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 ?


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

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

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

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