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Math Help - Wave equation help.

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    Wave equation help.

    Question: Solve the wave equation given the following boundary conditions.
    u(0,t)=0
    u(L,t)=0
    u(x,0)=0

    du/dt|t=0= x(L-x)


    Attempt:
    ok i'm not gonna write out all the steps because i can't use latex and it will be really messy, but i have
    un= [Ancos(nπαt/L) + Bnsin(nπαt/L)]sin(nπx/L)
    u(x,t) = Σ un

    u(x,0) = ΣAnsin(nπx/L) = 0

    Question: Does this make An=0?




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    Quote Originally Posted by linalg123 View Post
    Question: Solve the wave equation given the following boundary conditions.
    u(0,t)=0
    u(L,t)=0
    u(x,0)=0

    du/dt|t=0= x(L-x)


    Attempt:
    ok i'm not gonna write out all the steps because i can't use latex and it will be really messy, but i have
    un= [Ancos(nπαt/L) + Bnsin(nπαt/L)]sin(nπx/L)
    u(x,t) = Σ un

    u(x,0) = ΣAnsin(nπx/L) = 0

    Question: Does this make An=0?




    Yes it does and the form of your solution is correct so far.
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    ok so

    du/dt= Σ Bn(nπα/L)sin(nπx/L)

    du/dt|t=0 = x(L-x) = Σ Bn(nπα/L)sin(nπx/L)

    Bn(nπα/L) = 2/L ∫ x(L-x)sin(nπx/L).dx

    Bn= 2/(nπα) ∫ x(L-x)sin(nπx/L).dx

    i'm getting stuck here.. do i expand the bracket or can i integrate this from here?
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    Quote Originally Posted by linalg123 View Post
    ok so

    du/dt= Σ Bn(nπα/L)sin(nπx/L)

    du/dt|t=0 = x(L-x) = Σ Bn(nπα/L)sin(nπx/L)

    Bn(nπα/L) = 2/L ∫ x(L-x)sin(nπx/L).dx

    Bn= 2/(nπα) ∫ x(L-x)sin(nπx/L).dx

    i'm getting stuck here.. do i expand the bracket or can i integrate this from here?
    Okay so you have

    u_{n}(x,t)=\sum B_n\sin\left( \frac{n\pi \alpha t}{l} \right)\sin\left(\frac{n \pi x}{l} \right)

    Now if you take a partial with respect to t you get

    \frac{\partial u_n}{\partial t}\bigg|_{(x,t)}=\sum B_n\cdot \frac{n \pi \alpha}{L}  \cdot \cos\left( \frac{n\pi \alpha t}{L} \right)\sin\left(\frac{n \pi x}{L} \right)

    evaluating when t=0 and for each n this gives

    x(L-x)= B_n\cdot \frac{n \pi \alpha}{L}  \cdot \sin\left(\frac{n \pi x}{L} \right)

    Now just use orthogonality to solve for the B's.
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    Quote Originally Posted by TheEmptySet View Post

    x(L-x)= B_n\cdot \frac{n \pi \alpha}{L}  \cdot \sin\left(\frac{n \pi x}{L} \right)

    Now just use orthogonality to solve for the B's.
    In the book it says from here you calculate the B's using a half range sine expansion.
    I know orthogonality means perpendicular, but i'm not sure what you mean by 'use orthogonality' in this context.
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    Quote Originally Posted by linalg123 View Post
    In the book it says from here you calculate the B's using a half range sine expansion.
    I know orthogonality means perpendicular, but i'm not sure what you mean by 'use orthogonality' in this context.
    I am just saying the same thing as your book. We know that

    \int_{0}^{L}\sin\left(\frac{n \pi x}{L}\right) \sin\left(\frac{m \pi x}{L}\right)dx=\frac{L}{2}\delta_{m,n}

    So multiply both sides of the equation by

    \sin\left(\frac{n \pi x}{L}\right)

    and integrate from 0 to L to get

    \int_{0}^{l}x(L-x)\sin\left(\frac{n \pi x}{L}\right)dx= B_n\cdot \frac{n \pi \alpha}{L}  \frac{L}{2}=B_n\cdot \frac{n \pi \alpha}{2}

    Now you can solve for all of the B's
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    Quote Originally Posted by TheEmptySet View Post

    \sin\left(\frac{n \pi x}{L}\right)

    and integrate from 0 to L to get

    \int_{0}^{l}x(L-x)\sin\left(\frac{n \pi x}{L}\right)dx= B_n\cdot \frac{n \pi \alpha}{L}  \frac{L}{2}=B_n\cdot \frac{n \pi \alpha}{2}

    Now you can solve for all of the B's
    for this integral i got  \frac{2L^3}{n^3 \pi^3} - \frac{2L^3(-1)^n}{n^3 \pi^3} = B_n\cdot \frac{n \pi \alpha}{2}

     B_n= \frac{4L^3}{n^4 \pi^4 \alpha}\cdot ( 1 - (-1)^n)
    does this look right?
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    Quote Originally Posted by linalg123 View Post
    for this integral i got  \frac{2L^3}{n^3 \pi^3} - \frac{2L^3(-1)^n}{n^3 \pi^3} = B_n\cdot \frac{n \pi \alpha}{2}

     B_n= \frac{4L^3}{n^4 \pi^4 \alpha}\cdot ( 1 - (-1)^n)
    does this look right?
    Yes that is what I get.
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