Does anyone have an example of a linear parabolic PDE that blows up in finite time in a Sobolev space setting? How does one show blow up for that particular example?

The one in Evans unfortunately is nonlinear. I want to study linear ones first. Thanks.

  • $\begingroup$ Section 7.1.4 of Evans gives a pretty good reason why you don't expect to be able to find such equations easily. Usually blow-up for parabolic PDEs uses strongly the nonlinearity to defeat the maximum principle and the dissipation; therefore I don't see how studying "linear ones" will help your understanding in general. $\endgroup$ – Willie Wong Feb 4 '14 at 14:43
  • $\begingroup$ (That said, you can always force the solution to blow-up by introducing a singularity in the coefficients, but I suspect that's not what you are looking for.) $\endgroup$ – Willie Wong Feb 4 '14 at 14:43

There are many.  Take for instance, a Cauchy problem for the heat equation $$ \begin{cases} u_t=u_{xx}\,\,,\quad x\in\mathbb{R},\; t>0,\\ u|_{t=0}=e^{x^2},\; x\in\mathbb{R}, \end{cases} $$ that does possess a unique classical solution $$ u(x,t)=\frac{e^{\frac{x^2}{1-4t}}}{\sqrt{1-4t}} $$ on $\,Q=\mathbb{R}\times [0,1/4)\,$.  Nowadays, checking the uniquenes of this solution is seen as nearly a routine exercise.  Though academically, it is well to remember that uniqueness in this Cauchy problem is guaranteed by the classical uniqueness theorem discovered by A. Tychonov:
If need be, this classical blowup solution may well be treated as a weak solution in some weighted Sobolev space.

With little or no change, this example can be modified to fit initial boundary value problems for the heat equation on the half-axis $\mathbb{R}_{+}=\{x\in\mathbb{R}\,\colon\, x>0\}$. Indeed, the initial boundary value problem $$ \begin{cases} u_t=u_{xx}\,\,,\quad x>0,\; t>0,\\ u|_{x=0}=0,\quad t\geqslant 0,\\ u|_{t=0}=xe^{x^2},\; x\geqslant 0, \end{cases} $$ with the first boundary condition does possess a unique classical solution $$ u(x,t)=\frac{xe^{\frac{x^2}{1-4t}}}{(1-4t)^{3/2}} $$ on $\,Q_{+}\!=\mathbb{R}_{+}\!\times [0,1/4)\,$. While for the initial boundary value problem $$ \begin{cases} u_t=u_{xx}\,\,,\quad x>0,\; t>0,\\ u_x|_{x=0}=0,\quad t\geqslant 0,\\ u|_{t=0}=e^{x^2},\; x\geqslant 0, \end{cases} $$ with the second boundary condition, its unique classical solution on $\,Q_{+}\!=\mathbb{R}_{+}\!\times [0,1/4)\,$ coincides with that of the Cauchy problem: $$ u(x,t)=\frac{e^{\frac{x^2}{1-4t}}}{\sqrt{1-4t}}\,. $$ As to the third boundary condition, constructing a somewhat similar blowup example for the heat equation might prove a good question on math.stackexchange.com .

  • $\begingroup$ Does this equation do not need some boundary conditions? $\endgroup$ – Lion Feb 6 '14 at 10:56
  • $\begingroup$ @Lion: A Cauchy problem for the heat equation requires only one initial condition, and nothing else, since real axis $\mathbb{R}$ has no boundary. Meanwhile, it is a good idea to modify this example to fit initial boundary value problems. I will edit my answer to include such examples. $\endgroup$ – mkl314 Feb 6 '14 at 12:19
  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ – weasd Feb 7 '14 at 15:34

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