1
$\begingroup$

Let $u(x,t)$ solve the partial differential equation

$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} - u$$

where $x,t\in\mathbb{R}$ with $t>0$ and initial condition $u(x,0) = v(x)$.

Also, let $\hat{u} = \mathcal{F}[u]$, the Fourier transform of $u$ wrt to $x$. Show that $\hat{u}$ satisfies

$$ \frac{\partial \hat{u}}{\partial t} = -(k^2 +1)\hat{u} $$

with initial condition $\hat{u}(k,0) = \mathcal{F}[v]$ and also the solution of $\hat{u}$ of the equation in Fourier space and thereby the solution $u$ of the original equation.

Can anybody show the steps to take ?

$\endgroup$
1
$\begingroup$

Take the fourier transform of your differential equation

$$\mathcal{F}\left(\frac{\partial u}{\partial t}\right) = \mathcal{F}\left(\frac{\partial^2 u}{\partial x^2} -u\right) = \mathcal{F}\left(\frac{\partial^2 u}{\partial x^2}\right) -\mathcal{F}(u)$$

by the linearity of the fourier transform. The first term is just $\frac{\partial \mathcal{F}(u)}{\partial t}$ since the fourier transform is wrt $x$. For the second term use

$$\mathcal{F}\left(\frac{\partial u}{\partial x}\right) = (ik)\mathcal{F}(u)$$

which implies $\mathcal{F}\left(\frac{\partial^2u}{\partial x^2}\right) = (ik)^2\mathcal{F}(u)$.

The initial conditions for $\mathcal{F}(u)$ follows by taking the fourier transform of the initial condition: $$u(x,0)=v(x) \to \mathcal{F}(u(x,0)) = \mathcal{F}(v(x))$$

You should be able to solve the final equation, a hint: if $f = Ae^{\lambda t}$ what is $\frac{df}{dt}$ ? Relate $\lambda$ to your equation and find $A$ from applying the initial condition.

Having found the solution in fourier space then the real space solution follows by taking the inverse transform

$$u(x,t) = \mathcal{F}^{-1}(\hat{u}(t,k))$$

$\endgroup$
  • $\begingroup$ Ah sweet got it...thanx so much! $\endgroup$ – auriga123 Oct 22 '14 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.