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Math Help - Linear PDE u_x+u_y=0

  1. #1
    Senior Member bugatti79's Avatar
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    Linear PDE u_x+u_y=0

    Folks,

    trying 2 methods, neither working out for me

    1)Method of Characteristics

     \displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0 given u(x,0)=x
     \frac{dy}{dx}=1 \implies c=y-x,  \frac{du}{dx}=0 \implies u=f(c)

    Using IC's, x=f(-x), let  t=-x \implies x=-t=f(t) \therefore u(x,y)=y-x

    but the correct is u(x,y)=-y+x

    2) Parameterise x and y with s with the given IC's

     x(s); x(0)=x, y(s);y(0)=0

    \frac{d}{ds} u(x(s),y(s))= \frac{dx}{ds}\frac{\partial u}{\partial x}+\frac{dy}{ds}\frac{\partial u}{\partial y} = 0

    \frac{dx}{ds}=1 \implies x=s+A, \frac{dy}{ds}=1 \implies y=s+B, \frac{du}{ds}=0 \implies u=f(c)

    Not sure what is next, when I use the IC's, it doesnt look right.

    THanks
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    I am not familiar with the method in part one, but for the 2nd one you have

    x=s+a \quad y=s+b \quad u=c

    Now using the intial conditions gives

    x_0=0+a \quad 0=0+b \quad u(x_0,0)=x_0

    This gives

    x=s+x_0 \quad y=s \implies x_0=x-s=x-y

    This gives

    u(x,y)=x-y
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  3. #3
    MHF Contributor
    Jester's Avatar
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    You have a sign error in part (i). From what you have

    u = f(c) = f(y-x) so when y = 0 then u = x so f(-x) = x or f(t) = -t

    this is everything you have.

    Thus u = \color{red}{-}\color \;(y-x) = x - y (note the red negative sign)
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  4. #4
    Senior Member bugatti79's Avatar
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    Perfect,
    For (1), I used x=f(t) which threw me off.

    Thanks guys
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