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Let us compile a list of counterexamples in PDE, similar in spirit to the books Counterexamples in topology and Counterexamples in analysis. Eventually I plan to type up the examples with their detailed derivations.

Please give one example per answer, preferably with clear descriptions and pointers to literature.

A related question:

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I suppot having the question open – Gil Kalai Jun 24 '11 at 16:29

12 Answers 12

Scheffer has shown that there is a nontrivial weak solution $u(x,t)\in L^2(\mathbb R^2\times\mathbb R)$ to the incompressible Euler equations in 2D

$$\begin{cases} \frac{\partial u}{\partial t}+\nabla\cdot(u\otimes u) +\nabla p=0, \\[5pt] \nabla\cdot u=0 . \end{cases}$$ such that $u(x,t)\equiv 0$ for $|x|^2+|t|^2>1$. In other words, the solution is identically zero for $t<-1$, then "something happens" and the solution becomes non-zero, and for all $t>1$ the solution vanishes again. In the real world, this would look like if the water suddenly started to move in a cup that stands firmly on a table.

See V. Scheffer, "An inviscid flow with compact support in space-time", Journal of Geometric Analysis, vol. 3 (1993), pp. 343-401.

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Amazing! This kind of thing among standard equations is quite rare... – Glen Wheeler Jun 24 '11 at 9:20
This has lead to a huge theory afterwards, with contributions by A. Shnirelman, C. DeLellis and L. Szekehilidy. – Denis Serre Jun 24 '11 at 12:55
And this is presented by C. Villani in a recent Bourbaki report ( see ) – Denis Chaperon de Lauzières Jun 26 '11 at 11:26
This (amazing) result is more a statement about the definition of weak solutions than about the Euler equation as a model. In fact, in 2D, the Euler equation is well posed in the classical sense. – Luis Silvestre Jan 8 '13 at 1:27

In a paper of 2001, N. Filonov has constructed a second order uniformly elliptic operator in divergence form on $\mathbb R^n$ (where $n\ge3$) with Hölder continuous coefficients and compactly supported eigenfunctions. (It is known that this can't happen for Lipschitz continuous coefficients since then one has the unique continuation property.)

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Oh, that's pretty neat. Thanks for the reference. – Willie Wong Jun 24 '11 at 15:59

Lewy's Example gives a PDE where local solvability fails.

Fritz John's PDE book has a detailed discussion.

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Is it obvious that the equation from this example is formally integrable? I guess that equations without solutions abound if one allows "hidden" integrability conditions. – Michael Bächtold Jun 24 '11 at 9:38
What does formally integrable mean in this context? – Yakov Shlapentokh-Rothman Jun 24 '11 at 13:33
The Lewy example obeys the conditions of the Cauchy-Kovalewski theorem (it has analytic solutions from analytic data), so there are no formal obstructions to solvability. – Terry Tao Jun 24 '11 at 15:41
Thanks Prof Tao, i'll have to think about that. @Yakov Shlapentokh-Rothman: formal integrability means roughly that you can not derive a contradiction from the equations by taking total derivatives and algebraic manipulations. A simple example of an (overdetermined) system of equations which is not formally integrable is $\partial_x f=y, \partial_y f=0$ – Michael Bächtold Jun 24 '11 at 19:37
@terrytao The Lewy example may be eligible to a CK treatment, it has terrible pathologies: there exists a second category (Baire sense) subset of $\C^\infty$ rhs $f$ such that the equation $(Lewy) u=f$ has no distribution solutions, even locally. Moreover the CK solutions are extremely unstable: take an analytic data $u_0$ on some analytic hypersurface. Then in any $\C^\infty$ neighborhood of $u_0$, there exists a $C^\infty$ function $w_0$ such that the IVP has no solutions. Best, Bazin. – Bazin May 19 '12 at 15:01

In his classic 1935 work Tikhonov showed that the Cauchy problem for the heat equation with 0 initial data has nonzero solutions. He also identified uniqueness classes, of course.

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It's noteworthy that these solutions $u$ have "0 initial data" in about the strongest possible sense: if you extend them to negative times by setting $u(t,y) = 0$ for $t \le 0$, the resulting extension is $C^\infty$. If you only require $u(t,y) \to 0$ as $t \to 0$, then simpler counterexamples are possible. – Nate Eldredge May 17 '12 at 20:00

Here are three nontrivial examples with uniformly elliptic equations.

1 Nadirashvilli and Valduts example of a solution to a fully nonlinear uniformly elliptic PDE with constant coefficients which is not $C^2$.

Nadirashvili, N. and Vlăduţ, S. Nonclassical solutions of fully nonlinear elliptic equations. Geom. Funct. Anal. 17 (2007), no. 4, 1283–1296.

The original example was in dimension 12. In a later work (joint with Vladimir Tkachev) they brought the dimension down to 5. The example is known to be impossible in dimension 2. Dimensions 3 and 4 are still open.

2 Safonov's example of a uniformly elliptic equation in 3D whose solution cannot be Lipschitz continuous.

Safonov, M. V. Unimprovability of estimates of Hölder constants for solutions of linear elliptic equations with measurable coefficients. (Russian) Mat. Sb. (N.S.) 132(174) (1987), no. 2, 275--288; translation in Math. USSR-Sb. 60 (1988), no. 1, 269–281

Such example is known to be impossible in dimension 2.

3. Plis' example of a uniformly elliptic equation with Holder coefficients for which the unique continuation property does not hold.

Pliś, A. On non-uniqueness in Cauchy problem for an elliptic second order differential equation. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 11 1963 95–100. 35.42

In particular, he proves that there exists a non zero solution to some uniformly elliptic PDE in 3D with Holder coefficients which is identically zero outside of a ball.

It is related to the example that Hendrik Vogt suggested. Again, in dimension 2 it would not be possible.

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in 1955 Ennio De Giorgi constructed an example of parabolic-type linear equation, whose Cauchy problem has non-unique solution. An English translation of this paper appears in De Giorgi's collected works.

To be more specific, he constructs 4 smooth functions $a(x,t)$, $b(x,t)$, $c(x,t)$, and $u(x,t)$ defined on the strip $\mathbb{R}\times[0,1]$, such that

$$ \frac{\partial^8u}{\partial t^8} = a\frac{\partial^4u}{\partial x^4} + b\frac{\partial^2u}{\partial x^2}+cu, $$

on the whole strip,

$$ \frac{\partial^nu}{\partial t^n} = 0, $$

identically on the line $t=0$ for $n=0,\ldots,7$, and $u$ not identically zero.

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In what sense is that equation parabolic? Shouldn't there be half as many time derivatives as spatial derivatives? Also what growth assumptions are you imposing. There us non-uniqueness for the Cauchy problem for the standard heat equation after all if you don't impose any growth restrictions. – Rbega Jun 24 '11 at 1:37
@RBega Checking the proof shows that u is bounded. About the parabolicity, I don't know, De Giorgi himself calls it "of parabolic type" in the paper. Do you mean there is non-uniqueness for the heat equation bounded in time for each spacial point but unbounded in space? – timur Jun 24 '11 at 1:57
I mean that there is a solution $u$ to the heat equation on $R^n$ so that $u|_{t=0}=0$ but $u\neq 0$ for $t>0$. However, this can't happen if for $t>0$ $u=(e^{|x|})$ (by using the backwards heat kernel) and so one does have uniqueness under some spatial growth assumptions. – Rbega Jun 24 '11 at 2:45

In the paper "One cannot hear the shape of a drum", Carolyn Gordon, David L. Webb and Scott Wolpert give an example of two simply connected regions which are isospectral but not isometric. The article is available here. It was in answer to the article "Can you hear the shape of a drum by Mark Kac. Also see this article for more information about this problem.

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While that is a good article, once we start putting spectral theory and Riemannian geometry as a subset of PDEs, we'd run the risk of getting on the slippery slope toward my thesis advisor's grand vision of everything being a subset of PDEs.... :-) – Willie Wong Jun 26 '11 at 12:13

There is an example of non-uniqueness of solutions of the Cauchy problem for the heat equation in a class of functions with possible rapid growth at infinity. The example is constructed using the theory of quasi-analytic classes. See, for example, Section 1.9 in A. Friedman, Partial Differential Equations of Parabolic Type, Prentice-Hall, Englewood Cliffs, NJ, 1964.

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Such an example was first constructed by Tikhonov around 1935. It can be found in Lectures on Elliptic and Parabolic Equations in Holder Spaces (Graduate Studies in Mathematics, V. 12) by N. V. Krylov. – Andrew Jun 24 '11 at 6:18

It seems that non-uniqueness is the main source of counter-example, at least in the above answers. So, one more:

Consider the heat equation for harmonic maps: $$ u_t-\Delta u+|\nabla u|^2u=0,\qquad |u(x,t)|\equiv1,\qquad(1) $$ with prescribed boundary data $u=g$. A steady solution is a $\phi$ (a harmonic map) such that $$-\Delta \phi+|\nabla u|^2\phi=0,\qquad |\phi(x,t)|\equiv1$$ and $\phi=g$ on the boundary. It is a critical point of the functional $$I[z]:=\int_\Omega|\nabla z|^2dx$$ under the constraints that $|z|\equiv1$ in $\Omega$ and $z=g$ on the boundary.

One may choose $g$ such that there exists a harmonic map $\phi$ that does not minimize locally $I[z]$. In this case, the Cauchy problem for (1), with initial data $\phi$, has two solutions. One is $\phi$, and the other one is time-dependent, with $I[u(t)]$ non-constant (it decays).

This result was due to Bethuel, Coron, Ghidaglia, and Soyeur. See also the work of Coron and later Bertsch, Dal Passo, and van der Hout.

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Lewy's example and most other allied examples of non-solvable operators have only complex coefficients. Leaving open the question of unsolvable operators with real coefficient. F.Treves in 1962 has constructed a fourth order operator iterating Lewy operator.

If $P$ is Lewy operator then $PP\bar{P}\bar{P}$ is a real coefficient fourth order operator which is not local solvable.

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A particularly simple example is Norton's dome, with height given as a function of radial distance on the surface of the dome by

$h = \frac{2}{3g}r^{3/2}$

where $g$ is the gravitational constant near the surface of the earth. The dome has a curvature singularity at the apex. And, if we model a mass at the ($r=0$) apex of this dome with zero velocity, we find that Newton's equation does not have a unique solution; the mass can "fall" at any arbitrary time $t$ for no reason at all.

Norton's paper about the dome:

A helpful reply:

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How does it relate to PDE? – timur Jan 8 '13 at 4:42
@timur: I guess technically an ODE is just a very special PDE :) – Willie Wong Jan 8 '13 at 8:36

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