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Thread: Orthogonal Trajectories and Cauchy-Riemann Equations

  1. #1
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    Orthogonal Trajectories and Cauchy-Riemann Equations

    Hello,

    This is a snippet from a book:

    Orthogonal Trajectories and Cauchy-Riemann Equations-ortho_traject_cauchy.png

    I am trying to apply the same principle to:

    $\displaystyle x^2+(y-c)^2=1+c^2$

    Rearranging:
    $\displaystyle u(x,y)=\frac{x^2}{y}+y-\frac{1}{y}=2c$

    Then:
    $\displaystyle \frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=\frac{2x}{y}$
    $\displaystyle \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=\frac{x^2}{y}-1-\frac{1}{y^2}$

    But from another thread, a function that satisfies both partial derivatives cannot be found.

    I don't know the best way to link to a another thread but here it is anyway:


    I am wondering if using the Cauchy-Riemann equations for Orthogonal Trajectories doesn't work in general or if I am missing something?
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  2. #2
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    Ok, this is how I'm interpreting your question: If you have a complex-analytic function $\displaystyle f(z)=u(x,y)+i v(x,y)$ then the families $\displaystyle u(x,y)=c$ are orthogonal to the families $\displaystyle v(x,y)=k$. And if you're given a harmonic function $\displaystyle g(x,y)$ then you can find it's complex conjugate $\displaystyle h(x,y)$ so that the function $\displaystyle f(z)=g(x,y)+i h(x,y)$ is analytic and so the families $\displaystyle g(x,y)=c$ are orthogonal to the families $\displaystyle h(x,y)=k$.

    Your function $\displaystyle f(x,y)=x^2+(y-c)^2-c^2-1=0$ is not harmonic so you can't use the CR equations to find the families orthogonal to it. In your exercise, you're given $\displaystyle e^x\cos(y)$ which is harmonic so that the CR equations could be used to find the orthogonal families to it. However, you can still find the orthogonal famililies to $\displaystyle x^2+(y-c)^2-c^2-1=0$ and it turns out to be an interesting problem in DEs: differentiate throughout to get:

    $\displaystyle \frac{dy}{dx}=\frac{2 x y}{x^2-y^2-1}$

    and therefore the orthogonal families satisfy:

    $\displaystyle \frac{dy}{dx}=-\frac{x^2-y^2-1}{2xy}$

    which you can solve by finding an integrating factor which I think is $\displaystyle 1/x^2$.

    . . . did it kinda' quick. Work out the bugs if any ok.
    Last edited by shawsend; Mar 18th 2010 at 06:21 AM. Reason: corrected de and orthogonal de
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  3. #3
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    The integrating factor of $\displaystyle 1/x^2$ is spot on

    I get the solution $\displaystyle y^2+x^2+1=cx$ (c a constant).
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