I need to prove that this function is harmonic! [Solved]

Question

I need to prove that $u:\mathbb{R}\times(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow\mathbb{R}$

$$u(x,y)=\sum_{n \ \text{ is odd}}\cos(ny)e^{n(x-n)}$$

is harmonic. I have no idea which theorem or result to use.

Answer 1

Im gonna show that there exists $n_{0}\in\mathbb{N}$ such that for $|x|\leq\delta$ and $\forall \ y$, we have $$\Big|\sum_{k=n_{0}}^{\infty}\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big|<\epsilon$$ for some $\epsilon$.

In fact we have

\begin{eqnarray} \Big|\sum_{k=n_{0}}^{\infty}\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big| &\leq& \sum_{k=n_{0}}^{\infty}\Big|\cos((2k-1)y)e^{(2k-1)(x-(2k-1))}\Big| \nonumber \\ &\leq& \sum_{k=n_{0}}^{\infty}\Big|e^{(2k-1)(x-(2k-1))}\Big| \nonumber \end{eqnarray}

Now take $x=\delta$. It is easy to see that there exists $n_{0}$ such that the sum above is less than $\epsilon$ (because exponential decay too fast). If $x<\delta$, as exponential is increasing, you still have the same thing.

With a little adaptation of the argument above, you can show that the convergence is uniform and then you can use the result posted by Nate Eldredge.

Answer 2

The function has to satisfy the equation $ u_{xx} + u_{yy}=0$.