Solve partial differential equation

[skrat]

Homework Statement

Solve PDE $\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0$ for $0\leq x \leq \pi$ and $t\geq 0$.

Also $\frac{\partial u}{\partial x}(0,t)=\frac{\partial u}{\partial t}(\pi ,t)=0$ and $u(x,0)=sin(x)$.

Homework Equations

The Attempt at a Solution

Differential operator $L$ for $ku_{xx}=u_t$ is symmetrical, therefore $Lu=ku_{xx}=\lambda u$.

$ku_{xx}=\lambda u$

$ku_{xx}-\lambda u=0$ so

$u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}$ and

$u(x)_x=-A\sqrt{\frac{\lambda}{k}} e^{-\sqrt{\frac{\lambda}{k}}x}+B\sqrt{\frac{\lambda}{k}} e^{\sqrt{\frac{\lambda}{k}}x}$

Using $\frac{\partial u}{\partial x}(0,t)=0$ gives me that $A=B$ and the second condition gives me:

$u(\pi )_x=\sqrt{\frac{\lambda}{k}}Ae^{-\sqrt{\frac{\lambda}{k}}\pi }(e^{2\sqrt{\frac{\lambda}{k}}\pi }-1)$

therefore

$2\sqrt{\frac{\lambda}{k}}\pi =2n\pi i$ and finally $\lambda =-n^2k$

So eigenfunctions of differential operator $L$ are :

$u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}$ where we already know that $A=B$ and inserting $\lambda =-n^2k$ leaves me with:

$u_n(x)=A_ncos(nx)$

IF everything so far is ok, than I shall continue:

$ku_{xx}=u_t$

$\sum_{n=0}^{\infty }k\lambda u(x)f(t)=\sum_{n=0}^{\infty }u(x){f}'(t)$ so

$\frac{{f}'(t)}{f(t)}=\lambda k$ and therefore

$f_n(t)=De^{-n^2k^2t}$So $U(x,t)=\sum_{n=0}^{\infty }u(x)f(t)=\sum_{n=0}^{\infty }Acos(nx)e^{-n^2k^2t}$

Know I think I should somehow write that $sin(x)$ in $u(x,0)=sin(x)$ in terms of eigenfunctions of $L$ but something has to be wrong, because:

$<sin(x),cos(nx)>=<\sum_{n=0}^{\infty }A_ncos(nx),cos(kx)>$ where the first scalar product is 0, which means what exactly? Where did I get it wrong?

 

[LCKurtz]

Homework Statement

Solve PDE $\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0$ for $0\leq x \leq \pi$ and $t\geq 0$.

Also $\frac{\partial u}{\partial x}(0,t)=\frac{\partial u}{\partial t}(\pi ,t)=0$ and $u(x,0)=sin(x)$.

Homework Equations

The Attempt at a Solution

Differential operator $L$ for $ku_{xx}=u_t$ is symmetrical, therefore $Lu=ku_{xx}=\lambda u$.

$ku_{xx}=\lambda u$

$ku_{xx}-\lambda u=0$ so

$u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}$ and

$u(x)_x=-A\sqrt{\frac{\lambda}{k}} e^{-\sqrt{\frac{\lambda}{k}}x}+B\sqrt{\frac{\lambda}{k}} e^{\sqrt{\frac{\lambda}{k}}x}$

Using $\frac{\partial u}{\partial x}(0,t)=0$ gives me that $A=B$ and the second condition gives me:

$u(\pi )_x=\sqrt{\frac{\lambda}{k}}Ae^{-\sqrt{\frac{\lambda}{k}}\pi }(e^{2\sqrt{\frac{\lambda}{k}}\pi }-1)$

therefore

$2\sqrt{\frac{\lambda}{k}}\pi =2n\pi i$ and finally $\lambda =-n^2k$

So eigenfunctions of differential operator $L$ are :

$u(x)=Ae^{-\sqrt{\frac{\lambda}{k}}x}+Be^{\sqrt{\frac{\lambda}{k}}x}$ where we already know that $A=B$ and inserting $\lambda =-n^2k$ leaves me with:

$u_n(x)=A_ncos(nx)$

IF everything so far is ok, than I shall continue:

$ku_{xx}=u_t$

$\sum_{n=0}^{\infty }k\lambda u(x)f(t)=\sum_{n=0}^{\infty }u(x){f}'(t)$ so

$\frac{{f}'(t)}{f(t)}=\lambda k$ and therefore

$f_n(t)=De^{-n^2k^2t}$So $U(x,t)=\sum_{n=0}^{\infty }u(x)f(t)=\sum_{n=0}^{\infty }Acos(nx)e^{-n^2k^2t}$

Know I think I should somehow write that $sin(x)$ in $u(x,0)=sin(x)$ in terms of eigenfunctions of $L$ but something has to be wrong, because:

$<sin(x),cos(nx)>=<\sum_{n=0}^{\infty }A_ncos(nx),cos(kx)>$ where the first scalar product is 0, which means what exactly? Where did I get it wrong?

Nothing is wrong. You want $$
\sin x = A_0 + \sum_{n=1}^\infty A_n \cos(nx)$$You want to extend $\sin x$ to $(-\pi,0)$ as an even function and use the half-range formulas for the $A_n$.

[Edit, added] Alternatively if you are thinking in terms of inner products, your orthogonal set on $(0,\pi)$ is the set $\{\cos(nx)\}$. $\sin(x)$ is not orthogonal to them.