smoothness of solution of heat equation

Suppose $$g(x)\in C^k(\mathbb R^N)$$ with $$D^\alpha g$$ uniformly bounded on $$\mathbb R^N$$ for each $$|\alpha|\le k$$. Show that

$$u(x,t) : =\frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^N}e^{\frac{-1}{4t}|x-y|^2}g(y)~dy$$

satisfies $$u\in C^k(\mathbb R^N*[0,\infty))$$.

I am able to show that $$u\in C^k(\mathbb R^N*(0,\infty))$$ but I am not getting any idea for $$t=0$$.

Any type of help will be appreciated. Thanks in advance.

1 Answer

Here is a proof from Evans: To show $$u$$ is continuous $$t=0$$, first note that $$\int \Phi(x,t)dx = 1, \text{ where } \Phi(x,t) := \frac{1}{(4\pi t)^{n/2}}e^{\frac{-|x|^2}{4t}}$$ and $$u(x,t) = \int \Phi(x-y,t)g(y)dy$$ Since $$g$$ is continuous, fix $$x_0\in \mathbb{R}^n$$, let $$\epsilon>0$$ and let $$\delta>0$$ such that $$|g(y) - g(x_0)| < \epsilon \implies |y-x_0| \leq \delta$$ and consider $$|u(x,t) - g(x_0)| \leq \int_{B(x_0,\delta)}\Phi(x-y,t)|g(y) - g(x_0)|dy$$ $$+ \int_{\mathbb{R}^n-B(x_0,\delta)}\Phi(x-y,t)|g(y) - g(x_0)|dy$$ The first term is bounded by $$\epsilon$$. Then if $$|x - x_0| \leq \delta/2$$, and any $$y\in \mathbb{R}^n-B(x_0,\delta)$$, we have $$|y - x_0| = |y - x + x - x_0|\leq |y-x| + \frac{1}{2}|y - x_0|$$ In particular $$|y -x_0| \leq 2|x-y|$$. So that the second term is bounded by $$2 ||g||_{L^\infty} \int_{\mathbb{R}^n-B(x_0,\delta)}\Phi(x-y,t)dy \leq \frac{C}{t^{n/2}}\int_{\mathbb{R}^n-B(x_0,\delta)}e^{-\frac{|y - x_0|^2}{16t}}dy$$ $$= C\int_{\mathbb{R}^n - B(0,\delta/\sqrt{t})}e^{-\frac{|z|^2}{16}}dz \rightarrow 0 \text{ as } t\rightarrow 0^+$$ Where we have made the variable change $$z = \frac{y -x_0}{\sqrt{t}}$$. So that for $$t$$ small enough and $$|x-x_0|\leq \delta/2$$ we have $$|u(x,t) - g(x_0)|\leq 2\epsilon$$. It follows that $$\lim_{(x,t)\rightarrow (x_0,0^+)}u(x,t) = g(x_0)$$. A similar result holds for derivatives of $$u$$ and taking the limit.

• i still did not get it even for the $1$st derivative as say $g\in C^1(\mathbb R^N)$ then what will be $u_t(x,0)$. With what one should approximate it? – bunny Apr 25 at 12:35
• It is a convolution of $g$ and $\Phi$ in $x$ so $$\partial_{x_i} u = \int \Phi(x-y,t)\partial_{y_i}g(y)dy$$ and $\partial_{y_i}g \in L^\infty$. Also $\partial_t u = \Delta u$, so this works provided $g\in C^2(\mathbb{R}^n)$. However if you only have $g\in C^1$ you need to consider $$\int\partial_{x_i}\Phi(x-y,t)\partial_{y_i}g(y)dy$$ and justify the convergence. – Dayton Apr 25 at 13:28