We are interested in the following initial-value problem for the Convection-Diffusion Equation:
$\partial_t u(t,x) – \Delta u(t,x) = a\cdot\nabla u(t,x)^3 $, for all $t \in (0,T)$, $x \in \mathbb{R}^N$
$u(0,x) = u_0(x) \in L^1(\mathbb{R}^N) \cap L^\infty(\mathbb{R}^N) $.
Note this is similar to the Heat Equation, but we have replaced the usual $0$ on the right-hand side with the term $a\cdot\nabla u(t,x)^3$, where $a \in \mathbb{R}^N $ is a constant.
So far, we have successfully found a solution to the following integral equation:
$u(t) = G(t) \ast u_0 + \int_{0}^{t} a\cdot \nabla G(t-s) \ast u(s)^3 \,ds$,
where $G(t,x) = \dfrac{1}{(4 \pi t)^{N/2}} \,\exp\left(\dfrac{-|x|^2}{4t}\right) $ is the Heat Kernel. $\ast$ denotes convolution over space: $\displaystyle (G(t) \ast u_0)(x) = \int_{\mathbb{R}^N} G(t, x-y)\,u_0(y) \,dy.$
We have shown that the solution to the above integral solution is unique in the set $C([0,T] ; L^1(\mathbb{R}^N) \cap L^\infty(\mathbb{R}^N) )$ for some sufficiently small $T > 0$, and is bounded in the following sense: $\displaystyle\sup_{t\in [0,T]} \|u(t)\|_{L^1} + \|u(t)\|_{L^\infty} < R$, for some constant $R > \|u_0\|_{L^1} + \|u_0\|_{L^\infty}$.
Finally, we have successfully shown some regularity results in this integral solution $u$. We have shown that $ u \in C([0,T] ; L^1(\mathbb{R}^N) \cap L^\infty(\mathbb{R}^N) ) \cap C([0,T] ; W^{2,p}(\mathbb{R}^N) ) $.
Our claim is that this solution $u$ to the integral equation is also a solution to the differential equation.
In order to prove this, we try to take the time-derivative of $u$ directly.
Using the integral equation, we know that $\partial_t u(t) = \partial_t G(t) \ast u_0 + \partial_t \int^{t}_{0} a\cdot\nabla G(t-s) \ast u(s)^3 \,ds$.
Using the definition of the Heat Kernel, it's clear to see that $\partial_t G(t) \ast u_0 = \Delta_x G(t) \ast u_0$.
We are concerned with the second term $\partial_t \int^{t}_{0} a\cdot\nabla G(t-s) \ast u(s)^3 \,ds$. First, using the property of convolution $f \ast g = g \ast f$, we can move the gradient $\partial_t \int^{t}_{0} a\cdot\nabla G(t-s) \ast u(s)^3 \,ds = \partial_t \int^{t}_{0} G(t-s) \ast a\cdot\nabla u(s)^3 \,ds$
Next, we know the following property of the heat kernel $\displaystyle\lim_{t \rightarrow 0} \int_{\mathbb{R}^N} G(t,y) f(x-y) \,dy = \int_{\mathbb{R}^N} \delta(x) f(x-y) \,dy = f(x)$. I have been told this is enough to say that:
$ \displaystyle\partial_t \int^{t}_{0} G(t-s) \ast a\cdot\nabla u(s)^3 \,ds = a\cdot\nabla u(s)^3 + \int^{t}_{0} \partial_t G(t-s) \ast a\cdot\nabla u(s)^3 \,ds $.
I can't quite understand this last step. Could someone please explain it to me? Thank you very much.
Best Answer
The last step is the application of the Leibniz integral rule $$\partial_t\int_0^tf(t,s)\,\mathrm{d}s=f(t,s)+\int_0^t\partial_tf(t,s)\,\mathrm{d}s.$$ Hence, in this case $$ \begin{aligned} \partial_t \int^{t}_{0} G(t-s) \ast a\cdot\nabla u(s)^3 \,\mathrm{d}s &= \lim_{s\to t} \left(G(t-s) \ast a\cdot\nabla u(s)^3\right) \\&+ \int^{t}_{0} \partial_t G(t-s) \ast a\cdot\nabla u(s)^3 \,\mathrm{d}s, \end{aligned} $$ which gives the desired result by virtue of the property of the heat kernel you quoted. It is also worth pointing out that "moving" the gradient in $$\partial_t \int^{t}_{0} a\cdot\nabla G(t-s) \ast u(s)^3 \,\mathrm{d}s = \partial_t \int^{t}_{0} G(t-s) \ast a\cdot\nabla u(s)^3 \,\mathrm{d}s$$ is possible not by the symmetry of the convolution but upon an integration by parts.