Fourier Transform Real Function Wave Equation

[VVS]

Hello,

I hope somebody can help me with this.

1. Homework Statement
I am supposed to show that if there is a function $\phi(x,t)$ which is real, satisfies a linear wave equation and which satisfies $\phi(x,0)=0$ for $x<0$ then the Fourier Transform $\tilde{\phi}(k)$ of $\phi(x,0)$ is in the lower complex plane.

Homework Equations

The Fourier Transform of $\phi(x,0)$ is $\tilde{\phi}(k)=\frac{1}{\sqrt{2\pi}}\int\phi(x,0)e^{-ikx}dx$

The Attempt at a Solution

Basically my idea was to multiply $\phi(x,0)$ by the Heaviside function, since that would make no difference and then take the Fourier Transform: $\tilde{\phi}(k)=\frac{1}{\sqrt{2\pi}}\int\phi(x,0)u(x)e^{-ikx}dx$. Then you can use the Convolution property and write: $\tilde{\phi}(k)=\mathbb{F}(\phi(x,0))*\mathbb{F}(u(x))=\mathbb{F}(\phi(x,0))*\frac{1}{2}\left(\delta(k)+\frac{1}{ik}\right)$. Then you can use the linearity of the convolution and you obtain an integral equation:
$\tilde{\phi}(k)=\tilde{\phi}(k)*\frac{1}{ik}$
Then I can split $\tilde{\phi}(k)=\tilde{\phi_{re}}(k)+i\tilde{\phi_{im}}(k)$
So basically the real part is the imaginary part convolved and the imaginary part is the real part convolved.
Now I am stuck. Somehow if I can show that the real part is positive then it will always be in the lower complex plane.

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[VVS]

Hey thanks for the reply.
We solved the problem in class so there is no more need for help.