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Math Help - PDE-Characteristic Curves

  1. #1
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    PDE-Characteristic Curves

    Let  f(x,y) be the soloution of xu_x +yu_y = u^4 that is defined in the whole plane. Prove that  f = 0 .
    Hint: Think of the characteristic curves of this PDE.

    HOPE You'll be able to help me
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    Let  f(x,y) be the soloution of xu_x +yu_y = u^4 that is defined in the whole plane. Prove that  f = 0 .
    Hint: Think of the characteristic curves of this PDE.

    HOPE You'll be able to help me
    Thanks in advance!


    Solve the characteristic system dx/x=dy/y=du/u^4 and assure that any nonzero solution on the whole plane will develop a discontinuity at the origin.
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  3. #3
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    I've already done this...We'll get:
     y=c_1 \cdot x and:  u^3 = -\frac{1}{3ln(x) +3k(y)} where is k is a function of y ...
    We'll have a discontinuity every time that  ln(x) = -k(y) and every time that x=0...
    How did you deduced the concloution that the discontinuities are at the origin?

    Thanks !
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  4. #4
    Super Member Rebesques's Avatar
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    From y=c_1x. Thus there will always be a discontinuity at the origin, but the expression lnx+k(y) may be nonzero for a suitable choice of initial data.
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  5. #5
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     y=c_1 \cdot x is a line segmant in  R^2 ...Why there is a discontinuity at the origin?
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  6. #6
    Super Member Rebesques's Avatar
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    It is one of the characteristics though. The constant c_1 is the slope and will vary along each characteristic, while being (explicitly) determined by c_1=y/x. So at x=0 (for which y=0) the slope will become infinite and the solution will develop a shock. check your notes.
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  7. #7
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    actually, at x=0, the expression ln(x) has a discontinuity... So we can deduce the result from this (I think...)...

    Thanks...I think I understand it
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