Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For the heat equation $(\partial_t-\partial_x^2)f(t,x)=0$ defined on $[0,T)\times(-\infty,\infty)$, to obtain uniqueness of the initial value problem, usually it is required to limit the growth of the potential solution at infinity, i.e. $|f(t,x)|<\exp(c\cdot x^2)$. My question is, if we do not impose any such conditions, is uniqueness no longer valid? In particular, is there a well known example of a function $f(t,x)$ that satisfies the heat equation on $[0,T)\times(-\infty,\infty)$, $f(0,x)=0$, but $f$ is not identically zero?

What if we relaxed the conditions a little bit, and only required that $f$ satisfies the heat equation in $(0,T)\times(-\infty,\infty)$ and is continuous on $[0,T)\times(-\infty,\infty)$, is there an example in this case?

share|improve this question
6  
There's loads of counterexamples. See my answer to the following question, and the comments. mathoverflow.net/questions/72195. Also, Terry Tao links to the paper by Tychonoff in the question (same as Richard Borcherds' link below) –  George Lowther Dec 1 '11 at 21:20
1  
I see that this question was also asked on math.stackexchange (math.stackexchange.com/q/87464). Posting the same question on both sites at the same time is not generally considered to be a good thing to do. –  George Lowther Dec 2 '11 at 1:04
    
Sorry, I'll avoid doing that in the future. In my defense, I wasn't sure which site was more appropriate for this question. –  Ivan Dec 2 '11 at 5:47
add comment

1 Answer

up vote 10 down vote accepted

Tychonoff in his 1935 paper Théorèmes d'unicité pour l'équation de la chaleur proved uniqueness if the solutions are not too large, and gave an example to show that the solution is not unique in general. His counterexample grows extremely rapidly for large x.

share|improve this answer
add comment

Your Answer

 
discard

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.