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Math Help - Initial Value Problems with Delta Functions

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Initial Value Problems with Delta Functions

    I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it!

    Solve the initial value problem : x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0.

    Also, I don't get how to do this problem:

    Apply Duhamel's principle to write an integral formula for the solution of the initial value problem x^{//}+4x^{/}+8x=f(t); x(0)=x^{/}(0)=0
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    Quote Originally Posted by Chris L T521 View Post
    I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it!

    Solve the initial value problem : x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0.

    [snip]
    I'd try solving it using Laplace Transforms - something that, if you don't know much about, you might consider learning more as a pre-requisite to solving differential equations involving generalised functions such as the Dirac Delta function .......

    (Although I'm sure I saw something somewhere that suggested you loved Laplace transforms ....?)
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    Quote Originally Posted by Chris L T521 View Post
    [snip]
    Also, I don't get how to do this problem:

    Apply Duhamel's principle to write an integral formula for the solution of the initial value problem x^{//}+4x^{/}+8x=f(t); x(0)=x^{/}(0)=0
    Isn't Duhamel's Principle used to solve certain types of partial differential equations .......?

    Or do you mean Duhamel's Convolution Principle - used to invert a Laplace transform (by using a convolution integral) when using Laplace Transform to solve IVPís with general forcing functions?
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    Rhymes with Orange Chris L T521's Avatar
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    I'm looking in my book, and it appears to be Duhamel's Convolution Principle (I see \tau 's in there, so I think may have to do with convolution...)
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    Quote Originally Posted by Chris L T521 View Post
    Apply Duhamel's principle to write an integral formula for the solution of the initial value problem x''+4x'+8x=f(t); x(0)=x'(0)=0
    I don't know what Duhamel's principle, but it can be solved using the Laplace transform and the convolution theorem.

    Assuming that the Laplace transform of f(t) exists we get

    s^2X+4sX+8X=F(s) \iff X=\frac{F(s)}{s^2+4s+8}

    X=\frac{F(s)}{(s+2)^2+4}

    The Covolution theorem states that

    \mathcal{L}^{-1}(F(s)G(s))=\int_{0}^{t}f(t-\tau)g(\tau)d\tau

    so G(s)=\frac{1}{(s+2)^2+4}

    The inverse transform of the s-axis translation is

    g(t)=\frac{1}{2}e^{-2t}\sin(2t)

    Now using the above and the convolution theorem we get

    x(t)=\mathcal{L}^{-1}\frac{F(s)}{(s+2)^2+4}=\int_{0}^{t}f(t-\tau)(\frac{1}{2}e^{-2\tau}\sin(2\tau))d\tau

    I hope this helps.
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    Quote Originally Posted by Chris L T521 View Post
    I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it!

    Solve the initial value problem : x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0.
    Taking the Laplace transform of both sides gives

    s^2X+9X=e^{3s\pi }+\frac{s}{s^2+9}

    Solving for X gives

    X=\frac{e^{3s\pi}}{s^2+9}+\frac{s}{(s^2+9)^2}

    To avoid using partial fractions on the 2nd note that

    \frac{s}{(s^2+9)^2}=\frac{1}{6}(-1)\frac{d}{ds}\left( \frac{3}{s^2+9}\right)

    We can now use the derivative of a transform theorem backwards

    \mathcal{L}(t^nf(t))=(-1)^n\frac{d^n}{ds^n}F(s)

    so we get the inverse transform is \frac{1}{6}t\sin(3t)

    so finally our solution is

    x(t)=\frac{1}{3}\mathcal{U}(t-3\pi)\sin(3(t-3\pi)+\frac{1}{6}t\sin(3t)
    Last edited by TheEmptySet; May 10th 2008 at 11:19 PM. Reason: left off an s
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