Given $u$ and $v$ two harmonic functions, is it true that $e^{u+v}$ is a harmonic function?.
Using the Laplace equation for proving that $e^{u+v}$ is harmonic, I get: Setting $I=e^{u+v}$ $$I_{xx}+I_{yy}=e^{u+v}\left[(u_{x}+v_{x})^2+(u_{y}+v_{y})^2+u_{xx}+u_{yy}+v_{xx}+v_{yy}\right]$$ It reduces to: $$I_{xx}+I_{yy}=e^{u+v}\left[(u_{x}+v_{x})^2+(u_{y}+v_{y})^2\right]$$ my question is if this expression can be reduced to zero. Thanks.
If $u,v$ are harmonic functions then $u+v$ is a harmonic function too, and we just need to prove (or disprove) that the exponential map sends harmonic functions into harmonic functions. We have: $$ \frac{\partial^2}{\partial x^2} e^{f(x,y)} = e^{f(x,y)}\left[\left(\frac{\partial f}{\partial x}\right)^2+\frac{\partial^2 f}{\partial x^2}\right],\quad \frac{\partial^2}{\partial y^2} e^{f(x,y)} = e^{f(x,y)}\left[\left(\frac{\partial f}{\partial y}\right)^2+\frac{\partial^2 f}{\partial y^2}\right]$$ hence $$ \Delta e^{f(x,y)} = e^{f(x,y)}\left[\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2\right]$$ which equals zero only if $f$ is constant.