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Math Help - Partial Differential eqn - method of characteristics

  1. #1
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    Partial Differential eqn - method of characteristics

    Am going through problems from book and have come across a section of my lecture notes that i don't understand, i'm sure it's something small i'm missing but can't get my head around it.

    In lecture notes we solve the wave eqn;

    it gets to the General solution which is;

    Q(x,t) = F(x-t) +G(x-t)

    and then as one of my initial conditions is;

    dQ/dx(0,t) = 2 (where this is the partial derivative of Q wrt. x)

    i need to find an expression for the dQ/dx, which was written as;

    dQ/dx = F'(x+t) + G'(x-t)
    i dont understand how this works?
    can anyone help?
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  2. #2
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    Quote Originally Posted by monster View Post
    Am going through problems from book and have come across a section of my lecture notes that i don't understand, i'm sure it's something small i'm missing but can't get my head around it.

    In lecture notes we solve the wave eqn;

    it gets to the General solution which is;

    Q(x,t) = F(x-t) +G(x-t)

    and then as one of my initial conditions is;

    dQ/dx(0,t) = 2 (where this is the partial derivative of Q wrt. x)

    i need to find an expression for the dQ/dx, which was written as;

    dQ/dx = F'(x+t) + G'(x-t)
    i dont understand how this works?
    can anyone help?
    It's the chain rule. Let r = x + t and s = x- t so Q = F(r) + G(s) so

     <br />
\frac{\partial Q}{\partial x} = \frac{\partial Q}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial Q}{\partial s}\frac{\partial s}{\partial x}<br />
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  3. #3
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    Consider the IVP
    a(x,t,u) u_t + b(x,t,u) u_x = c(x,t,u),\; u(x,0) = f(x)
    Parameterizing your variable as x=r,\; t=s,\; u=z and assuming that you start on a characteristic curve with initial conditions \Gamma: x = r,\; t = 0,\; u = f(s) then the solution to the IVP can be found by the ODE
    <br />
\dfrac{dx}{ds} = b(r,s,z)<br />
    <br />
\dfrac{dt}{ds} = a(r,s,z)<br />
    <br />
\dfrac{dz}{ds} = c(r,s,z)<br />
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