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Proof of orthogonality of harmonic conjugates

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data

    My textbook (Churchill) is asking me to prove that the contours $$u(x,y) = c_1$$ and $$v(x, y) = c_2$$ where $$u$$ and $$v$$ are the real and imaginary components of an analytic function $$f(z)$$ are orthogonal at any point by noting that $$u_x + u_y \frac{dy}{dx} = 0 $$ and then showing that the tangent lines are orthogonal.


    2. Relevant equations



    3. The attempt at a solution

    I can do this via taking the dot product of $$\nabla u$$ and $$\nabla v$$ and using the Cauchy-Riemann equations to show that $$\nabla u·\nabla v = 0$$ hence the normal vectors of the contours are orthogonal, so the contours themselves are as well. But they asked to find the tangent line first, so here's what I tried, I haven't done any vector calculus in a few years though so I'm not sure if it's valid:

    $$u_x + u_y \frac{dy}{dx} = \langle u_x,u_y \rangle · \langle 1, \tfrac{dy}{dx} \rangle = 0$$

    $$\langle u_x, u_y \rangle$$ is $$\nabla u$$ which is normal to the contour, so $$\langle 1, \frac{dy}{dx} \rangle$$ must be tangent to the $$u$$ contour. From the above, $$\frac{dy}{dx} = -\frac{u_x}{u_y}$$ so $$\langle 1, -\frac{u_x}{u_y} \rangle$$ is tangent to the $$u$$ contour.

    Similarly,
    $$v_x + v_y \frac{dy}{dx} = \langle v_x, v_y \rangle · \langle 1, \tfrac{dy}{dx} \rangle = 0$$

    So $$\langle 1, \frac{dy}{dx} \rangle$$ is tangent to the $$v$$ contour. In this case $$\frac{dy}{dx} = -\frac{v_x}{v_y}$$ so the tangent vector is $$\langle 1,-\frac{v_x}{v_y} \rangle$$ which is equal to $$\langle 1,\frac{u_y}{u_x} \rangle$$ due to the Cauchy-Riemann equations. Then $$\langle 1, -\frac{u_x}{u_y} \rangle · \langle 1, \frac{u_y}{u_x} \rangle = 1-1 = 0$$ so the $$u$$ and $$v$$ contours are orthogonal.

    Is this reasoning valid?
     
  2. jcsd
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