This problem came up in a PDE where I used separation of variables to formally get a solution. Now I need to know whether that formal solution is sensible.

Let $a_k >0$ be an increasing sequence of real numbers. Let $f_k$ be real numbers.

Let $R \in (0,\infty)$ and $y \in (0,\infty)$. Let $$v(y) = \sum_{k=1}^\infty f_k\left(B_1(k,R)e^{-a_ky} -B_2(k,R)e^{a_ky}\right)$$ where $$B_1(k,R) = \frac{e^{a_kR}}{e^{a_kR}-e^{-a_kR}}\quad\text{and}\quad B_2(k,R) = \frac{e^{-a_kR}}{e^{a_kR}-e^{-a_kR}}.$$ We also know $\sum_{k=1}^\infty f_k < \infty$.

Is $v(y)$ finite for all $y$ and $R$ (i.e. does the infinite sum exist)? Also I would like to know whether the series is uniformly convergent (so whether the partial sums uniformly converge) so that I can do term by term integration.

I don't see why the sum even exists...

Remark: In fact, $a_k^2$ are the eigenvalues of the Neumann Laplacian, and $f_k := (u,\varphi_k)_{L^2}\varphi_k(x)$ on some bounded domain $\Omega$ where $\varphi_k$ are the eigenvectors of the Neumann Laplacian.


For fixed $R > 0$, $B_1(k,R) \sim 1$ as $k$ goes to infinity, while $B_2(k,R) \sim e^{- 2 a_k R}$. The bound on $B_1$ combined with the summability of the $f_k$ takes care of one half of your sum.

The other half has a general term of order $f_k e^{-2a_kR + a_ky}$. If the $f_k$ are only generic summable reals as your post seems to imply, then your sum is convergent only for $y \leq 2R$, divergent otherwise. But hopefully, you are working on some bounded domain such that $y$ never exceeds $2R$, aren't you ?

Also, uniform (in $y$) convergence holds on $[0,2R]$.

  • $\begingroup$ Thanks for this answer. You are right, in fact $y \in (0,R)$, I was mistaken in my OP. One question: if I want to send $R$ off to infinty in the infinite sum denoted by $v(y)$ in my post, can I interchange the limits $\lim_{R}\sum_{1}^\infty (\cdot) = \sum_1^\infty \lim_{R}(\cdot)$? I need uniform convergence wrt. $R$, but I don't know how. $\endgroup$ – RealMax Apr 21 '15 at 11:46
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    $\begingroup$ With $y$ restricted to be $< R$, your general term is now $\lesssim f_k + e^{-a_k R}$, which is definitely more than enough to apply whatever interversion theorem you like. The first half is bounded uniformly of $R$, the second half has its bound improving as $R$ increases. Your situation is much comfortable. EDIT : Actually, your whole sum is now controlled by the $f_k$, independant of $R$. $\endgroup$ – Hachino Apr 21 '15 at 11:51
  • $\begingroup$ Hachino, aren't you assuming that $\sum |f_k| < \infty$ in your comment? The $f_k$ are not necessarily positive so I don't see how you can bound it like that to get the uniform convergence (related math.stackexchange.com/questions/1244956/…). $\endgroup$ – RealMax Apr 21 '15 at 18:47
  • $\begingroup$ Oh sorry, I indeed thought so after reading that $a_k > 0$. Well, fortunately this is not really harmful, since you still easily have uniform convergence on $[\varepsilon, R]$ ; the first half will be controlled by $e^{- \varepsilon a_k}$, the other by $e^{- a_k R_0}$ for $R \geq R_0$. Perhaps you could push the uniformity up to $0$ by carefully decomposing $B_1 = 1 + o(1)$ with an exponential $o(1)$, but that would require a pen and some paper to be sure. $\endgroup$ – Hachino Apr 22 '15 at 7:16
  • $\begingroup$ Actually I think it is easy to show the uniform convergence by splitting the sum into two and using Abel's uniform convergence test (and then recombining the sums). $\endgroup$ – RealMax Apr 22 '15 at 13:52

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