Prove that $\phi(x,y)=e^{u(x,y)}cos(v(x,y))$ is harmonic

Question

Suppose that $u,v$ are harmonic functions on doman $D$, and they are harmonic conjugate. Prove that function $\phi(x,y)=e^{u(x,y)}cos(v(x,y))$ is harmonic on $D$.

What I've done was to take the second order derivative of $\phi$ with respect to $x$ and $y$, but then I end up with a bunch of ugliness.

$\frac{\partial^2\phi}{\partial x^2}=e^u(-sin(x))\frac{\partial^2v}{\partial x^2}+\frac{\partial v}{\partial x}\frac{\partial}{\partial x}(e^u(-sin(v))))+e^u cos(v)\frac{\partial^2 u}{\partial x^2}+\frac{\partial u}{\partial x}\frac{\partial}{\partial x}(cos(v)e^u)$

And I got something similar for $\frac{\partial^2\phi}{\partial y^2}$. Since $u,v$ are harmonic, two of the terms will cancel out, but I have no idea how to deal with the rest. Am I actually heading toward the right direction?

Answer 1

hammer and chisel method $$ \frac{\partial}{\partial x}e^u\cos v=e^u\cos v \frac{\partial u}{\partial x}-e^u\sin v \frac{\partial v}{\partial x} $$ so $$ \frac{\partial^2}{\partial x^2}e^u\cos v=e^u\cos v \frac{\partial^2 u}{\partial x^2}-e^u\sin v \frac{\partial^2 v}{\partial x^2} \\ +e^u\cos v (\frac{\partial u}{\partial x})^2-e^u\sin v \frac{\partial v}{\partial x}\frac{\partial u}{\partial x} \\ -e^u\sin v \frac{\partial u}{\partial x}\frac{\partial v}{\partial x}-e^u\cos v (\frac{\partial v}{\partial x})^2 $$ and similarly $$ \frac{\partial^2}{\partial y^2}e^u\cos v=e^u\cos v \frac{\partial^2 u}{\partial y^2}-e^u\sin v \frac{\partial^2 v}{\partial y^2} \\ +e^u\cos v (\frac{\partial u}{\partial y})^2-e^u\sin v \frac{\partial v}{\partial y}\frac{\partial u}{\partial y} \\ -e^u\sin v \frac{\partial u}{\partial y}\frac{\partial v}{\partial y}-e^u\cos v (\frac{\partial v}{\partial y})^2 $$ and all the required cancellations happen when we substitute using the harmonic conditions on $u$ and $v$ and the Cauchy-Riemann equations expressing conjugacy

Answer 2

since $u,v$ are harmonic conjugates, $u+iv$ is an analytic function of $z=x+iy$, so the function $e^{u+iv}$ is also analytic. hence its real part, which is $\phi$, is harmonic

Answer 3

If $u(x, y)$ and $v(x, y)$ are harmonic conjugates, then $u(x, y) + iv(x, y)$ is holomorphic, hence $e^{u + iv}$ is also holomorphic. Thus $\Re(e^{u + iv})$ is harmonic; but $\Re(e^{u + iv}) = \Re(e^u e^{iv}) = \Re(e^u(\cos v + i\sin v)) = e^{u(x, y)}\cos v(x, y)$, showing this latter function is indeed harmonic. QED.

Hope this helps. Cheers,

and as ever,

Fiat Lux!!!