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The Vinogradov mean value isBy orthogonality of integer Fourier modes, it counts the ordered integer solutions of ${\sum }_{i=1}^k{x}_i^j={\sum }_{i=1}^k{y}_i^j$ for $1\le j\le r$, with every coordinate between one and $\lfloor Z\rfloor$. The diagonal solutions $y_i=x_i$ give $J_{k,r}(Z)\ge \lfloor Z{\rfloor }^k$.

Put $D=r(r+1)/2$. The moment vector $({\sum }_i{x}_i^j{)}_{j=1}^r$ has at most ${\prod }_{j=1}^r(k\lfloor Z{\rfloor }^j+1){\ll }_{k,r}{Z}^D$ possible values. If $R(v)$ counts the tuples with vector $v$, then ${\sum }_{v}R(v)=\lfloor Z{\rfloor }^k$ and $J_{k,r}(Z)={\sum }_{v}R(v{)}^2$. The Cauchy-Schwarz inequality gives $J_{k,r}(Z){\gg }_{k,r}{Z}^{2k-D}$. Combining the two lower bounds, with one common positive constant for large $Z$, yields

Here is an explicit way that upper bounds enter the Vinogradov mean-value method for a bilinear exponential sum. Write $U={\sum }_{x,y\le Z}e({\sum }_j{\alpha }_j{x}^j{y}^j)$ and $L_j=k\lfloor Z{\rfloor }^j$. Two applications of the Holder inequality, followed by grouping equal differences of moment vectors, giveAt integer ${\alpha }_{j}v$ the minimum is defined as $2L_j+1$; $\Vert u\Vert$ means distance to the nearest integer. To explain the mean-value factor, let $r(v)$ count pairs of $k$-tuples with prescribed moment difference. It is an autocorrelation of $R$, so $r(v)\le {\sum }_{w}R(w{)}^2=J_{k,r}(Z)$ by Cauchy-Schwarz inequality. The first Holder inequality groups the $y$ tuples, and the second groups the $x$ tuples, providing the two factors $J$. The remaining sums over moment differences are bounded by the displayed finite geometric series estimates. Good Vinogradov mean value upper bounds, together with rational approximation or spacing bounds for ${\alpha }_j$, therefore give cancellation in $U$. The mean-value estimate alone does not force cancellation for arbitrary coefficients: when all ${\alpha }_j$ are integers, $U=\lfloor Z{\rfloor }^2$. For the paper take $Z=N^{2/5}$ and the specified ${\alpha }_j$.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

By orthogonality of integer Fourier modes, this integral counts pairs of $k$-tuples with the same first $r$ power sums. Diagonal pairs give $J_{k,r}(Z)\gg Z^k$. There are $O_{k,r}(Z^{r(r+1)/2})$ possible moment vectors; the Cauchy-Schwarz inequality gives $J_{k,r}(Z)\gg Z^{2k-r(r+1)/2}$. Upper bounds measure how much arithmetic coincidence remains beyond these necessary contributions, and enter the Vinogradov mean-value method for a bilinear exponential sum.