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For $u_{tt}=\Delta u+\alpha u$ with a homogeneous Dirichlet boundary condition, let ${\lambda }_1>0$ be the first Dirichlet Laplacian eigenvalue. On a bounded one-dimensional interval, finite-energy displacement remains bounded for all time and all initial data exactly when $\alpha <{\lambda }_1$. The conserved energy $\frac{1}{2}(\Vert u_t{\Vert }_2^2+\Vert \nabla u{\Vert }_2^2-\alpha \Vert u{\Vert }_2^2)$ controls the gradient and, by the Poincaré inequality and one-dimensional integration, the maximum norm. At equality a mode can grow linearly; above it a mode can grow exponentially. Bounded oscillations need not have a long-time limit.

At zero growth rate put $P=\psi +iq\theta$, $q=\sqrt{R}$. The coupled Darcy equations become $\Delta P+iq{P}_x=0$. Substituting $P=e^{-iq(x-1/2)/2}F$ gives $\Delta F+(q^2/4)F=0$ with homogeneous Dirichlet boundary conditions. The square Dirichlet Laplacian eigenvalues give $q^2=4{\pi }^2(m^2+n^2)$, $m,n\ge 1$, whose minimum is $8{\pi }^2$. For $S=\sin \pi x\sin \pi z$, $a=q_c/2$, the complex multiples $F=S,iS$ give real pairs $(S\cos (a\xi ),-S\sin (a\xi )/q_c)$ and $(S\sin (a\xi ),S\cos (a\xi )/q_c)$, $\xi =x-1/2$. Their reflection parities are opposite, yielding two independent physical modes.

A scalar $\lambda$ for which a nonzero Laplacian eigenfunction satisfies this equation, with specified boundary conditions or on a closed manifold, is an eigenvalue of the positive Laplacian. A Dirichlet Laplacian eigenvalue imposes zero boundary values; other boundary conditions generally change the spectrum.

Identical oriented Dirichlet boundary data imply a rotation-invariant solution whenever the Dirichlet problem is unique. For $\lambda \ge 0$, uniqueness follows directly: the difference $h$ of two solutions has zero boundary trace, and integration by parts gives ${\int }_D|\nabla h{|}^2+4\lambda {\int }_D|h{|}^2=0$. Thus $h=0$. The same conclusion holds for a nonresonant negative parameter by invertibility of the Dirichlet Laplacian shift. Then $q(az)=q(z)$, all three normal traces equal a common $q_N(s)$, and ${\Psi }_j=\Psi$.

SetDivide the global relation by $E(-iak)$. Using $1+a+\bar{a}=0$ yields $E(-ik)/E(-iak)=e(\bar{a}k)$ and $E(-i\bar{a}k)/E(-iak)=e(-k)$. Hencewhere the requested known function isFor an explicitly derivative-free form, put $H(k)={\int }_{-\ell /2}^{\ell /2}{e}^{(k+\lambda /k)s}f(s)ds$ and $f_{\ast }=f(-\ell /2)=f(\ell /2)$. An integration by parts givesThe uniqueness qualification cannot be removed for every real $\lambda$: at a Dirichlet Laplacian eigenvalue a homogeneous addition need not preserve rotation invariance. The scalar reduced relation then applies to the rotation-averaged solution, but not necessarily to the originally chosen solution. Part (d) treats this genuine exceptional case.

Use the Dirichlet Laplacian eigenvalues and eigenfunctions on $(-1,1)$:An expansion $u={\sum }_j{q}_j(t)e_j$ reduces the wave equation to independent ordinary differential equationsWhen $\alpha <{\lambda }_1$, put ${\omega }_j=\sqrt{{\lambda }_j-\alpha }$; each coefficient isFor finite-energy data $\varphi \in H_0^1(-1,1)$ and $\psi \in L^2(-1,1)$, a uniform spatial bound follows from the energy method, rather than from unjustified absolute summation of the Fourier series. The conserved quantity isThe Poincaré inequality gives $\Vert u{\Vert }_2^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_2^2$. Hence $2E\ge c_{\alpha }\Vert u_x{\Vert }_2^2$ with $c_{\alpha }=1-\alpha /{\lambda }_1>0$ for $0\le \alpha <{\lambda }_1$, and with $c_{\alpha }=1$ for $\alpha <0$. Since the Dirichlet boundary condition gives $u(x,t)={\int }_{-1}^x{u}_x(s,t)ds$, the Cauchy-Schwarz inequality yieldsAt $\alpha ={\lambda }_1$, the first coefficient is $q_1(t)={\varphi }_1+t{\psi }_1$, which is unbounded for some admissible data. For $\alpha >{\lambda }_1$, that coefficient has an exponentially growing component for some admissible data. Thus all-time boundedness of a wave equation with a reaction term requiresAt equality, boundedness for a particular data set requires ${\psi }_1=0$; the remaining modes are bounded by their spectral gap. Above the threshold, every unstable mode must have its growing component canceled, and any zero-frequency mode must have zero initial velocity. There is no condition on $\alpha$ alone for arbitrary specially chosen data.

The printed reference to a limit needs qualification. Bounded oscillations generally have no limit as $t\to \infty$. If actual existence of that limit for every admissible initial datum is required, no real $\alpha$ works: for every $\alpha$, choose $j$ with ${\lambda }_j>\alpha$ and nonzero pure oscillatory data in that eigenfunction. The boxed inequality answers the intended long-time boundedness question.

There is a genuine conflict in the printed coefficient assumptions: no uniformly positive coefficient in a zero-boundary Sobolev space exists. A function $a\in H_0^1(\Omega )$ cannot also obey $a\ge a_{-}>0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min (\max (s,0),a_{-})$ has $g(0)=0$, so Lipschitz truncation preserves zero-boundary Sobolev spaces, giving $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$. Its zero gradient contradicts the Poincaré inequality in the zero-boundary Sobolev space. Thus the literal coefficient class is empty.

For the meaningful uniformly elliptic problem, take $a\in L^{\infty }(\Omega )$ with $0<a_{-}\le a\le a_{+}$, and impose the Dirichlet boundary condition on the unknown and test functions: $V=H_0^1((0,1{)}^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The divergence-form elliptic operator is $L=-\mathrm{div}(a\nabla )$. Integration by parts defines the symmetric bounded bilinear formIn particular,Here $2{\pi }^2$ is the first Dirichlet Laplacian eigenvalue on the unit square; the Poincaré inequality follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the gradient norm on $V$, and the Dirichlet realization of $L$ is a positive definite symmetric operator. On its operator domain, $⟨Lv,v⟩=B(v,v)$.

The required functional and weak equation areBecause $f\in L^2$, the right side is a bounded linear functional by the Cauchy-Schwarz inequality and Poincaré inequality. The Lax-Milgram theorem gives a unique weak solution, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.

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.

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 Poincaré 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.

There is a sign issue in the printed problem. Part (c) constructs the inverse of $\Delta -1$, whereas the second equation printed in part (e) contains $\Delta +1$. The latter operator is invertible with homogeneous Dirichlet data only when $1$ is not a Dirichlet Laplacian eigenvalue. Thus the assertion as printed needs this nonresonance hypothesis; with a minus sign it follows directly from parts (c) and (d).

Under either the intended minus sign or the stated nonresonance condition, let $G_0$ and $G_1$ be the bounded Dirichlet solution operators for the two linear equations. Choose $s>n/2$ and work with $u,v\in H^{s+2}(\Omega )\cap H_0^1(\Omega )$. Since $H^s(\Omega )$ is a Sobolev algebra,DefineElliptic regularity gives, on a ball of radius $R$,and the difference estimate has Lipschitz constant at most $CR$. Choose $R$ small and then ${\epsilon }_0$ so that $C(R^2+{\epsilon }_0\Vert f{\Vert }_{H^s})\le R$. The contraction mapping theorem gives a solution for $0\le \epsilon <{\epsilon }_0$. Repeated elliptic regularity and smoothness of $f$ bootstrap the solution to $C^{\infty }$.