Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be the bounded solution to \begin{gather*} \Delta u -u_t=0 \text{ for } t>0, x\in\mathbb{R}^n, \\ u(x,0)=g(x) \text{ for } x\in\mathbb{R}^n \end{gather*} Show

  1. $\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)|=0$
  2. $\int\limits_{\mathbb{R}^n} u(x,t)\,dx=\int\limits_{\mathbb{R}^n} g(x)\,dx$ for $t>0$

Proof of (a): Assume $g\in C(\mathbb{R}^n)$, $g\in L^1(\mathbb{R}^n)$, $|g|\leq M$. Let $u$ be a solution to the PDE above. Recall the solution to this PDE is $$u(x,t)=\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy,$$ where $x\in \mathbb{R}^n$ and $t>0$. So, \begin{equation*} \begin{aligned} |u(x,t)| & =\left|\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy\right| \\ & \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \left|\exp\left[ -\dfrac{|x-y|^2}{4t}\right]\right| \cdot \left|g(y)\right|\,dy \\ & \leq \dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right]\cdot M\,dy \end{aligned} \end{equation*} Recall also the formula for the integral of a Gaussian function $$\int_{\mathbb{R}} a\exp\left[-\dfrac{(x-b)^2}{2c^2}\right]\,dx=\sqrt{2a} \cdot |c|\cdot \sqrt{\pi}.$$ We can extend this result to $\mathbb{R}^n$ and use it in our inequality above. \begin{equation*} \begin{aligned} |u(x,t)| & \leq \dfrac{M}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|y-x|^2}{2(\sqrt{2t})^2}\right]\,dy \\ & = \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{2\pi} \cdot|\sqrt{2t}| = \dfrac{M}{(4\pi t)^{n/2}} \cdot \sqrt{4\pi t} = \dfrac{M}{(4\pi t)^{(n-1)/2}} \end{aligned} \end{equation*}

Hence $$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| \leq \lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} \dfrac{M}{(4\pi t)^{(n-1)/2}} = \lim\limits_{t\to\infty} \dfrac{M}{(4\pi t)^{(n-1)/2}} =0 $$ i.e. $$\lim\limits_{t\to\infty} \sup\limits_{x\in\mathbb{R}^n} |u(x,t)| =0 $$

Proof of (b): \begin{equation*} \begin{aligned} \int\limits_{\mathbb{R}^n} u(x,t)\,dx & = \int\limits_{\mathbb{R}^n}\left[\dfrac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] g(y)\,dy \right]\,dx \\ & = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \int_{\mathbb{R}^n} \exp\left[ -\dfrac{|x-y|^2}{4t}\right] \,dx \right]g(y)\,dy \\ & = \dfrac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}\left[ \sqrt{2\pi} \cdot|\sqrt{2t}|\right]g(y)\,dy \\ & = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n} g(y)\,dy \\ & = \dfrac{1}{(4\pi t)^{(n-1)/2}} \int\limits_{\mathbb{R}^n} g(x)\,dx \\ \end{aligned} \end{equation*}


  • Is my proof of (a) correct?
  • I know I am very close to finishing the proof of b. What am I missing?

If $g(x) = M$ then you can easily check that the solution is $u(x,t) \equiv M$ for which $u\not\to 0$ as $t\to\infty$ which contradicts your result. This does not contradict the original claim since $g\not\in L^1$, but this assumption is not used in your derivation. Your mistake is in computing the integral, and if you do the calculation right you will find $$|u| \leq \int K(x,y,t)|g(y)|\,{\rm d}y \leq M\int K(x,y,t)\,{\rm d}y = M$$ since the integral of the Heat kernel is unity, i.e. $\int K(x,y,t){\rm d}y = 1$. Thus this approach is not good enough to show $u \to 0$, only that $u$ remains bounded.

Since both $g\in L^1$ and the properties of the heat kernel is critical for the claim to hold true it's natural to apply some integral inequality to try to bound the integral. Applying Cauchy-Schwarz we find

$$|u|^2 = \left(\int K(x,y,t)g(y)\,{\rm d}y\right)^2\leq \left(\int K^2(x,y,t)\,{\rm d}y\right)\left(\int g^2(y)\,{\rm d}y\right)$$ Now you can check that $K^2(x,y,t)= \frac{1}{\sqrt{8\pi t}^{n/2}}K(x,y,t/2)$ which gives you the first term and for the second term note that $g^2(y) \leq M|g(y)|$ and use the fact that $g\in L^1$.

  • $\begingroup$ Can you add more details to your proof? $\endgroup$ – Username Unknown Feb 27 '18 at 17:14
  • $\begingroup$ @UsernameUnknown What is it that is not clear? It's not meant to be a proof, it's meant to be a sketch that allows you to write the proof. $\endgroup$ – Winther Feb 27 '18 at 17:19
  • $\begingroup$ Just the sudden details. Also, how would I do the proof of part (b) if my integral formula is wrong $\endgroup$ – Username Unknown Feb 27 '18 at 17:33
  • $\begingroup$ @UsernameUnknown You have to be more specific than that or I'm just going to leave it as it is. You are integrating the gaussian wrongly: you are treating a $n$-dimensional integral as if it was an $1$-dimensiona integral. To evaluate the $n$ dimensional integral note that (here for a simplified integrand, but the same method works for your case): $\int e^{-|x|^2}{\rm d}^nx = \int e^{-x_1^2-x_2^2 - \ldots - x_n^2}{\rm d}x_1\cdots {\rm d}x_n = (\int e^{-x_1^2}{\rm d}x_1)\cdots (\int e^{-x_n^2}{\rm d}x_n) = (\int e^{-x^2}{\rm d}x)^n$. The integral reduces to a product of $n$ Gaussian integrals. $\endgroup$ – Winther Feb 27 '18 at 17:35
  • $\begingroup$ Ok. How do I show that $K^2(x,y,t)=\dfrac{1}{(8\pi t)^{n/2}} K(x,y,t/2)$? $\endgroup$ – Username Unknown Feb 27 '18 at 18:15

Hint: For both parts, you are forgetting the fact that the integration is taking place over $\mathbb{R}^n.$ This is why you end up having the extra exponent $n-1$ on the bottom. Thus, the Gaussian formula in $\mathbb{R}^n$ takes a slightly different form. Other than that, your proof has no flaw.

Also, for part $(b)$, an easier way of doing it would be integrating the LHS with respect to $t$ and use integration under the integral. That might need some justification but usually trivial with the help of Lebesgue's differentiating under integral theorem.

  • $\begingroup$ So what is the form of the Gaussian formula under $\mathbb{R}^n$? $\endgroup$ – Username Unknown Feb 25 '18 at 20:10

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