Help Using Heat Kernel and Convolutions

[erok81]

My text gives the following definition for the solution of the heat equation with initial temperature distribution f is the convolution of f with the heat kernel.


$$u(x,t)= \frac{1}{c \sqrt{2t}}e^{-x^{2}/4c^{2}t} \ast f~=~ \frac{1}{2c \sqrt{\pi t}} \int_{-\infty}^{\infty}f(s)e^{-(x-s)^{2}/4c^{2}t}~ds$$

I am confused how to use this. Most of the practice problems I tried only used the RHS of the above equation. I tried solving the same problems only using the LHS but didn't get anything close (but I am sure I was doing that part incorrectly). There is another example in the solution manual where they seem to use only the constants from the LHS and add them into the RHS equation. I've attached a screen shot of that case.

So does one use the stuff on the LHS? Maybe there are cases I haven't seen that use it, but for now I've done most of the problems in the text and haven't use it yet.

I may have messed up somewhere but in my example one image I was able to solve that and get the correct answer without using the convolution. Only the stuff on the RHS of my given formula.

Example two is the most confusion. Here they say to use (4) which i gave above. With this one they seem to pull out the constants and drop the rest.

Example two the actual problem shows...

$$\frac{\partial u}{\partial t}=\frac{\partial ^{2} u}{\partial ^{2} x}, ~u(x,0)=x^2$$

So I thought that maybe they take the LHS stuff and use t=0 but then I get a singularity in my exponent stuff.

Hopefully this post makes sense. I might be trying to cram too much into a single post. Let me know if you need any clarification and I'll do my best.

 

[jvicens]

Thanks!

Thank you for your question. The solution of the heat equation with initial temperature distribution f is indeed given by the convolution of f with the heat kernel. However, it is important to note that the LHS of the equation you provided is the general solution for the heat equation, while the RHS is the specific solution for the given initial temperature distribution f.

In order to solve a specific problem using the heat equation, you will need to use the RHS of the equation, which involves the convolution integral. The constants on the LHS are used to determine the specific form of the heat kernel for a given problem, but they are not used in the actual convolution calculation.

In the example you provided, the problem involves a specific initial temperature distribution (x^2) and the constants from the LHS are used to determine the specific form of the heat kernel. However, in order to actually solve the problem, you will need to use the convolution integral on the RHS.

It is important to note that the convolution integral may seem daunting at first, but with practice it becomes easier to use. I would suggest practicing with more problems that involve different initial temperature distributions to become more comfortable with the convolution process.

I hope this clarifies your confusion. If you have any further questions, please don't hesitate to ask. Good luck with your studies!