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For $n\ge 3$ and a bounded smooth domain, a Dirichlet Laplacian eigenfunction with eigenvalue $\lambda$ satisfies the displayed estimate. A signed power test for a Laplacian eigenfunction and the zero-boundary Sobolev inequality give a Moser iteration with exponents $p_j=2[n/(n-2){]}^j$. The geometric series ${\sum }_{j}1/p_j=n/4$ gives the exact power of $\lambda$, while ${\sum }_j(\log p_j)/p_j<\infty$ controls the remaining constant. The constant depends only on dimension, because the zero-boundary Sobolev inequality follows by extension by zero to the whole space.

For $Qu=\Delta u-V(u)$, a subsolution and supersolution satisfy $Q{\phi }^{-}\ge 0$ and $Q{\phi }^{+}\le 0$, respectively. An ordered pair additionally satisfies ${\phi }^{-}\le {\phi }^{+}$ throughout the domain, with boundary data $\psi$ between their boundary traces. This interior condition is separate from boundary ordering when $V$ is non-increasing: for $V(s)=-{\lambda }_{1}s$, a positive first Dirichlet Laplacian eigenfunction $e_1$ and $-e_1$ both solve the homogeneous equation and vanish on the boundary, but taking ${\phi }^{-}=e_1$ and ${\phi }^{+}=-e_1$ produces reversed interior barriers.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

Multiply the Dirichlet Laplacian eigenfunction equation by $u$ and use integration by parts. The zero Dirichlet boundary condition removes the boundary term and givesThe denominator is positive because the eigenfunction is nonzero. The numerator is also positive: if it were zero, the given Sobolev inequality for $u\in W_0^{1,2}(\Omega )$ would force $u=0$. Therefore the Dirichlet Laplacian eigenvalue has the positive Rayleigh quotientEvery nonzero Dirichlet eigenfunction in this problem has a positive eigenvalue.

Assume real bounded coefficients, with a principal part defining a uniformly elliptic operator, say $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ for some $\lambda >0$, and assume $c\le 0$. No smoothness of the boundary is needed for this classical weak maximum principle for elliptic operators. Its conclusion isIf $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the Hessian matrix contraction.

For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^{2}u$ is negative semidefinite, and $cu\le 0$, so $Lu\le 0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large thatThus $L(u+\epsilon q)>0$. The preceding maximum argument and continuity on the compact closure giveLet $\epsilon \downarrow 0$; the exponential is bounded on the bounded domain. This is the exponential perturbation proof of the weak maximum principle with drift, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first Dirichlet Laplacian eigenfunction violates the conclusion for $L=\Delta +{\lambda }_1$.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincare inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

As printed, the hypotheses are insufficient. They order the barriers only on the boundary; the intended monotone iteration for a semilinear elliptic equation needs an ordered subsolution and supersolution throughout the domain. On $B_1$, take a positive first Dirichlet Laplacian eigenfunction $e_1$ with $-\Delta e_1={\lambda }_1{e}_1$, and setBoth barriers satisfy $Q{\phi }^{\pm }=0$ and vanish on the boundary, but ${\phi }^{-}>{\phi }^{+}$ inside. No function lies between them. This also disproves the unconditional conclusions in (b) and (d).

With the intended extra hypothesis ${\phi }^{-}\le {\phi }^{+}$ on $\overline{\Omega }$, solvability of the Poisson equation and the Schauder estimate give a unique $u_1\in C^{2,\alpha }$ with forcing $V({\phi }^{-})$ and boundary data $\psi$. Smooth $V$ composed with the $C^{0,\alpha }$ barrier has the required forcing regularity. The barrier inequalities giveBoth differences have nonnegative boundary values. The weak maximum principle for elliptic operators applied to their negatives gives $\boxed{{\phi }^{-}\le u_1\le {\phi }^{+}}$ under the corrected hypothesis.

For a smooth Dirichlet Laplacian eigenfunction, testing $-\Delta u=\lambda u$ with $|u{|}^{\gamma -2}u$ for $\gamma \ge 2$ gives the displayed energy estimate. The Sobolev chain rule for $v=|u{|}^{\gamma /2}$ then yields $\int |Dv{|}^2=\lambda {\gamma }^2/[4(\gamma -1)]\int |u{|}^{\gamma }$. The absolute value and the sign in the test function permit nodal changes. At $\gamma =2$, use the Lipschitz absolute-value chain rule and the fact that a gradient of a Sobolev function vanishes on a level set.