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- 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.

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.