MathOverflow is a question and answer site for professional mathematicians. 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

Consider the problem of finding a bounded classical solution $u:\mathbb{R}\times [0,T]\to\mathbb{R}$ (such that $u$ is continuous and $u_t$, $u_x$ and $u_{xx}$ exist and are continuous on $\mathbb{R}\times (0,T]$) to a Cauchy problem for a reaction-diffusion equation $$ u_t−u_{xx}=f(u)\ \text{ on }\ \mathbb{R}\times (0,T] $$ $$ u(x,0)=u_0 (x)\ \ ∀x\in \mathbb{R}.$$ Here the initial data $u_0:\mathbb{R}\to\mathbb{R}$ is bounded and continuous and the reaction function $f:\mathbb{R}\to\mathbb{R}$ is bounded and locally Holder continuous of degree $\alpha \in (0,1)$.

Uniqueness of solutions is not guaranteed for this type of problem, for example, consider the problem with $$f(u) = |u|^\alpha\ \ \ \forall u\in\mathbb{R},\ \ u_0(x) = 0 \ \ \ \forall x\in\mathbb{R}, $$ which has solutions given by $$u_1(x,t) = 0,\ \ \ u_2(x,t) = ((1-\alpha )t)^{1/(1-\alpha )}\ \ \ \forall (x,t)\in\mathbb{R}\times [0,T] .$$

Now, the question I ask is does there exist a Cauchy problem for a reaction diffusion equation (as detailed above), which has distinct solutions $u_1$ and $u_2$, for which $$ f(u_1(x,t)) \not= 0 ,\ \ f(u_2(x,t))\not= 0 \ \ \ \forall (x,t)\in\mathbb{R}\times [0,T]\ ?$$

share|cite|improve this question
In your example, $u_{xx}=0$ for both solutions. So you might start by asking about the ODE $u''(x) = f(u)$ instead of the PDE above. – David Ketcheson Nov 18 '15 at 15:50
Reasonable idea. I'll have a think about it (sorry for the delay in response). – JCM Apr 4 at 12:40
On second thought, the example you give seems to be exactly what you are looking for. I get $f(u_2)\ne 0$. – David Ketcheson Apr 4 at 13:41
Check $u_2$ at $t=0$, the time where nonuniqueness for $u_2$ is an issue. – JCM Apr 5 at 12:09

Your Answer


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

Browse other questions tagged or ask your own question.