= Solution
Use the <Fourier coefficients>
$$
b_n(y)=\frac2\pi\int_0^\pi u(x,y)\sin(nx)\,dx,\qquad n\geq1.
$$
They are continuous for $0\leq y\leq1$ and bounded by $2\|u\|_\infty$. On compact subintervals of $0<y<1$, the equation is <uniformly elliptic> with smooth coefficients. Local <elliptic regularity> at the flat vertical sides with zero boundary values permits differentiation and <integration by parts> there. Thus
$$
y^2b_n''(y)=-\frac2\pi\int_0^\pi u_{xx}(x,y)\sin(nx)\,dx=n^2b_n(y).
$$
For completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\varepsilon,\pi-\varepsilon)$ against $\sin(nx)$ and a <test function> of $y$ with support away from zero and one. The <interior gradient estimate near a zero Dirichlet side> gives $|u_x|\leq C\varepsilon^{-1}\sup|u|$ on a nearby ball of radius comparable to $\varepsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin(nx)$ and $u\cos(nx)$ tend to zero. Move the $y$ derivatives to the <test function> before taking $\varepsilon\downarrow0$. This proves the <ordinary differential equation>, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.
The <Cauchy-Euler differential equation> has <indicial roots>
$$
p_n^\pm=\frac{1\pm\sqrt{1+4n^2}}2,\qquad b_n(y)=A_n y^{p_n^+}+B_n y^{p_n^-}.
$$
For $n\geq1$ the negative root is strictly negative. Boundedness as $y\downarrow0$ forces $B_n=0$. Continuity at $y=1$ identifies
$$
A_n=\frac2\pi\int_0^\pi u(x,1)\sin(nx)\,dx,\qquad |A_n|\leq2\|u\|_\infty.
$$
Because $p_n^+>n$, the series $\sum_{n\geq1}A_ny^{p_n^+}\sin(nx)$ converges absolutely and uniformly for $0\leq y\leq\rho<1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its <Fourier coefficients> equal those of $u(\cdot,y)$; completeness of the <Fourier sine basis> in $L^2(0,\pi)$ and continuity in $x$ make the two functions equal. Therefore
$$
\boxed{u(x,y)=\sum_{n=1}^\infty A_ny^{(1+\sqrt{1+4n^2})/2}\sin(nx),\qquad (x,y)\in R.}
$$
This is also the requested sum over integers: take all coefficients with $n\leq0$ to be zero. Negative indices give redundant <Fourier modes>, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary <Fourier series> on the top boundary is needed.
An important consequence is a <forced zero trace at a quadratically degenerate boundary>. Indeed the absolute sum is bounded by $2\|u\|_\infty\sum_{n\geq1}y^n=2\|u\|_\infty y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow0$, uniformly in $x$. Thus \b[$u(x,0)=0$] on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.
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