PiRho31416
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The problem is as follows:
\frac{\partial u}{\partial t}=k\frac{\partial^{2}u}{\partial x^{2}}+c\frac{\partial u}{\partial x},
-\infty<x<\infty
u(x,0)=f(x)
Fourier Transform is defined as:
F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx
So, I took the Fourier Transform which brought me to
\frac{\partial F}{\partial t}=-\omega^{2}F-ci\omega F=-F(\omega^{2}+ci\omega)
Solving the first order differential equation brought me to
F(\omega)=e^{-\frac{1}{6}(3ic+2\omega)\omega^{2}}
When I try to integrate using the inverse Fourier transform
f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega
I get stuck. Did I do the steps right?
\frac{\partial u}{\partial t}=k\frac{\partial^{2}u}{\partial x^{2}}+c\frac{\partial u}{\partial x},
-\infty<x<\infty
u(x,0)=f(x)
Fourier Transform is defined as:
F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx
So, I took the Fourier Transform which brought me to
\frac{\partial F}{\partial t}=-\omega^{2}F-ci\omega F=-F(\omega^{2}+ci\omega)
Solving the first order differential equation brought me to
F(\omega)=e^{-\frac{1}{6}(3ic+2\omega)\omega^{2}}
When I try to integrate using the inverse Fourier transform
f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega
I get stuck. Did I do the steps right?
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