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