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Math Help - 2nd Order Linear PDE - Harmonic Function

  1. #1
    Senior Member bugatti79's Avatar
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    2nd Order Linear PDE - Harmonic Function

    Folks,

    Can anyone assist in solving the attached question?

    I believe the PDE is ellipitic becasue B^2-AC = 0 -(1)(1)<0 implies elliptic.
    Therefore we can write



     \displaystyle \frac{dy}{dx}=\frac{0\pm \sqrt{0-AC}}{A}=\pm i

    Therefore I choose y-ix = Const. From this I choose new variables with s(x,y)= y and t(x,y)=-x

    Not sure what to do after this
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  2. #2
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    Not sure what you're trying to achieve. You have Laplace's equation

    \dfrac{\partial^2 u}{\partial x^2}  + \dfrac{\partial^2 u}{\partial y^2}  = 0.

    It won't reduce to more than that.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Not sure what you're trying to achieve. You have Laplace's equation

    \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0.

    It won't reduce to more than that.
    Well I am trying to answer part b i and b ii in the attachment. Obviously my attempt is wrong...

    Any suggestions?
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  4. #4
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    Well, if you consider the Cauchy-Riemann equations

    u_x = v_y, u_y = - v_x

    then eliminating v gives Laplace's equation. If you substitute your u into these and solve for v, this is the imaginery part that you are seeking.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Well, if you consider the Cauchy-Riemann equations

    u_x = v_y, u_y = - v_x

    then eliminating v gives Laplace's equation. Is this for part i?

    If you substitute your u into these and solve for v, this is the imaginery part that you are seeking. I presume this for part ii
    For part i, I attempted the following to find v(x,y)

    \displaystyle U_x=\frac{\partial }{\partial x} \left[\frac{x}{x^2+y^2}\right]=\frac{y^2-x^2}{(x^2+y^2)^2}=\upsilon_y (1)

    \displaystyle U_y=\frac{\partial }{\partial y} \left[\frac{x}{x^2+y^2}\right]=\frac{-2xy}{(x^2+y^2)^2}=-\upsilon_x (2)

    I attempt to find v(x,y) from (1) and (2) by integrating and hence arriving at

    from (2)

    \displaystyle v(x,y)=-\frac{y}{2(x^2+y^2)}+f(y)
    from (1)

    \displaystyle v(x,y)=-\frac{y}{(x^2+y^2)}+f(x)

    Not sure if this right approach or what to do next?

    Thanks
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  6. #6
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    For part I, directly substitute U = \dfrac{x}{x^2+y^2} into U_{xx} + U_{yy} = 0 to show it is identically satisfied. For part ii, yes what you have done is correct except for the 2 in the denominator of the first one and those arbitrary functions must both be a constant (the same constant as we want the same answer).

    Another way to do this is to consider

    f(x+iy) = \dfrac{x}{x^2+y^2} + v(x,y) i

    Then set y = 0 giving, thus f(x) = \dfrac{1}{x} (if f(x) is real then v(x,0)=0). So f(w) = \frac{1}{w}.

    Thus f(x+iy) = \dfrac{1}{x+iy} simplify and the imaginery part is your v.
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Another way to do this is to consider

    f(x+iy) = \dfrac{x}{x^2+y^2} + v(x,y) i

    Then set y = 0 giving, thus f(x) = \dfrac{1}{x} (if f(x) is real then v(x,0)=0). So f(w) = \frac{1}{w}.

    Thus f(x+iy) = \dfrac{1}{x+iy} simplify and the imaginery part is your v.
    Hmmm..Im afraid I dont follow this. How does one get the imaginary part v(x) from this last expression?
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  8. #8
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    Multiply top and bottom by the conjugate x - iy and simplify.
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  9. #9
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Multiply top and bottom by the conjugate x - iy and simplify.
    Perfect! Thank you!
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