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Solving PDF by using Fourier transform

Markov

Member
Feb 1, 2012
149
$u_t-u_{xx}=0,$ $x\in\mathbb R,$ $t>0$ and $u(x,0)=e^{-x^2}.$

By applying Fourier transform on $t$ I have $\dfrac{\partial }{\partial t}F(u)+{{\omega }^{2}}F(u)=0,$ the solution of the latter equation is $F(u)(\omega,t)=ce^{-\omega^2t},$ now by applying the initial condition I have $F(u)(x,0)=c=e^{-x^2},$ so $F(u)(\omega,t)=e^{-x^2}e^{-\omega^2t}.$ So I use the convolution property to get $u(x,t)=\dfrac1{\sqrt{2\pi}}(e^{-x^2}*e^{-\omega^2t}).$

Is this correct?
Thanks!
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
インテグラルキラー;438 said:
$u_t-u_{xx}=0,$ $x\in\mathbb R,$ $t>0$ and $u(x,0)=e^{-x^2}.$

By applying Fourier transform on $t$ I have $\frac{\partial }{\partial t}F(u)+{{\omega }^{2}}F(u)=0,$ the solution of the latter equation is $F(u)(\omega,t)=ce^{-\omega^2t},$ now by applying the initial condition I have $F(u)(x,0)=c=e^{-x^2},$ so $F(u)(\omega,t)=e^{-x^2}e^{-\omega^2t}.$ So I use the convolution property to get $u(x,t)=\frac1{\sqrt{2\pi}}(e^{-x^2}*e^{-\omega^2t}).$

Is this correct?
Thanks!
No this is not correct.

You need to find the Fourier transform of $e^{-x^2}$ and that becomes your new initial conditions that you substitute.
 

Markov

Member
Feb 1, 2012
149
Okay, so the solution is actually $u(x,t)=\dfrac1{\sqrt{2\pi}}(F^{-1}(e^{-x^2})*e^{-\omega^2t}),$ or is still incorrect? How to do it then?
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
You should check that,
$$ \frac{1}{\sqrt{4\pi t}} \int \limits_{-\infty}^{\infty} \exp \left( - \frac{x^2}{4t} \right) e^{-i\omega x}~ dx = e^{-\omega^2 t} $$

Can you write down the solution now?
 

Markov

Member
Feb 1, 2012
149
Okay, I'm sorry, I'm a bit confused, on my post #3, does make sense the thing I wrote? I think it's not correct.
I suppose to use $F(f*g)(\xi)=\sqrt{2\pi}F(f )(\xi)\cdot F(g)(\xi).$

Thanks!