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Math Help - Parametric form of Quasi Linear PDE

  1. #1
    Senior Member bugatti79's Avatar
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    Parametric form of Quasi Linear PDE

    Hi,

    Could anyone check the attached and see where i have gone wrong on this problem. Asked to find parametric form of pde and hence relationship between u and x,y. The general solution does not match the solution for initial conditions

    Thanks
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  2. #2
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    I believe you changed sign going from

    - \dfrac{1}{x} = t + A \; \text{to}\; x = \dfrac{1}{-t - A}

    Edit. No mistake - my bad.
    Last edited by Jester; December 23rd 2010 at 05:47 AM.
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  3. #3
    A Plied Mathematician
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    But isn't it true that if

    -\dfrac{1}{x}=t+A, then

    \dfrac{1}{x}=-t-A, and hence

    x=\dfrac{1}{-t-A}? So, I think that's all right.

    I think the mistake might be in applying the initial conditions. Are you told that when u=0, it's also true that t=0? I don't see that in the original problem statement. So, you can only say that

    s=\dfrac{1}{-t-A}.

    You've parametrized the equation in terms of t but there's no guarantee that that parametrization has its origin just anywhere you want it, right? I would go through and re-apply the initial conditions after your integrations, and just leave t in there and see what happens.
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    But isn't it true that if

    -\dfrac{1}{x}=t+A, then

    \dfrac{1}{x}=-t-A, and hence

    x=\dfrac{1}{-t-A}? So, I think that's all right.

    I think the mistake might be in applying the initial conditions. Are you told that when u=0, it's also true that t=0? I don't see that in the original problem statement. So, you can only say that

    s=\dfrac{1}{-t-A}.

    You've parametrized the equation in terms of t but there's no guarantee that that parametrization has its origin just anywhere you want it, right? I would go through and re-apply the initial conditions after your integrations, and just leave t in there and see what happens.
    Danny and Ackbeet,

    The problem statement is all that was given however examples in the lectures notes are based on the idea that we parametrise the initial conditions so that they correspond to t=0. So I am assuming this for the above problem, ie

    A characteristic passes through \Gamma at point P (x=s, y=1-s, u=0) at t=0

    This is the only way we have been taught. I spotted an error where s should be equal to \frac{1}{t+\frac{1}{x}} Ie, the minus is changed to a plus. This still doesnt solve the problem....
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  5. #5
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    So your final equation there is

    \displaystyle y=\frac{u^{2}}{2}+1-\left[\frac{1}{u+\frac{1}{x}}\right], incorporating the sign correction.

    Evaluating at u=0, x=s yields

    \displaystyle y=1-\left[\frac{1}{\frac{1}{s}}\right]=1-s,

    as required. Which initial condition is not being satisfied here?
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    So your final equation there is

    \displaystyle y=\frac{u^{2}}{2}+1-\left[\frac{1}{u+\frac{1}{x}}\right], incorporating the sign correction.

    Evaluating at u=0, x=s yields

    \displaystyle y=1-\left[\frac{1}{\frac{1}{s}}\right]=1-s,

    as required. Which initial condition is not being satisfied here?
    mmmm...that makes perfect sense what you did but im going the other way, ie I want to see if u goes to 0 when i sub x=s and y=1-s into general solution.
    Afer some manipulation I get
    -s=\frac{u{^2}}{2}-\frac{s}{us+1} I cant see how this gets u=0. Is this to do with the general solution being defined implicitly?

    Thanks
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  7. #7
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    Your equation is correct. You can keep going, you know. Get a common denominator on the RHS, multiply through by that common denominator, get all the u's over to one side, etc. I get the following:

    u^{3}s+u^{2}+2us^{2}=0, which implies either that u=0, or that

    u=\dfrac{-1\pm\sqrt{1-8s^{3}}}{2s}.

    Does u have to be real for all s? If so, then you can definitely rule out both of these solutions, and you're forced to conclude that u=0. Otherwise, I'm not sure you can.
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Your equation is correct. You can keep going, you know. Get a common denominator on the RHS, multiply through by that common denominator, get all the u's over to one side, etc. I get the following:

    u^{3}s+u^{2}+2us^{2}=0, which implies either that u=0, or that

    u=\dfrac{-1\pm\sqrt{1-8s^{3}}}{2s}.

    Does u have to be real for all s? If so, then you can definitely rule out both of these solutions, and you're forced to conclude that u=0. Otherwise, I'm not sure you can.
    I got that expression but I didnt 'see' how to go on further. I need to remind myself that solutions to u can appear quadratically. To answer your question, it doesnt say whether u has to be real for all s.
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  9. #9
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    Hmm. Well then, there's no a priori reason to rule out the nonzero solutions. Incidentally, for those nonzero solutions, you'll get real u if and only if s \le 1/2. I don't know how much more can be said on the matter.
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