Verify the Appell transformed solution also solves the PDE

[docnet]

Homework Statement

verify $u_\delta$ solves the PDE!

Relevant Equations

$\partial_t u-\Delta u=0$

I feel that my reasoning becomes shaky near the conclusion. So, someone should tell me why it is weak, and suggest how to make it stronger. Thanks.

For $\delta>0$ we define the Appell transform of $u$ by $$u_\delta=(1+\delta t)^{-\frac{n}{2}}exp\Big(-\frac{\delta|x|^2}{4(1+\delta t)}\Big)u\Big(\frac{t}{1+\delta t}, \frac{x}{1+\delta t}\Big)$$ Verify that if $u(t,x)$ solves the PDE, then the Appel transform of u also solves it.

We operate on $u_\delta$ with $\mathcal{L}:=\partial_t-\Delta$ after introducing the following shorthand notation
$$\Rightarrow ABC$$
where
$$A=(1+\delta t)^{-\frac{n}{2}}$$
$$B=exp\Big(-\frac{\delta|x|^2}{4(1+\delta t)}\Big)$$
$$C=u\Big(\frac{t}{1+\delta t}, \frac{x}{1+\delta t}\Big)$$
The partial derivatives
$$\partial_t A=-\frac{\delta n}{2(1+\delta t)}A$$
$$\partial_{x_i}A=\partial_{x_i}^2A=0$$
$$\partial_tB=\frac{\delta^2|x|^2}{4(1+\delta t)^2}B$$
$$\partial_{x_i}B=-\frac{\delta x_i}{2(1+\delta t)}B$$
$$\partial_{x_i}^2B=-\frac{\delta}{2(1+\delta t)}B+\frac{\delta^2 x_i^2}{4(1+\delta t)^2}B$$
$$\partial_t C=\partial_tC\partial_t\frac{t}{1+\delta t}+\partial_xC\partial_t\frac{x}{1+\delta t}=\partial_t C \frac{1}{(1+\delta t)^2}-\partial_x C\frac{\delta x}{(1+\delta t)^2}$$
$$\partial_{x_i}C=\partial_{x_i}C\partial_{x_i}\frac{x}{1+\delta t}+\partial_tC\partial_{x_i}\frac{t}{1+\delta t} =\partial_{x_i}C\frac{1}{1+\delta t}$$
$$\partial_{x_i}^2 C=\partial_{x_i}^2C\frac{1}{(a+\delta t)^2}$$
To show $u_\delta$ is also a solution of the heat equation, first we operate on $u_\delta$ with $\partial_t$ using the chain rule
$$\partial_t(ABC)=(B\partial_tA+A\partial_tB)C+AB\partial_tC$$
$$=(B\cdot-\frac{\delta n}{2(1+\delta t)}A+A\frac{\delta^2|x|^2}{4(1+\delta t)^2}B)C+AB(\partial_tC\frac{1}{(1+\delta t)^2}-\partial_x C\frac{\delta x}{(1+\delta t)^2})$$
And then on $u_\delta$ with $\partial_{x_i}$
$$\partial_{x_i}(ABC)=(B\partial_{x_i}A+A\partial_{x_i}B)C+AB\partial_{x_i}C$$
$$=A\cdot -\frac{\delta x_i}{2(1+\delta t)}BC+AB\partial_{x_i}C\frac{1}{1+\delta t}$$
operate once more using $\partial_{x_i}$
$$\partial_{x_i}\Big[A\cdot -\frac{\delta x_i}{2(1+\delta t)}BC+AB\partial_{x_i}C\frac{1}{1+\delta t}\Big]$$
$$=\partial_{x_i}(A\cdot -\frac{\delta x_i}{2(1+\delta t)})BC+A\cdot-\frac{\delta x_i}{2(1+\delta t)}\partial_{x_i}(BC)$$$$+\partial_{x_i}(AB)\partial_{x_i}C\frac{1}{1+\delta t}+AB\partial_{x_i}^2C\frac{1}{1+\delta t}$$
$$=(A\cdot\frac{\delta}{2(1+\delta t)}+0\cdot)BC+A\cdot -\frac{\delta x_i}{2(1+\delta t)}(-\frac{\delta x_i}{2(1+\delta t)}BC+B\partial_{x_i}C\frac{1}{1+\delta t})$$
$$+(0\cdot+A\cdot -\frac{\delta x_i}{2(1+\delta t)}B)\partial_{x_i}C\frac{1}{1+\delta t}+AB\partial_{x_i}^2C\frac{1}{(1+\delta t)^2}$$
$$=ABC\frac{\delta}{2(1+\delta t)}+ABC\frac{\delta^2x_i^2}{4(1+\delta t)^2}-AB\partial_{x_i}C\frac{\delta x_i}{2(1+\delta t)^2}$$
$$-AB\partial_{x_i}C\frac{\delta x_i}{2(1+\delta t)^2}+AB\partial_{x_i}^2C\frac{1}{(1+\delta t)^2}$$
sum over i
$$=ABC\frac{\delta n}{2(1+\delta t)}+ABC\frac{\delta^2|x|^2}{4(1+\delta t)^2}-2AB\partial_x C\frac{\delta \sum x_i}{2(1+\delta t)^2}+AB\Delta C\frac{1}{(1+\delta t)^2}$$
compute $\partial_t u_\delta -\Delta u_\delta$
$$(B\cdot-\frac{\delta n}{2(1+\delta t)}A+A\frac{\delta^2|x|^2}{4(1+\delta t)^2}B)C+AB(\partial_tC\frac{1}{(1+\delta t)^2}-\partial_x C\frac{\delta x}{(1+\delta t)^2})$$$$+ABC\frac{\delta n}{2(1+\delta t)}-ABC\frac{\delta^2|x|^2}{4(1+\delta t)^2}+2AB\partial_x C\frac{\delta \sum x_i}{2(1+\delta t)^2}-AB\Delta C\frac{1}{(1+\delta t)^2}$$
$$=\frac{1}{(1+\delta t)^2}AB\Big[\partial_tC-\Delta C\Big]$$
$$=\frac{1}{(1+\delta t)^2}AB\Big[\partial_tu\Big(\frac{t}{1+\delta t}, \frac{x}{1+\delta t}\Big)-\Delta u\Big(\frac{t}{1+\delta t}, \frac{x}{1+\delta t}\Big)\Big]$$
We make a change of variables using $t'=\frac{t}{1+\delta t}$ and $x'=\frac{x}{1+\delta t}$ and by the heat equation, the terms inside the parenthesis vanish to zero
$$\partial_tu(t',x')-\Delta u(t',x')=0$$
We conclude that if $u(t',x')$ solves the heat equation, $u_\delta$ also solves the heat equation.

 

[lippyka]

Your reasoning is weak near the conclusion because you don't explain why the change of variables results in the terms inside the parentheses going to zero. To make it stronger, explain why the change of variables results in the terms inside the parenthesis going to zero, or provide an additional step of computation to show this directly.