than the problem of ODE's. Non-linear PDE's are worse still. – TrialAndError Feb 10 '14 at 6:15
up vote 1 down vote accepted

In general, the sense of solution (and the space) is a very delicate topic. Maybe, this is the reason for some authors to avoid a detailed explanation in a textbook.

Let me point out the difficulty of finding a solution of a linear pde. If a ODE has a finite dimensional space of solutions, the vector space of solutions oa given pde has infinitely many dimensions. Of course, with a nonlinear pde the compactness problem is even harder.

I will try to explain it with an example. There are "many" different concepts of solutions. Let's consider the equation $$ y''(t)=-y, y(-\pi)=y(\pi)=0. $$ The solution $y(t)=\sin(x)$ is a classical solution because it verifies the boundary conditions (in the classical sense, i.e., by evaluating) and verifies the equation in the classical sense (i.e. by taking the classical derivatives and ealuation at every point).

Let's denote the Fourier transform by $\hat{v}$. Then, we can define the space $$ H=\{v | \int_{-\pi}^\pi (1+|\xi|^2)\hat{v}^2dx\}. $$ We can multiply the equation by $v$ and integrate by parts. We get $$ \int u'v'=\int uvdx\;\;\forall v\in H. $$ Well, if $u$ verifies the previous integral equality for every $v$ in $H$, we call that function a weak solution. It's a weak solution because in general it has not the required regularity to verify the equation pointwise in the classical sense. Of course, without a minimum regularity the boundary conditions doesn't make sense too.

There are some more sense of solution. These could be related with uniqueness. For instance, if we consider the equation $$ |y'|=1,y(0)=y(1)=0, $$ we see that there is not a unique (weak) solution (think on different triangles). It's now when we use what is called "viscosity" solution. But I think that this is a quite long response, so I will stop here.

Even if the previous ideas are stated for boundary values of an ode (thus, some sort of natural analogous of elliptic pse), these ideas remains the same for evolutionary pde. For an evolutionary pde, the usual approach, is to regularize (in order one can find approximate solutions by ode techniques in Banach spaces), and then one finds estimates that ensures a minimum time of existence and a maximum growth of the (spatial) norm (or in general "energy"). With these ingredients (the approximate solutions + the good estimates) one invokes some compactness theorem and verify that the limit verifies the equation according to the classical/weak/distributional sense.

Hopefully this helps you a little bit.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.