Using the Fourier Transform, solve:

$u_t=u_{xx}+\alpha u$ with $\alpha>0$, for $x \in \mathbb{R}, t>0$

with initial data $u(x,0)=f(x)$, with $f$ continuous in $\mathbb{R}$

Apllying Fourier transform in equation and initial data, we obtain

$\partial_t^2 \hat u(\xi)=-\xi^2\hat u + \alpha \hat u $

$\hat u (\xi,0)=\hat f(\xi)$

Solving, we obtain $\hat u(\xi,t)=\hat f(\xi)e^{(\alpha-\xi^2)t}$.

To get $u$, we have to apply the inverse Fourier transform, but I'm not getting no useful result.


You need to make use of the fact that if $f(x)=e^{-ax^2}$, $a>0$ constant, then its Fourier transform is $$\hat{f}(\xi)=\sqrt{{\pi\over a}}\exp\left({-\xi^2\over 4a}\right)$$ which is commonly found in textbooks on the subject. (Note: I am using $\hat{f}(\xi):=\int_{-\infty}^{\infty} f(x)e^{-i\xi x}dx$.)

That, together with the Convolution Theorem, will get you the solution you seek:

$$ u(x,t)=\sqrt{\pi}\exp\left(-{1\over 4}+\alpha t\right)f(x)*e^{-x^2}. $$

  • $\begingroup$ Thank you ! Good solution $\endgroup$ – Renato Moreira Nov 25 '13 at 20:13

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