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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.

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The problem is to prove that the sum of the partial derivates vanishes. Because we have an infinity sum of functions. I think that need some thoerem about convergence before anything. –  José Carlos Oct 1 '12 at 23:48
    
Well, for any $x$; $n>x$ at some point, so that $e^{n(x-n)}$ will go to zero pretty fast, so don't worry about that. –  Pedro Tamaroff Oct 1 '12 at 23:55
    
How about this theorem: a uniform limit of harmonic functions is harmonic. –  Nate Eldredge Oct 1 '12 at 23:59
    
Setting $u_j=\sum_{n \ is \ odd}^j\cos(ny)e^{n(x-n)}$, for each $j$ the term of the sequence $(u_j)_j$ is harmonic. Why this sequence has uniform convergence? Proving it, we use the theorem that nate comented and we have the result. –  José Carlos Oct 2 '12 at 2:59

2 Answers 2

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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.

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The OP says "I have no idea which theorem or result to use." Maybe you can add what the definition of harmonic function is, so as to help a little more. –  Pedro Tamaroff Oct 1 '12 at 23:43
    
You are right, Benghorbal did it. –  Tomás Oct 1 '12 at 23:56

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

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