Verify the Appell transformed solution also solves the PDE

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In summary, the Appell transform of u also solves the heat equation. To show this, we operate on u with the differential operator and use the chain rule to introduce new variables. By simplifying the resulting expressions and making a change of variables, we can see that the terms inside the parentheses vanish, resulting in the conclusion that if u solves the heat equation, then the Appell transform of u also solves it.
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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.
 
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  • #2
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.
 

Related to Verify the Appell transformed solution also solves the PDE

1. What is the Appell transformed solution?

The Appell transformed solution is a mathematical technique used to solve partial differential equations (PDEs). It involves transforming the original PDE into a simpler form, known as the Appell equation, which can then be solved using standard methods.

2. How does the Appell transformed solution work?

The Appell transformed solution works by using a transformation function to convert the original PDE into an Appell equation, which is a simpler form of the PDE. This allows for easier solution using standard methods, such as separation of variables or Fourier transforms.

3. Why is it important to verify that the Appell transformed solution also solves the PDE?

Verifying that the Appell transformed solution also solves the PDE is important because it ensures the accuracy and validity of the solution. It also provides a way to check for any errors or mistakes in the solution process.

4. What are the benefits of using the Appell transformed solution?

The Appell transformed solution offers several benefits, including simplifying the solution process for PDEs, allowing for the use of standard solution methods, and providing a way to check for errors in the solution. It also allows for the solution of more complex PDEs that may not be solvable using other methods.

5. Are there any limitations to using the Appell transformed solution?

While the Appell transformed solution is a useful technique, it does have some limitations. It may not be applicable to all types of PDEs, and the transformation process can be complex and time-consuming. Additionally, the solution may not always be unique, and further analysis may be required to determine the most appropriate solution.

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