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Let $u \in C^\infty (\mathbb R^n \times (0,\infty)) \cap C^0(\mathbb R^n \times [0,\infty))$ be the only bounded solution for the problem

$$ \partial_t u = \partial_x ^2 u, \quad \forall (x,t )\in \mathbb R^n \times (0,\infty) \quad \text{and} \quad u (x,0) = u_0(x), \forall x \in \mathbb R^n, $$

where $u_0 \in C^\infty(\mathbb R^n)$ is bounded.

Suppose that exists $M, \delta > 0$ such that

$$ |u(x,0)| \leq M \, \exp {(-\delta \|x\|^2)}, \quad \forall x \in \mathbb R^n $$

I'm trying to proof that

$$|u(x,t)| \leq \frac{M}{(1 + 4\delta t)^n/2} \, \exp{\left (\frac{\delta \| x\| ^2}{1+4\delta t} \right )}, \quad \forall (x,t) \in \mathbb R^n \times [0,\infty) \quad \quad (I).$$

My attempt:

Since the only solution for this problem is $$ u(x,t) = \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}} \, u_0(y)\, dy$$ we have

$$ |u(x,t)| \leq \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}} |u_0(y)|\, dy \leq \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}}Me^{-\delta\|y\|^2} \, dy $$ $$ \therefore \quad |u(x,t)| \leq \frac{M}{(4\pi t)^{n/2}}\int_{\mathbb R^n} \exp{\left (\frac{-\| x-y \|^2}{4t} - \delta\|y\|^2\right )} \, dy $$

How can I conclude the argument, i.e showing $(I)$?

Help?

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  • 1
    $\begingroup$ You missed a square...the heat kernel is Gaussian not exponential. $\endgroup$ – Ian Jun 26 '17 at 14:15
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    $\begingroup$ Correct the formula for the fundamental solution. Then proceed as in your solution. Complete the square in the last step, integrate, and your estimate should follow. $\endgroup$ – Hans Engler Jun 26 '17 at 14:18

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