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windsylph

  • 4 years ago

PDE's: For the partial differential equation u_tt = 9u_xx, a general solution is u(x,t) = F(x+3t)+G(x-3t). Find a particular solution with initial conditions u(x,0)=0, u_t(x,0)=sin(pi*x), u(0,t)=u(1,t)=0.

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  1. windsylph
    • 4 years ago
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    I'm having trouble especially using/interpreting the last initial condition.

  2. myininaya
    • 4 years ago
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    before we look at initial conditions do you know how to do this part: solve for particular solution \[u_{t t}=9u_{xx}\] ? is it separation of variables?

  3. windsylph
    • 4 years ago
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    If the problem asked me to solve it, yeah I would use separation of variables since that's the only method I know so far. but the problem only stated that the given solution is indeed a solution. actually, the first part of this problem was to show that the given solution is indeed a solution, the second part is this.

  4. myininaya
    • 4 years ago
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    ok

  5. myininaya
    • 4 years ago
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    \[u(x,0)=F(x+3(0))+G(x-3(0))=F(x)+G(x)=0\] f(x)+g(x)=0 \[u_t=3f(x+3t)-3g(x-3t)\] \[u_t(x,0)=3f(x)-3g(x)=\sin(\pi x)\] \[u(0,t)=u(1,t)=0\] \[u(0,t)=F(0+3t)+G(0-3t)=F(3t)+G(-3t)=0\] \[u(1,t)=F(1+3t)+G(1-3t)=0\]

  6. myininaya
    • 4 years ago
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    so we have \[F(x)+G(x)=0\] \[f(x)+g(x)=0\] \[3f(x)-3g(x)=\sin(\pi x)\] \[F(3t)+G(-3t)=0\] \[F(1+3t)+G(1-3t)=0\]

  7. myininaya
    • 4 years ago
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    \[f(x)+g(x)=0 => f(x)=-g(x) =>3f(x)+3f(x)=\sin(\pi x)\]

  8. myininaya
    • 4 years ago
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    \[f(x)=\frac{1}{6} \sin(\pi x)\]

  9. myininaya
    • 4 years ago
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    F(x)=\[\int\limits_{}^{}f dx=\int\limits_{}^{}\frac{1}{6} \sin(\pi x) dx=-\frac{1}{6} \cos(\pi x) +c_1\]

  10. myininaya
    • 4 years ago
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    \[3f(3t)-3g(-3t)=0 => f(3t)-g(-3t)=0 => f(3t)=g(-3t)\] \[3f(1+3t)-3g(1-3t)=0 => f(1+3t)=g(1-3t)\] i hope some of this will help i have to leave

  11. windsylph
    • 4 years ago
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    thanks still! ill look at it

  12. myininaya
    • 4 years ago
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    you might want to check what i wrote for mistakes gl

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