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Math Help - Determining a function

  1. #1
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    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 u: [0,\pi] \times [0,\infty[-> R <br />
such that  <br />
    \frac{du}{dt} (x,t)-\frac{d^2u}{dx^2}(x,t) =0
    , t<0,x\in]0,\pi[  <br />
  u(x,0)=sin(x)cos^2(x), x \in [0,\pi] <br />
u(0,t)=0, u(\pi,t)=0, t>0

    Any help/tips/hints or guides is appreciated
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Zaph View Post
    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 <br />
such that  <br />
    \frac{du}{dt} (x,t)-\frac{d^2u}{dx^2}(x,t) =0
    , t<0,x\in]0,\pi[  <br />
  u(x,0)=sin(x)cos^2(x), x \in [0,\pi] <br />
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
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  3. #3
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    I might add that it might be helpful to know that

     <br />
\sin x \cos^2 x = \frac{1}{4} \left(\sin x + \sin 3x \right)<br />
.
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  4. #4
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    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?
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  5. #5
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    Quote Originally Posted by Zaph View Post
    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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