- #1
PiRho31416
- 19
- 0
The problem is as follows:
[tex]\frac{\partial u}{\partial t}=k\frac{\partial^{2}u}{\partial x^{2}}+c\frac{\partial u}{\partial x},[/tex]
[tex] -\infty<x<\infty[/tex]
[tex]u(x,0)=f(x)[/tex]
Fourier Transform is defined as:
[tex]F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx[/tex]
So, I took the Fourier Transform which brought me to
[tex] \frac{\partial F}{\partial t}=-\omega^{2}F-ci\omega F=-F(\omega^{2}+ci\omega)[/tex]
Solving the first order differential equation brought me to
[tex]F(\omega)=e^{-\frac{1}{6}(3ic+2\omega)\omega^{2}}[/tex]
When I try to integrate using the inverse Fourier transform
[tex]f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega[/tex]
I get stuck. Did I do the steps right?
[tex]\frac{\partial u}{\partial t}=k\frac{\partial^{2}u}{\partial x^{2}}+c\frac{\partial u}{\partial x},[/tex]
[tex] -\infty<x<\infty[/tex]
[tex]u(x,0)=f(x)[/tex]
Fourier Transform is defined as:
[tex]F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx[/tex]
So, I took the Fourier Transform which brought me to
[tex] \frac{\partial F}{\partial t}=-\omega^{2}F-ci\omega F=-F(\omega^{2}+ci\omega)[/tex]
Solving the first order differential equation brought me to
[tex]F(\omega)=e^{-\frac{1}{6}(3ic+2\omega)\omega^{2}}[/tex]
When I try to integrate using the inverse Fourier transform
[tex]f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega[/tex]
I get stuck. Did I do the steps right?
Last edited: