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Math Help - Help with DE

  1. #1
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    Help with DE

    Ive got this DE: 8(du/dx)+4(du/dy)=9u anyone could help me?
    I separate it with u=X(x)Y(y)
    then I get: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) +9X(x)Y(y)
    If it would be just: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) I could write = constant, but with that 9X(x)Y(y) how to solve?

    Then I got conditions: u(0,y)=33*e^-y +50*e^4y


    Please help me solve this.
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  2. #2
    Super Member Rebesques's Avatar
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    It is of the form au_x+bu_y+cu=0, so solve it with characteristics.

    We are looking for curves (x(s),y(s)) such that x'(s)=8, \ y'(s)=4 and x(0)=0, \ y(0)=y_0 (since the initial conditions are of the form u(0,y)=...). We easily get x=8s, \ y=4s+y_0 (1). On these curves (the characteristics) the pde becomes the ode u'(s)-9u(s)=0, which we solve to get u=f(y_0){\rm e}^{9s}, for some differentiable f. Using (1), we return to the initial variables: u(x,y)=f(y-(x/2)){\rm e}^{9x/8}. Now, use the initial condition u(0,y)=\ldots to find an expression for f
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DanielM99 View Post
    Ive got this DE: 8(du/dx)+4(du/dy)=9u anyone could help me?
    I separate it with u=X(x)Y(y)
    then I get: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) +9X(x)Y(y)
    If it would be just: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) I could write = constant, but with that 9X(x)Y(y) how to solve?

    Then I got conditions: u(0,y)=33*e^-y +50*e^4y


    Please help me solve this.
    I can solve the equation using separation of variables, but I can't get the initial condition to work out. Is there a typo?

    8X^{\prime}Y + 4 XY^{\prime} = 9 XY <-- Divide both sides by XY

    8 \frac{X^{\prime}}{X} + 4 \frac{Y^{\prime}}{Y} = 9

    We can either solve for the X terms or the Y terms. I'll solve for the X terms:
    8 \frac{X^{\prime}}{X} = 9 - 4 \frac{Y^{\prime}}{Y}

    Now, since the LHS is a function only of y and the RHS is a function only of x the two sides must be equal to the same constant. I'll set
    8 \frac{X^{\prime}}{X} = k

    X^{\prime} = \frac{k}{8}X

    Thus
    X(x) = Ae^{kx/8}

    Again:
    9 - 4 \frac{Y^{\prime}}{Y} = k

    Y^{\prime} = \frac{9 - k}{4}Y

    Thus
    Y(y) = Be^{(9-k)y/4}

    So
    u(x,y) = ABe^{(kx/8) + (9-k)y/4} = Ce^{(kx/8) + (9-k)y/4}
    (Setting C = AB since this is just a single arbitrary constant anyway.)

    Your initial condition says that u(0, y) = 33e^{-y} +50e^{4y}

    So we must have:
    Ce^{(k \cdot 0/8) + (9-k)y/4} = 33e^{-y} +50e^{4y}

    Ce^{(9-k)y/4} = 33e^{-y} +50e^{4y}

    Ce^{9y/4}e^{-ky/4} = 33e^{-y} +50e^{4y}

    As you can see we can't possibly match the initial condition from this equation. As I understand it, the solution to this differential equation is unique, so I don't understand how the boundary conditions can be correct.

    -Dan
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    Super Member Rebesques's Avatar
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    As I understand it, the solution to this differential equation is unique, so I don't understand how the boundary conditions can be correct.
    It's just that the solution here cannot be of the form X(x)Y(y), because (as you confirmed) the initial conditions do not allow it.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    It's just that the solution here cannot be of the form X(x)Y(y), because (as you confirmed) the initial conditions do not allow it.
    Okay, but I have a question about the form of the solution then. For a linear ODE two solutions (not considering initial conditions) can only differ by a constant. Is there a similar theorem for linear PDEs?

    -Dan
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  6. #6
    Super Member Rebesques's Avatar
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    Only in special cases that would be true.
    The reason is dependance of the equation on u.

    Consider this one, which is fairly simple. If u is a solution, then u*=u+c is not.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Only in special cases that would be true.
    The reason is dependance of the equation on u.

    Consider this one, which is fairly simple. If u is a solution, then u*=u+c is not.
    Ah wait! The uniqueness theorem I was trying to remember applies to equations in wave mechanics of the form \nabla^2 \Psi + V(\vec{x})\Psi = E \Psi. (Disregarding some constants.) So my uniqueness thought probably doesn't apply anyway.

    -Dan
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  8. #8
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    The best approach here is with charachteristics, like Rebesques did.


    I do not think that Seperation of Variables works here. We need to have a Dirichlet problem which the poster did not supply.
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