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Math Help - u_{xx}-u=0

  1. #1
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    u_{xx}-u=0

    Find the general solution to u_{xx}-u=0

    The solution is \displaystyle u(x,y)=f(y)e^x+g(y)e^{-x}

    How is this obtained?

    Thanks.

    I solved the first part:

    \displaystyle P(x)=-1\Rightarrow e^{\int P(x)dx}=e^{-\int dx}=e^{-x}

    \displaystyle \int\frac{\partial}{\partial x}\left[e^{-x}u_x\right]dx=\int 0dx\Rightarrow e^{-x}u_x(x,y)=f(y)

    Should that be in the integral the partial derivative or just derivative?

    \displaystyle\Rightarrow u_x(x,y)=e^xf(y)
    Last edited by dwsmith; December 23rd 2010 at 03:45 PM.
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  2. #2
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    I think of this problem as find the solution u(x,y) such that u(x,y) = u_{xx}

    As you only have a partial with respect to the same variable then the solution is similar to that of a 2nd order ODE. I.e y'' = y has the solution y=e^x and y= e^{-x}

    The difference in your problem is that f(y) and g(y) need to be introduced but are constant with respect to x
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    \displaystyle P(x)=-1\Rightarrow e^{\int P(x)dx}=e^{-\int dx}=e^{-x}
    Why not try separation of variables?
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  4. #4
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    Quote Originally Posted by pickslides View Post
    I think of this problem as find the solution u(x,y) such that u(x,y) = u_{xx}

    As you only have a partial with respect to the same variable then the solution is similar to that of a 2nd order ODE. I.e y'' = y has the solution y=e^x and y= e^{-x}

    The difference in your problem is that f(y) and g(y) need to be introduced but are constant with respect to x
    I just updated my post as you replied.

    My problem is should it be partial derivatives or written like an ODE with just the derivative?

    Should I use the same method for the integration again?
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Why not try separation of variables?
    That was what I thought of when the light bulb turned on.
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  6. #6
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    I understand your method which is simple but I would like to finish the problem how I started it. If you wouldn't mind, how should I finish it off?

    Thanks.

    If I just integrate, I wouldn't obtain the e^{-x} so I am not sure how to proceed.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    I understand your method which is simple but I would like to finish the problem how I started it. If you wouldn't mind, how should I finish it off?

    Thanks.

    If I just integrate, I wouldn't obtain the e^{-x} so I am not sure how to proceed.
    Well your attempt suggests finding an integrating factor in a 2nd order DE, will that work? Should you not be finding a characteristic equation?

    ;D
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  8. #8
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    I just looked it up. Integrating factors aren't restricted to first order linear equations.
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  9. #9
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    I would say \displaystyle u_{xx}-u(x,y)=0

    Gives characteristic equation in the form \displaystyle r^2-1= (r-1)(r+1)=0 \implies r_{1,2}=\pm 1 and \displaystyle y=c_1e^{r_1x}+c_2e^{r_2x} therefore seek a solution in the form of \displaystyle y=c_1e^x+c_2e^{-x}

    As we have a PDE \displaystyle c_1 and \displaystyle c_2 can also be separate functions of \displaystyle y.
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  10. #10
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    Now, I am looking to find the sol. of u_{xx}-u=0 which satisfies the auxiliary conditions

    u(0,y)=\varphi (y) \ , \ u_x(0,y)=\psi (y)

    u(0,y)=f(y)+g(y)=\varphi (y)

    u_x(0,y)=f(y)-g(y)=\psi (y)

    How do I incorporate this into the solution?

    u(x,y)=e^xf(y)+e^{-x}g(y)
    Last edited by dwsmith; December 23rd 2010 at 05:04 PM.
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  11. #11
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    Is this the correct procedure?

    u(0,y)=f(y)+g(y)=\varphi (y)

    u_x(0,y)=f(y)-g(y)=\psi (y)

    By adding the two equations and simplifying, we obtain \displaystyle f(y)=\frac{\varphi (y)+\psi (y)}{2}

    Now, solving for g(y) I get \displaystyle g(y)=\frac{\varphi (y)-\psi (y)}{2}

    \displaystyle u(x,y)=e^x\left[\frac{\varphi (y)+\psi (y)}{2}\right]+e^{-x}\left[\frac{\varphi (y)-\psi (y)}{2}\right]
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