To be honest I do not know how I should define this question. Apologies


Let $u(x,t)$ be continuous on $(-\infty, \infty) \times [0,T]$ and suppose that $$u_t = u_{xx}$$ with $u(x,0) = 0, \lvert u(x,t) \rvert \le \exp(ax^2)$ for $-\infty\lt x\lt\infty$ and $0\lt t\lt T$. The aim is to conclude that $u(x,t) = 0$ everywhere.

Firstly, I pick an $s \gt 0$ such that $a \lt {1\over4s}$, and let $$b(x,t) = {1\over\sqrt{1-{s\over t}}}{\exp({x^2\over 4(s-t)}})$$

I have already verified that $b_t = b_{xx}$, the question wants me to show:

a)For any $\epsilon \gt 0$, there will be a sufficiently large $l$ such that $|u(\pm l,t)| \le \epsilon b(\pm l, t), 0\le t\lt s.$ b)Conclude that $u(x,t) = 0,-\infty\lt x\lt \infty, 0 \le t\le s$.

c)Finally induct that $u(x,t) = 0$ everywhere.

My first intuition about the first part is to take log on both sides since I thought that $\epsilon b(\pm l, t) \le \exp(al^2)$, but I do not know how to continue since after calculation that $l$ is cancelled out and I am not sure if this is a right direction. Hope for any suggestions, if you could give some hints or ideas about next 2 questions, then that would be perfect

  • $\begingroup$ Yes, we have covered maximum principles, but I do not quite know how I should apply that theorem $\endgroup$ – Cooper Oct 28 at 2:27
  • $\begingroup$ It seems like that maximum points are always on the boundary? I am really confused about this principle $\endgroup$ – Cooper Oct 28 at 2:32
  • $\begingroup$ I am trying to understand. But I am now stuck in part A $\endgroup$ – Cooper Oct 28 at 2:42
  • $\begingroup$ Something's up with your formula for $b$, in the region $t<s$, $s/t>1$ and the squareroot $\sqrt{1-s/t}$ gets a little funky. It should be more like $\frac{1}{\sqrt{s-t}}\exp(x^2/4(s-t))$. $\endgroup$ – Calvin Khor Oct 28 at 2:42
  • $\begingroup$ Yeah, I know. But question claimed like that. I do not have a clue about how to handle this $\endgroup$ – Cooper Oct 28 at 2:45

Your formula for $b(x,t)$ is a little odd, probably a typo and I'm not sure how you managed to check $b_t=b_{xx}$ when the square root is complex. You can complete the proof instead using $$b(x,t):=\frac{1}{\sqrt{s-t}}\exp(x^2/4(s-t)).$$ Note that $$ b_{xx} = \exp(x^2/4(s-t))\left[\frac{x^2}{4(s-t)^{5/2}} + \frac{1}{2(s-t)^{3/2}}\right],\\ b_t = \exp(x^2/4(s-t))\left[\frac{-1/2}{(s-t)^{3/2}}\times (-1) + \frac{-x^2(s-t)^{-2}}{4\sqrt{s-t}}\times (- 1)\right], $$ so $b_t = b_{xx}$.

A rough sketch that should lead you to the end of the question eventually. Once you have (a), then (b) will follow by maximum principle, and then for (c), you can just note that you can repeat the argument again for $\tilde u(x,t) := u(x,t-s)$.

For (a), this should be a direct calculation that follows from $a<\frac1{4s}$, using nothing more than the bound given.

$$|u(x,t)| - \epsilon b(x,t) \le \exp(ax^2)-\frac{\epsilon}{\sqrt{s-t}}\exp(x^2/4(s-t)) \le 0?$$ this question is equivalent to: $$ \sqrt{s-t}\exp\left(x^2\left[a-4/(s-t)\right]\right)\le \epsilon ?$$ We want control for all $t\in[0,s)$. So can we get rid of $t$ somehow? Yes. First, $\sqrt{s-t}\le \sqrt s$. Then, note that $$ a-\frac4{s-t} \le a - \frac{4}s < 0,$$

and these two things give $\sqrt{s-t}\exp\left(x^2\left[a-4/(s-t)\right]\right) \to 0$ uniformly in $t\in[0,s)$, as $|x|\to\infty$. This leads to (a).

  • $\begingroup$ Thanks a lot. I am now much clearer $\endgroup$ – Cooper Oct 28 at 3:53
  • $\begingroup$ @Cooper glad to help, please try hard to solve the problem fully, if there are still any questions, you can ask $\endgroup$ – Calvin Khor Oct 28 at 4:02

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