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Take $u\in W_0^{1,p}(U)$ and extend it by zero outside $U$. The zero extension of W01 belongs to $W^{1,p}({\mathbb{R}}^n)$ and the Sobolev inequality givesBecause $U$ has finite measure, the Holder inequality givesThusThe reverse estimate follows directly from $\Vert Du{\Vert }_{L^p}\le \Vert u{\Vert }_{W^{1,p}}$. After adjusting constants,This is the Poincaré inequality on $W_0^{1,p}(U)$.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincaré inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

Take a minimizing sequence $(X_k,Y_k)$ in the affine Sobolev class $W$. The standing bounds $C^{-1}\le X_k\le C$ make the energy uniformly equivalent toThe fixed boundary values and the Poincaré inequality therefore bound the sequence in $H^1(\Omega ;{\mathbb{R}}^2)$. By weak compactness in a reflexive Banach space, a subsequence converges weakly to $(X,Y)$. The assumed weak closedness keeps the limit in $W$, and the assumed weak lower semicontinuity givesThus the direct method in the calculus of variations produces a minimizer. Its first variation vanishes in every compactly supported direction, so part (b) makes it a weak solution of the system.

The boundary data are encoded by the Affine Sobolev spaceA function $u\in {\mathcal{A}}_{\phi }$ is a weak solution of the homogeneous p-Laplacian equation when

For existence, take a minimizing sequence for the p-energy on ${\mathcal{A}}_{\phi }$. The Poincaré inequality bounds $u-\phi$ in $W^{1,p}$ by its gradient, so the sequence is bounded in the reflexive Banach space $W^{1,p}$. A weakly convergent subsequence remains in the weakly closed affine space, and convexity of $|\eta {|}^p$ gives weak lower semicontinuity. The direct method in the calculus of variations therefore produces a minimizer, whose first variation is precisely the displayed weak equation.

For smooth periodic fields, the Holder inequality and $H^1\hookrightarrow L^6$ giveSimilarly,Adding the bounds proves that the skew-symmetrized transport form extends continuously from smooth fields to ${\tilde{V}}^3$ andThe periodic Poincaré inequality givesHence

The sobolev embedding theorem in three dimensions and the periodic elliptic estimate for the Stokes operator giveThe last step uses the absence of the zero Fourier mode: on mean-zero periodic fields, the Poincaré inequality makes the homogeneous $H^2$ norm controlled by $\Vert \Delta v{\Vert }_2=|Av|$.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincaré inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

Use on ${\tilde{H}}_m$ the inner product $(w,z{)}_1=(\nabla w,\nabla z)$. The map $F_m$ in the hint is continuous because it is a finite-dimensional polynomial map. Since the velocity $v={\nabla }^{\perp }\Phi$ is divergence-free,The Poincaré inequality bounds $\Vert w{\Vert }_2\le {\mu }_1^{-1/2}\Vert \nabla w{\Vert }_2$. Thus $(F_m(w),w{)}_1<0$ on every $H^1$ sphere whose radius is larger than $\Vert g{\Vert }_2/(\nu \sqrt{{\mu }_1})$. The Brouwer inward-pointing zero lemma supplies $w_{\ast }$ inside that sphere with $F_m(w_{\ast })=0$.

Set ${\omega }_m=w_{\ast }$, solve $-\Delta {\Psi }_m={\omega }_m$ in ${\tilde{H}}_m$, and put $u_m={\nabla }^{\perp }{\Psi }_m$. Expanding $F_m(w_{\ast })=0$ gives exactly the first equation of the Galerkin system, so $({\omega }_m,{\Psi }_m,u_m)$ is a solution.

Use a Lagrange multiplier $\mu$ for the norm constraint. For real variations $u+\delta \eta$ with $\eta (0)=\eta (\pi )=0$, integration by parts givesThus stationarity is equivalent to the Sturm-Liouville problemMultiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the eigenvalues, and the normalized stationary functions are their eigenfunctions. Conversely the eigenvalue equation makes the constrained first variation vanish. The self-adjoint operator has weight one.

For $ϵ=0$, zero and negative eigenvalues yield no nonzero solution with both Dirichlet boundary conditions. Positive eigenvalues have solutions $u=A\sin (\sqrt{\lambda }x)$ after the left condition, and the right condition gives $\sqrt{\lambda }=n$, $n\ge 1$. Hence the lowest eigenpair iswhere the amplitude follows from ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$.

The Rayleigh-Ritz variational principle bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi /2$ givesThus the sine trial bound for a linear potential on an interval isFor $ϵ>0$ the bound is strict, since $-Y_1^{\prime \prime }+ϵx{Y}_1=(1+ϵx)Y_1$ is not a constant multiple of $Y_1$. Also ${\lambda }_{ϵ}\ge 1$ by the Poincaré inequality and the nonnegative potential. Equality in the upper bound holds at $ϵ=0$.

A weak solution belongs to $V=H_0^1(\Omega )$ and satisfieswith the nonlinear right-hand side defining a bounded functional on $V$. Under the stated global Lipschitz hypothesis this requirement is automatic: $|f(s)|\le |f(0)|+ϵ|s|$ and boundedness of $\Omega$ give $f(u)\in L^2$ whenever $u\in V$. Continuity of $f$ alone, without growth control, would not guarantee this for an arbitrary Sobolev function.

For the semilinear Poisson contraction threshold, use the gradient norm on $V$, and let $C_P$ satisfy the Poincaré inequality $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$. For $g\in L^2$, Lax-Milgram gives the zero-Dirichlet Poisson solution $Pg\in V$. Testing with $Pg$ yields $\Vert \nabla Pg{\Vert }_2\le C_P\Vert g{\Vert }_2$. Define $T(v)=P(f(v))$, a self-map of the complete space $V$. For $v,w\in V$,Therefore $ϵC_P^2<1$ is a sufficient condition for Banach's theorem to give a unique fixed point and hence a unique weak solution. There is no need to assume $f(0)=0$ or to restrict to a small ball: the Lipschitz bound makes the map defined on all of $V$. The same estimates giveWith the optimal Poincaré constant, the threshold is $ϵ<{\lambda }_1$ for the First Dirichlet eigenvalue; the stated sufficient condition does not depend on knowing that optimal constant.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.

Give $H_0^1$ the equivalent Hilbert space norm $\Vert u{\Vert }_H=\Vert \nabla u{\Vert }_2$, using the Poincaré inequality. DefineThe Cauchy-Schwarz inequality gives $|a(u,v)|\le (1+C_P\Vert b{\Vert }_{\infty })\Vert u{\Vert }_H\Vert v{\Vert }_H$ and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_H$.

For $\phi \in C_0^{\infty }(\Omega )$, the definition of weak divergence, applied to ${\phi }^2/2$, givesThe left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by test functions and pass to the limit on the left to obtain $\int (b\cdot \nabla u)u\ge 0$. This density argument is essential: local integrability of $\mathrm{div}b$ alone does not justify treating $(\mathrm{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.

Consequently $a(u,u)\ge \Vert u{\Vert }_H^2$, so $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique weak solution with $\boxed{\Vert \nabla u{\Vert }_2\le C_P\Vert f{\Vert }_2}$. The reusable estimate is coercivity from a drift with nonpositive weak divergence.

A weak solution is $u\in H_0^1(\Omega )$ such thatLet $M_f=\Vert f{\Vert }_{\infty }$ and define the Nemytskii operator $N(v)=f\circ v$ on $L^2(\Omega )$. It satisfies $\Vert N(v){\Vert }_2\le M_f|\Omega {|}^{1/2}$. It is continuous in $L^2$: from any sequence $v_j\to v$ in $L^2$, every subsequence has a further subsequence converging almost everywhere; continuity and boundedness of $f$, followed by the dominated convergence theorem, give convergence of the corresponding $N(v_j)$ in $L^2$. This subsequence criterion proves continuity of the whole sequence.

For $h\in L^2$, the Lax-Milgram theorem defines the inverse Dirichlet solution map $Sh\in H_0^1$ by $\int \nabla (Sh)\cdot \nabla \phi =\int h\phi$. Testing with $Sh$ and using the Poincaré inequality yieldsThe Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. Therefore $T=S\circ N$ is a nonlinear compact operator and its image lies in the closed $L^2$ ball of radius $R=1+C_P^2{M}_f|\Omega {|}^{1/2}$. This ball is nonempty, closed, bounded and convex. The Schauder fixed-point theorem gives $u=Tu$. Since $Tu\in H_0^1$, its defining identity is exactly the required weak formulation. Thus a weak solution exists. This is the bounded semilinear Dirichlet existence theorem.

The displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\ge 0$, however,Use nonnegative truncation of a semilinear source: define $F(s)=f(\max \{s,0\})$. This is continuous and bounded on all of $\mathbb{R}$, with $F\ge 1/2$. Part (b) gives a weak solution of $-\Delta u=F(u)$, $u\in H_0^1$.

The weak maximum principle in the needed form says that if $u\in H_0^1$ and $\int \nabla u\cdot \nabla \phi \ge 0$ for every nonnegative $\phi \in H_0^1$, then $u\ge 0$ almost everywhere. Here this also follows directly by testing with the negative part $u_{-}=\max \{-u,0\}\in H_0^1$:The Poincaré inequality gives $u_{-}=0$. Hence $u\ge 0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one almost everywhere. Thus the original equation has a nonnegative weak solution, with bounded right side.

If $\lambda =0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda <0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda =k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive integer $n$. Thus $\boxed{{\lambda }_{\min }={\pi }^2}$.

The admissible trial function $y=\sin \pi x$ has quotientTherefore the minimum, or infimum before its attainment is established, is $\boxed{\le {\pi }^2}$ as requested. The sharp Poincaré inequality supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also ${\pi }^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.

Bounded coefficients and Cauchy-Schwarz give $|B[u,v]|\le C\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Uniform ellipticity gives $B[u,u]\ge \theta \Vert Du{\Vert }_2^2$ because $c\ge 0$. Poincaré inequality $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$ holds on $H_0^1$ of a bounded domain, soA weak solution is $u\in H_0^1$ satisfying $B[u,\varphi ]=\int f\varphi$ for every $\varphi \in H_0^1$. The energy is finite by boundedness of $B$ and the forcing functional, andIt is bounded below and a minimizing sequence is bounded in the Hilbert space $H_0^1$. Its closed linear span is separable, so part 1 supplies a weakly convergent subsequence with limit $u$. Symmetry and nonnegativity of $B$ showThe forcing term converges by weak convergence. Thus $I(u)\le \mathrm{liminf}I(v_j)$, proving attainment. First variation in every zero-trace direction gives the weak equation. If $u,w$ are two solutions, test their difference equation with $u-w$ and use coercivity to get $u=w$. Hence there is exactly one weak solution, and it is the unique energy minimizer. Indeed $I(v)-I(u)=\frac{1}{2}B[v-u,v-u]$ directly verifies the minimizing property.

The weak formulation isThe boundary condition is encoded by $u\in H_0^1$. The Sobolev embedding in dimension three gives $H_0^1\hookrightarrow L^6$, hence also $L^4,L^3,L^2$ on the bounded domain. All energy terms and all the displayed pairings are therefore finite.

Pointwise, $t^4/4-t^3/3\ge t^4/8-C$ for a constant $C$, since the difference is a quartic with positive leading coefficient. The Poincaré inequality and Young inequality give $\int fv\ge -\frac{1}{4}\Vert Dv{\Vert }_2^2-C_f$. ConsequentlyThus the infimum is finite and a minimizing sequence is bounded in $H_0^1$. Extract a weakly convergent subsequence there. The Rellich-Kondrachov theorem states that on a bounded $C^1$ domain in dimension three the embedding $H^1\hookrightarrow L^q$ is compact for $q<6$. A subsequence therefore converges strongly in $L^2,L^3,L^4$ to the same limit $u$. The potential terms and forcing converge, while the gradient norm is weakly lower semicontinuous. The weak limit retains zero trace. Hence $I(u)\le \mathrm{liminf}I(v_j)$ and $u$ attains the absolute minimum.

For any $\varphi \in H_0^1$, the function $t\mapsto I(u+t\varphi )$ has a minimum at zero. Differentiating its polynomial terms gives $\int [Du\cdot D\varphi +(u^3-u^2+f)\varphi ]=0$. These differentiations are justified by the Sobolev integrability already noted. This is the weak formulation, proving every energy minimizer is a weak solution.

At zero the quadratic form isFor $\lambda \ge 0$ it dominates $\int u^{\prime 2}$. For $-{\pi }^2<\lambda <0$ set $\mu =\sqrt{-\lambda }<\pi$ and $z(x)=\cos (\mu (x-1/2))$, positive on the closed interval and satisfying $z^{\prime \prime }+{\mu }^{2}z=0$. The Jacobi identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2$. To obtain a uniform lower bound, let $m=\min z>0$, $K=\max |z^{\prime }|$ and $v=u/z$. Since $\max z=1$ and $\int v^2\le \frac{1}{2}\int v^{\prime 2}$,so $Q_{\lambda }(u)\ge m^2\int u^{\prime 2}/(K^2+2)$.

Finally $I(u)=Q_{\lambda }(u)+O(\int u^4)$ near zero, because ${\sin }^{2}u=u^2+O(u^4)$. The remainder is at most a constant times $\Vert u{\Vert }_{\infty }^2\int u^2\le \frac{1}{2}\Vert u{\Vert }_{\infty }^2\int u^{\prime 2}$. It is dominated by the coercive bound for small perturbations. Thus the second variation is positive and zero is a strict weak local minimizer for every $\lambda >-{\pi }^2$. This argument has not assumed the sharp Poincaré inequality that is to be deduced next.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

For a solenoidal three-component velocity depending on two periodic coordinates and a zero-mean magnetic field with the same dependence, the in-plane flux potential satisfies a homogeneous advection-diffusion equation. The Poincaré inequality gives $⟨A^2⟩\le ⟨A_0^2⟩e^{-2\eta k^{2}t}$. Integrating the vertical stretching term by parts bounds it by $Q\sqrt{⟨A^2⟩⟨|\nabla B_z{|}^2⟩}$, where $Q=\Vert \nabla u_z{\Vert }_{\infty }$. Maximizing against resistive dissipation and integrating gives the displayed transient bound. Splitting off half the dissipation instead gives $E^{\prime }+\eta k^{2}E\le (Q^2/\eta )⟨A_0^2⟩e^{-2\eta k^{2}t}$, proving vertical energy decays. This permits transient generation by a nonzero vertical velocity while excluding sustained fields in this zero-mean, z-independent periodic class.