Determining a function

**Zaph** · June 3rd 2010, 09:27 AM

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

It is a follows:

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

Any help/tips/hints or guides is appreciated

**Amer** · June 3rd 2010, 12:45 PM

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

**Danny** · June 3rd 2010, 02:14 PM

I might add that it might be helpful to know that

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

**Zaph** · June 7th 2010, 09:52 AM

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

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

Am i correct?

**Danny** · June 7th 2010, 10:17 AM

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

Am i correct?

Looks good to me.

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