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Math Help - Charpits: Non Linear 1st order PDE: Particular IC

  1. #1
    Senior Member bugatti79's Avatar
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    Charpits: Non Linear 1st order PDE: Particular IC

    Folks,

    I am struggling to get the initial conditions @ t=0 for p and q.

    Given (u_x)^2+(u_y)^2-1=0 for u=0 on x^2+y^2=1
    My attempt:

    Parameterise x such that x=s and y=\sqrt{1-s^2} and differentiate the given IC

     \displaystyle \frac{\partial u}{\partial x} \frac{d(s)}{ds}+\frac{\partial u}{\partial y} \frac{d(\sqrt{1-s^2})}{ds}=\frac{d(0)}{ds} This gives

     \displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \frac{-s}{\sqrt{1-s^2}}=0

    Not sure if this right or how to proceed further to find pand q?

    Thanks
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  2. #2
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    So now you have one equation

    p - \frac{s}{\sqrt{1-s^2}}q = 0.

    Your second equation is the PDE itself p^2+q^2=1.

    Two equations for two unknowns. Now solve for p and q.

    BTW - what do you mean @ t= 0. There's no t in this problem.

    Because of the symmetry of the BC, I might also suggest switching to polar coordinates.
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    Senior Member bugatti79's Avatar
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Quote Originally Posted by Danny View Post
    So now you have one equation

    p - \frac{s}{\sqrt{1-s^2}}q = 0.

    Your second equation is the PDE itself p^2+q^2=1.

    Two equations for two unknowns. Now solve for p and q.

    BTW - what do you mean @ t= 0. There's no t in this problem.

    Because of the symmetry of the BC, I might also suggest switching to polar coordinates.
    All the problems involving Charpits I have been solving are in parametric form where the IC's are taken to be at t=0. I believe the curve is parametric in x(t), y(t) and u(t). ...?
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    Senior Member bugatti79's Avatar
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Maybe I should clarify that in this problem we have u=0 when x^2+y^2=1. These are the IC's I take at t=0. I hope that makes sense. This is the approach I take for all other charpit problems based on my notes.
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    I'm good. I usually use r and s and leave t for a time variable :-)
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Quote Originally Posted by Danny View Post
    I'm good. I usually use r and s and leave t for a time variable :-)
    Based on your post no 2, you mention about 2 eqns ans 2 unknowns. However, I dont believe I can use the second equation (actual PDE) because it does not represent an initial condition like the first equation does, so how does one proceed? Perhaps I am wrong :-)
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Actually you can! If u_x^2+u_y^2 = 1 for all x and y then certainly along some curve y = f(x) (as long as the derivatives exist).
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    Senior Member bugatti79's Avatar
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Quote Originally Posted by Danny View Post
    Actually you can! If u_x^2+u_y^2 = 1 for all x and y then certainly along some curve y = f(x) (as long as the derivatives exist).
    OK, thanks. I will give it a go and keep you posted. :-)
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  9. #9
    Senior Member bugatti79's Avatar
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Quote Originally Posted by Danny View Post
    Actually you can! If u_x^2+u_y^2 = 1 for all x and y then certainly along some curve y = f(x) (as long as the derivatives exist).
    OK, recall that I parameterise x as equal to s so that t=0 corresponds to the initial conditions.

    Based on this I calculate, using the 2 equations and 2 unknowns

    p= \sqrt{1-s^2} and q=s @ t=0.

    I should be able continue and find u(x,y)

    Are these right?
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  10. #10
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    Re: Charpits: Non Linear 1st order PDE: Particular IC

    Looks good be I might be careful on the \pm.
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