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In the process of solution of a PDE via Fourier cosine transform the author assumes at one step $$F_{c}^{-1}\exp(-tkw^2)=\frac{1}{\sqrt{2kt}}\exp(-\frac{x^{2}}{4kt}) $$ where Fc^{-1} is fourier cosine inverse operator. I know this is derived from formula of inverse fourier cosine transform which is $$F_{c}^{-1}[F_{c}(w)]=f(x)=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}cos(wx)F_{c}(w)dw$$ It means I have to prove $$\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}cos(wx)\exp(-tkw^2)dw=\frac{1}{\sqrt{2kt}}\exp(-\frac{x^{2}}{4kt})$$ There are similar questions like inverse fourier transform but I am unable to comprehend analogy between them and my question.

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    $\begingroup$ Possible duplicate of Quick question on Fourier Transform of $\exp(-\frac{x^2}{2})$ $\endgroup$ – Cameron Williams Mar 5 '16 at 21:59
  • $\begingroup$ @CameronWilliams You may be right that one can follow that question to understand this proof but I can't understand it because in formula of inverse fourier cosine we have cos(wx) multiplying with function while in normal fourier inverse we multiply exp(iwx) with function. It may be clear to you but I need more guidance. $\endgroup$ – Ather Cheema Mar 5 '16 at 22:07
  • $\begingroup$ Note that $\sin$ is an odd function and $e^{-\frac{x^2}{2}}$ is even. When you integrate an odd function times an even function (assuming things go to zero in the right way), you get zero. So really doing a cosine transform of the Gaussian is the same thing as doing the Fourier transform of the Gaussian. $\endgroup$ – Cameron Williams Mar 5 '16 at 22:13
  • $\begingroup$ @CameronWilliams you mean on LHS of my last equation above I can replace 'cos(wx)' with 'exp(iwx)' and then its solution will be same as in the link you provided?? $\endgroup$ – Ather Cheema Mar 5 '16 at 22:23
  • $\begingroup$ More or less, yep. $\endgroup$ – Cameron Williams Mar 5 '16 at 22:34
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You can use the same differentiation trick to find the transform: \begin{align} \frac{d}{dx}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw & = -\int_{0}^{\infty}w\sin(wx)e^{-tkw^2}dw \\ & = \int_{0}^{\infty}\sin(wx)\frac{1}{2kt}\frac{d}{dw}e^{-tkw^2}dw \\ & = \frac{1}{2kt}\left[\left.\sin(wx)e^{-tkw^2}\right|_{w=0}^{\infty}-x\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw\right] \\ & = -\frac{x}{2kt}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw. \end{align} That gives a first order differential equation for the Fourier cosine transform, resulting in a constant $C$ that does not depend on $x$ (but may depend on $k$, $t$) such that $$ \int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw=Ce^{-x^2/4kt} $$ Setting $x=0$ gives the constant $C(k,t)$: \begin{align} C(k,t)&=\int_{0}^{\infty}e^{-tkw^2}dw \\ &= \frac{1}{\sqrt{tk}}\int_{0}^{\infty}e^{-(w\sqrt{tk})^2}d(w\sqrt{tk}) \\ & = \frac{1}{\sqrt{tk}}\int_{0}^{\infty}e^{-u^2}du \\ & = \frac{1}{\sqrt{tk}}\frac{\sqrt{\pi}}{2} \end{align} Therefore the Fourier cosine transform is \begin{align} \sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\cos(wx)e^{-tkw^2}dw & = \sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{kt}}\frac{\sqrt{\pi}}{2}e^{-x^2/4kt} \\ & = \frac{1}{\sqrt{2kt}}e^{-x^2/4kt}. \end{align}

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