- #1
solanojedi
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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!
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!