1. ## Determining a function

Hey all, i got a question im struggling with, and as of yet, i havent got anywhere..

It is a follows:

Determine a function $\displaystyle u: [0,\pi] \times [0,\infty[-> R$ such that $\displaystyle \frac{du}{dt} (x,t)-\frac{d^2u}{dx^2}(x,t) =0$
$\displaystyle , t<0,x\in]0,\pi[$ $\displaystyle u(x,0)=sin(x)cos^2(x), x \in [0,\pi]$ $\displaystyle u(0,t)=0, u(\pi,t)=0, t>0$

Any help/tips/hints or guides is appreciated

2. Originally Posted by Zaph
Hey all, i got a question im struggling with, and as of yet, i havent got anywhere..

It is a follows:

Determine a function $\displaystyle u: [0,\pi] \times [0,\infty[-> R$ such that $\displaystyle \frac{du}{dt} (x,t)-\frac{d^2u}{dx^2}(x,t) =0$
$\displaystyle , t<0,x\in]0,\pi[$ $\displaystyle u(x,0)=sin(x)cos^2(x), x \in [0,\pi]$ $\displaystyle u(0,t)=0, u(\pi,t)=0, t>0$

Any help/tips/hints or guides is appreciated
this is a heat equation you can use separation of variables see this link

Heat equation - Wikipedia, the free encyclopedia

3. I might add that it might be helpful to know that

$\displaystyle \sin x \cos^2 x = \frac{1}{4} \left(\sin x + \sin 3x \right)$.

4. Ok i've been working on this on and off for some days now, here's what i came up with:

We know that $\displaystyle sin(x)cos^2(x)$ is a continous $\displaystyle C^\infty$ so we have that $\displaystyle \sum_{k=1}^{\infty}b_{k}e^-\frac{ck^2\pi^2}{L^2}$ then the series sumfunction $\displaystyle u(x,t)$ will be real solutions, also we got that $\displaystyle c=1, L=\pi$ and $\displaystyle b_{k}= 1/4$ for , $\displaystyle k={1,3}$ else its 0. This ultimately gives us : $\displaystyle u(x,t)=\frac{1}{4} e^{-t}sin(x)+ \frac{1}{4} e^{-9t}sin(3x)$

Am i correct?

5. Originally Posted by Zaph
Ok i've been working on this on and off for some days now, here's what i came up with:

We know that $\displaystyle sin(x)cos^2(x)$ is a continous $\displaystyle C^\infty$ so we have that $\displaystyle \sum_{k=1}^{\infty}b_{k}e^-\frac{ck^2\pi^2}{L^2}$ then the series sumfunction $\displaystyle u(x,t)$ will be real solutions, also we got that $\displaystyle c=1, L=\pi$ and $\displaystyle b_{k}= 1/4$ for , $\displaystyle k={1,3}$ else its 0. This ultimately gives us : $\displaystyle u(x,t)=\frac{1}{4} e^{-t}sin(x)+ \frac{1}{4} e^{-9t}sin(3x)$

Am i correct?
Looks good to me.