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Math Help - Using the Heat equation.

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    Using the Heat equation.

    Question: Find the temperature u(x,t) in a rod of length L if the initial temp is f(x) throughout and the ends are x=0 and x=L are insulated.

    Solve if L=2 and f(x) = {x , 0 < x <1
    {0 , 1 < x <2

    attempt:
    k d^2(u)/dx^2 = du/dt ; 0<x<L, t>0
    Insulated ends means: u(0,t)=0 u(L,t) = 0
    u(x,0)= f(x) ; 0 <x<L
    let u(x,t) = X(x)T(t)

    d^2(u)/d(x)^2 = TX'' ; du/dt = XT'
    kTX'' = XT'

    X''/X = T'/(kT) = -λ^2
    X'' + λ^2X = 0
    T' + λ^2kt = 0
    X'(0)=0 X'(L)=0

    for λ=0 , X(x) = ax+b
    but from the boundary condition a=0 , X(x)=b

    for λ^2 > 0
    X= acosλx + bsinλx
    X' = λ(-asinλx +bcosλx)

    this is where i am getting stuck. any help would be great, thanks
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    Quote Originally Posted by linalg123 View Post
    Question: Find the temperature u(x,t) in a rod of length L if the initial temp is f(x) throughout and the ends are x=0 and x=L are insulated.

    Solve if L=2 and f(x) = {x , 0 < x <1
    {0 , 1 < x <2

    attempt:
    k d^2(u)/dx^2 = du/dt ; 0<x<L, t>0
    Insulated ends means: u(0,t)=0 u(L,t) = 0
    u(x,0)= f(x) ; 0 <x<L
    let u(x,t) = X(x)T(t)

    d^2(u)/d(x)^2 = TX'' ; du/dt = XT'
    kTX'' = XT'

    X''/X = T'/(kT) = -λ^2
    X'' + λ^2X = 0
    T' + λ^2kt = 0
    X'(0)=0 X'(L)=0

    for λ=0 , X(x) = ax+b
    but from the boundary condition a=0 , X(x)=b

    for λ^2 > 0
    X= acosλx + bsinλx
    X' = λ(-asinλx +bcosλx)

    this is where i am getting stuck. any help would be great, thanks
    You just need to keep going

    X'(0)=0=\lambda (a\cdot 0 +b) \implies b=0


    Now using the other condition gives

    X'(L)=0=\lambda(-a \sin(\lamba L) \iff 0 = \sin(\lambda L) \iff \pi n = \lambda L \iff \lambda =\frac{\pi n}{L}, n \in \mathbb{N}

    Can you finish from here?
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    Is this from a course in fourier analysis? I can provide a solution, but it involves expanding f(x) in a fourier series with the basis e_n(x)=\sin(n\pi x/L)
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    so X(x)= acos(nπx/L)

    for λ=0 T'=0 so T=c
    u=bc

    for λ= nπ/L
    T' + n^2π^2/L = 0
    so T= ce^-(n^2π^2t/L)
    u= [sum of] a*c*e^-(n^2π^2t/L)cos(nπx/L)

    u(x,t) = bc + [sum of] a*c*e^-(n^2π^2t/L)cos(nπx/L)

    what about the second part where L=2
    and f(x) = {x , 0 < x <1
    {0 , 1 < x <2
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    Quote Originally Posted by linalg123 View Post
    so X(x)= acos(nπx/L)

    for λ=0 T'=0 so T=c
    u=bc

    for λ= nπ/L
    T' + n^2π^2/L = 0
    so T= ce^-(n^2π^2t/L)
    u= [sum of] a*c*e^-(n^2π^2t/L)cos(nπx/L)

    u(x,t) = bc + [sum of] a*c*e^-(n^2π^2t/L)cos(nπx/L)

    what about the second part where L=2
    and f(x) = {x , 0 < x <1
    {0 , 1 < x <2
    So what you have is

    u_0(x,t)=A_0

    u_n(x,t)=A_n\exp\left( -\frac{n^2\pi^2t}{2}\right)\cos\left( \frac{ n \pi x}{2} \right)

    Now use the inner product to find the constants A_n. So mutliply both sides by

    \cos(\left( \frac{n \pi x}{2}\right) evaluate at t=0 and integrate to get

    A_n\int_{0}^{2} \cos\left( \frac{ n \pi x}{2} \right)\cos\left( \frac{ n \pi x}{2} \right)dx=\int_{0}^{2} \cos\left( \frac{ n \pi x}{2} \right)f(x)dx

    This gives


    A_n=\int_{0}^{1} x\cos\left( \frac{ n \pi x}{2}  \right)dx
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    Both X(0)=0 and X'(0)= 0 can't be true though.

    We have that u(x,t)=X(x)T(t), so u(0,t)=0 \implies u(0,t)=X(0)T(t)=0 \implies X(0)=0 since we have assumed that T(t)\neq 0

    That leads us to
    X(0)=0=A\cos(\lambda \cdot 0)+B\sin(\lambda \cdot 0) = A \implies A=0\implies X(x)=B\sin(\lambda x)
    and
    X'(0)=\lambda B \cos(\lambda \cdot 0)=\lambda B \neq 0 unless B=0 which is a useless solution.
    Last edited by Mondreus; May 19th 2011 at 09:32 AM.
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    you lost me there. multiply both sides of which equation by cos(nπx/2)? and where did the exponential go.

    (p.s sorry if this is really obvious and i just can't see it)
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    Quote Originally Posted by linalg123 View Post
    you lost me there. multiply both sides of which equation by cos(nπx/2)? and where did the exponential go.

    (p.s sorry if this is really obvious and i just can't see it)

    u_n(x,t)=A_n\exp\left( -\frac{n^2\pi^2t}{2}\right)\cos\left( \frac{ n \pi x}{2} \right)

    Now use the initial condition


    u_n(x,0)=A_n\cos\left( \frac{ n \pi x}{2} \right)=f(x)

    Since your basis is a complete orthonormal set use the inner product

    <f,g>=\int_{0}^{2}f(x)g(x)dx

    and the fact that

    \bigg<\cos\left( \frac{ n \pi x}{2} \right),\cos\left( \frac{ n \pi x}{2} \right) \bigg>=\int_{0}^{2}\cos\left( \frac{ n \pi x}{2} \right)\cos\left( \frac{ n \pi x}{2} \right)dx=1
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    when i do this i get:

    Ao= ∫x dx from 0 to 1 = 1/2

    An= ∫xcos(nπx/2)dx
    = 2/nπ ∫x(sin(nπx/2))'dx
    = [2x/nπsin(nπx/2)] from 0 to 1 - 2/nπ ∫x'sin(nπx/2)dx
    = 0 + 4cos(nπx/2)/(n^2π^2)
    = 0
    this doesn't seem right.. have i made an error?
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    Quote Originally Posted by linalg123 View Post
    when i do this i get:

    Ao= ∫x dx from 0 to 1 = 1/2

    An= ∫xcos(nπx/2)dx
    = 2/nπ ∫x(sin(nπx/2))'dx
    = [2x/nπsin(nπx/2)] from 0 to 1 - 2/nπ ∫x'sin(nπx/2)dx
    = 0 + 4cos(nπx/2)/(n^2π^2)
    = 0
    this doesn't seem right.. have i made an error?
    I can really follow what you typed but I get the anti derivative

    integrate x&#42;cos&#40;n&#42;Pi&#42;x&#47;2&#41; from x&#61;0 to x&#61;1 - Wolfram|Alpha
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    ok i saw my mistake. i got

    An= 2sin(nπ/2)/πn - 4/(π^2n^2) , i think this is the same as yours because the cos(πn/2) term =0

    can sin(nπ/2) be simplified? i can't think of a way.
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    Quote Originally Posted by linalg123 View Post
    ok i saw my mistake. i got

    An= 2sin(nπ/2)/πn - 4/(π^2n^2) , i think this is the same as yours because the cos(πn/2) term =0

    can sin(nπ/2) be simplified? i can't think of a way.
    You should analyze the even and odd terms separately.

    \cos\left( \frac{n\pi}{2}\right)=0, \text{ when } n \text{ is odd and } \pm 1 \text{ when n is even}

    You will ge something similar with sine
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    ah, of course...

    so for even n,
    An= 4[(-1)^(n/2)-1]/(π^2n^2)

    and for odd n,
    An= 2[πn(-1)^((n-1)/2)-2]/(π^2n^2)

    how do i use proper notation to write this?

    (promise that's the last question i'll ask,lol)
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