A Heat equation on infinite domain

[solanojedi]

Hi everyone, I'm reading this paper about the solution of the heat equation inside an infinite domain: https://ocw.mit.edu/courses/mathema...quations-fall-2006/lecture-notes/fourtran.pdf
1) Please let me know if the following discussion is correct.
The solution $\Psi(x,t)$ reported is the following $$\Psi (x,t)= \int_{0}^{+\infty} [A(\omega) \cos(\omega x)+ B(\omega) sin( \omega x) ] e^{-\kappa \omega^{2}t} d\omega.$$ From second order ODE theory, the $x$ dependent part of the solution can be rewritten to get two complex exponentials $$\Psi (x,t)= \int_{0}^{+\infty} [c_{1}(\omega) e^{i\omega x}+ c_{2}(\omega) e^{-i\omega x} ] e^{-\kappa \omega^{2}t} d\omega.$$ The difference between the two expression should be that $c_1(\omega)$ and $c_2(\omega)$ are complex, while $A(\omega)$ and $B(\omega)$ are real (since the solution $\Psi(x,t)$ is real). Now, if I want to explicit the Fourier Transform inside the solution, I can decompose the solution in the sum of two integrals
$$\Psi (x,t)= \int_{0}^{+\infty} c_{1}(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega + \int_{0}^{+\infty} c_{2}(\omega) e^{-i\omega x} e^{-\kappa \omega^{2}t} d\omega$$ and changing the sign on the second integral we get $$\int_{0}^{+\infty} c_{1}(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega + \int_{-\infty}^{0} c_{2}(-\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega = \int_{-\infty}^{+\infty} C(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega,$$ where $C(\omega)=\begin{cases} c_{1}(\omega), ~~\omega \geq 0 \\ c_{2} (-\omega), ~~\omega<0 \end{cases}.$ Hence we get the full Fourier transform inside the solution. $C(\omega)$ will be evalued from the initial condition of the problem. Is this approach correct?

2) if this approach is correct, how can I now show that $C(\omega)$ is complex conjugate? From Fourier theory, supposing the signal $\Psi(x,t)$ and the initial conditions are real functions, the transform $C(\omega)$ should be complex conjugate, but in this general situation I'm stucked with the two functions $c_{1}$ and $c_2$ while I should get something like $C(-\omega)=C^{*}(\omega)$...

Thank you in advance!

 

[NFuller]

Is this approach correct?

Although what you have done is technically correct, I think it would have been more straight forward to just evaluate the real integrals as done in the paper you linked. You have shifted the work from evaluating those integrals to evaluating the form of $C(\omega)$ which will require a Fourier transform of the initial condition. Granted, this would be a clever approach if the initial condition was something simple like a delta function.

how can I now show that C(ω)C(ω)C(\omega) is complex conjugate?

Why do you want to show this? I don't see what it has to do with solving the heat equation.

 

[solanojedi]

Hi NFuller, thank you for your answer!

Why do you want to show this? I don't see what it has to do with solving the heat equation.

I have to admit that it has nothing to do with it. :)
However, when I got to the point of defining the $C(\omega)$ as $c_1(\omega)$ and $c_2(\omega)$ I questioned myself about how these two different constants could be rearranged to get the situation where $C(-\omega)=C^{*}(\omega)$ (since the initial conditions and hence $\Psi(x,t)$ are real functions) and I wasn't able to solve it.

However, maybe a possible demonstration would be that since the definition of $C(\omega)$ is the Fourier transform of the initial condition $\Psi(x,0)=f(x)$ we have $$C(\omega)=\int_{-\infty}^{+\infty} f(x) e^{-i\omega x} dx$$ and we can write $$C(-\omega)=\int_{-\infty}^{+\infty} f(x) e^{i\omega x} dx$$ and $$C^{*}(\omega)=\int_{-\infty}^{+\infty} f^{*}(x) e^{i\omega x} dx$$ that are equal having $f(x)=f^{*}(x)$, i.e. $f(x)$ is real.
It seems to work for me.

 

[NFuller]

that are equal having f(x)=f∗(x)f(x)=f∗(x)f(x)=f^{*}(x), i.e. f(x)f(x)f(x) is real.

Yes, it seems that if $f$ is real then this works.