# Thread: Heat equation? how to solve?

1. ## Heat equation? how to solve?

Hello
I have one problem with solve heat equation with integral Fourier,
This is problem
du/dt-d(2)u/dx^2= 2t when 0<x< $\pi$, 1 when x>0
u(x,0)=2x when o<x<1 , -1 when x>1
u(0,t)=t-1 when t $\geqslant$0

I write u(x,t)=F(x)G(t) and try to find answer!
but I can not find a(w) and b(w) and also G(t)

2. You should starting by taking the Fourier transform of $u(x,t)$ with respecto to $x$, this is $\mathcal{F}(u)(\xi,t)=U(\xi,t)$ and apply the properties of the Fourier transform of the derivative.

3. Originally Posted by Hamed
Hello
I have one problem with solve heat equation with integral Fourier,
This is problem
du/dt-d(2)u/dx^2= 2t when 0<x< $\pi$, 1 when x>0
u(x,0)=2x when o<x<1 , -1 when x>1
u(0,t)=t-1 when t $\geqslant$0

I write u(x,t)=F(x)G(t) and try to find answer!
but I can not find a(w) and b(w) and also G(t)
I cannot understand what it is you are trying to solve. It is very hard to read.

Do you mean

$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2 }=g(x,t)$

Where

$g(x,t)=\begin{cases} 2t, \text{ if } 0 \le x \le \pi \\ 1, \text{ if } x > \pi \end{cases}$

$u(x,0)=\begin{cases} 2x, \text{ if } 0 \le x \le 1 \\ -1, \text{ if } x > 1\end{cases}$

$u(0,t)=t-1$

So I am assuming that the function is not defined for x < 0. If that is the cases you should use the Fourier sine transform since you do not know the value of the normal derivative at x=0. Also note that the transform of the right hand side will be a distribution as the function is not in $L^1$

4. Thanks,
But I want to use integral of Fourier to solve this problem, I can solve it when
du/dt-d(2)u/dx^2=0
but now I don't know what should I do!!!

5. The procedure is similar. You can divide the problem in two parts, one for each domain of $x$.

When the problem is homogeneous yo take the Fourier transform of the equation $u(x,t)$, wich means just (using TheEmptySet notation for the right hand side)

$\mathcal{F}[$ Heat Equation $](\xi,t)=\mathcal{F}\left[\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2 }\right](\xi,t)=\mathcal{F}[g](\xi,t)$

and eventually you arrive at a First Order Ordinary Differential Equation for $U(\xi,t)$ treating $\xi$ as constant. In the Homogeneus problem $\mathcal{F}[g](\xi,t)=0$. In your problem this will not be true.

You solve both problems, one for each domain of $x$ and finally paste both solutions by continuity.

The point of using Foutier Transform is that your Heat Problem became an EDO, way easier to solve.