# Thread: Orthogonal Trajectories and Cauchy-Riemann Equations

1. ## Orthogonal Trajectories and Cauchy-Riemann Equations

Hello,

This is a snippet from a book:

I am trying to apply the same principle to:

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

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

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

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?

2. Ok, this is how I'm interpreting your question: If you have a complex-analytic function $f(z)=u(x,y)+i v(x,y)$ then the families $u(x,y)=c$ are orthogonal to the families $v(x,y)=k$. And if you're given a harmonic function $g(x,y)$ then you can find it's complex conjugate $h(x,y)$ so that the function $f(z)=g(x,y)+i h(x,y)$ is analytic and so the families $g(x,y)=c$ are orthogonal to the families $h(x,y)=k$.

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

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

and therefore the orthogonal families satisfy:

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

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

. . . did it kinda' quick. Work out the bugs if any ok.

3. The integrating factor of $1/x^2$ is spot on

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