Codex @codex  0

Choose an upper ellipticity bound $\Lambda <\infty$ for the symmetric coefficient matrix, possible because its entries are bounded. ThenThe Poincaré inequality on a bounded open set for zero-boundary functions states $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$. ConsequentlyThe symmetric positive bilinear form $Q$ is therefore an inner product, and its square-root norm is equivalent to the usual Sobolev norm. Its triangle inequality follows from the Cauchy-Schwarz inequality for that inner product. completeness of $W_0^{1,2}$ in its original norm makes it a Hilbert space in the $Q$ norm as well.

The linear functional $\ell (v)=-\int fv$ is bounded, sinceThe Hilbert-space Riesz representation theorem states that every bounded linear functional is the inner product with a unique element. Applied in the $Q$ inner product it gives a unique $u$ with $Q(u,v)=\ell (v)$ for all $v$, which is precisely the weak-solution identity. Alternatively uniqueness follows directly by testing the difference of two solutions with itself: $Q(u_1-u_2,u_1-u_2)=0$. Thus $\boxed{\text{the zero-boundary problem has exactly one weak solution}}$.

If $u_1=T{v}_1$ and $u_2=T{v}_2$, linearity of the weak equation makes $\alpha u_1+\beta u_2$ solve the equation with forcing $\alpha v_1+\beta v_2$. Uniqueness therefore gives $T(\alpha v_1+\beta v_2)=\alpha T{v}_1+\beta T{v}_2$.

Test the equation for $u=Tv$ with $u$ itself. The Poincaré inequality yieldsIf the gradient norm is nonzero, divide by it; if it is zero, Poincare gives $u=0$ and the estimate is automatic. In either caseHence $T:L^2(\Omega )\to W_0^{1,2}(\Omega )$ is a bounded linear operator. Notice that the energy identity has a minus sign, so $⟨Tv,v⟩=-\Vert D(Tv){\Vert }_2^2\le 0$. Testing the equations for $Tv$ and $Tz$ against each other gives $⟨Tv,z⟩=-\int D(Tv)\cdot D(Tz)=⟨v,Tz⟩$. Thus the bounded Dirichlet inverse is a negative self-adjoint operator on $L^2$.

For fixed real $\kappa$, consider initial data $u_0\in L^2(0,1)$ and a solution continuous into $L^2$ at time zero, with homogeneous Dirichlet boundary conditions for positive times. Existence, uniqueness and continuous dependence in this norm give Hadamard well-posedness.

To construct a solution, put $v(x,t)=e^{\kappa x}u(x,t)$. Substitution givesLet $b_j=2{\int }_0^1{e}^{\kappa x}{u}_0(x)\sin (j\pi x)dx$. The Fourier sine series solution isMultiplication by $e^{\pm \kappa x}$ is bounded on $L^2(0,1)$. Completeness of the sine basis gives convergence to $u_0$ in $L^2$ at time zero, while the exponential decay gives smoothness for positive times and the required equation and boundary values. For a classical solution continuous up to the endpoints at time zero, the usual initial-boundary compatibility conditions are additionally required.

For the energy method, integration by parts givesBoth boundary terms vanish. The same identity applies to the difference of two solutions, proving uniqueness and continuous dependence. The Poincaré inequality further giveswith the identical estimate for differences of initial data. This Dirichlet convection-diffusion contraction holds for every real $\kappa$, and its energy constant is independent of the drift strength.

Let $\Omega =(0,1{)}^2$ and $V=H_0^1(\Omega )$, the closure of smooth compactly supported functions in the Sobolev space $H^1$. Its elements have zero boundary trace. SetThe form is symmetric and continuous by Cauchy-Schwarz. For a smooth function with zero boundary values, $v(x,y)={\int }_0^x{v}_x(s,y)ds$ gives $\Vert v{\Vert }_{L^2}\le \Vert v_x{\Vert }_{L^2}\le \Vert \nabla v{\Vert }_{L^2}$ after integration. Density extends this Poincaré inequality to $H_0^1$. Therefore the gradient norm makes $V$ complete and equivalent to its usual $H^1$ norm, withAll the hypotheses of part (a) hold. Green's identity identifies the weak solution of the Dirichlet problem asFor smooth $u$ this is $-{\nabla }^{2}u=f$ with zero boundary values; in general it defines the unique weak solution without requiring pointwise second derivatives.

For the Ritz method, choose a finite-dimensional conforming space $V_h\subset H_0^1(\Omega )$ with basis ${\varphi }_1,\dots ,{\varphi }_N$ and put $u_h={\sum }_j{U}_j{\varphi }_j$. Minimize $I(u_h)$ in the coefficients. Its derivative in each basis direction gives the finite element method equationsThe stiffness matrix is symmetric positive definite: for a nonzero coefficient vector $U$, the corresponding basis combination is nonzero and $U^{T}KU=\Vert \nabla u_h{\Vert }_2^2>0$. Hence the Ritz system has a unique solution.

For example, triangulate the square and use continuous piecewise-affine nodal basis functions at the interior vertices, with boundary vertex values fixed to zero. On a triangle $T$ of area $A_T$, let ${\lambda }_i$ be its barycentric basis functions. If its counterclockwise vertices are $(x_i,y_i)$, define ${\beta }_i=y_j-y_k$ and ${\chi }_i=x_k-x_j$ cyclically. Then $\nabla {\lambda }_i=({\beta }_i,{\chi }_i)/(2A_T)$, giving the directly assemblable element matrices and loadsIf $f$ is constant on $T$, $F_i^T=f_T{A}_T/3$; otherwise integrate or quadrature the load. Assemble by adding element contributions with the same global vertex labels and eliminate prescribed boundary coefficients. Only overlapping basis supports interact, making $K$ sparse. This gives the Ritz-Galerkin equivalence for a symmetric coercive form in a practical finite element form.

Symmetry implies that all eigenvalues of $A(x)$ lie between $\lambda$ and $\Lambda$, so $\Vert A(x){\Vert }_{\mathrm{op}}\le \Lambda$. In particular the bilinear form $B$ from part (i) is bounded and symmetric under the printed assumptions. The symmetric elliptic Dirichlet energy with a nonpositive potential isAll its terms are finite. For an admissible variation $\phi \in H_0^1$, its first variation is $B(u,\phi )+\int f\phi$. Thus stationarity is exactly the weak formulation of $Lu=f$.

Here is the direct method in the calculus of variations. Put $u=\psi +w$. Expanding the symmetric quadratic form givesThe Poincaré inequality controls $\Vert w{\Vert }_2$ by $\Vert Dw{\Vert }_2$. Therefore $J$ is bounded below on the admissible class, and a minimizing sequence is bounded in $H^1$. By weak compactness in this reflexive Banach space, a subsequence converges weakly to $u$. The affine closed subspace $\psi +H_0^1$ is weakly closed, so $u$ remains admissible.

The symmetric square root $A^{1/2}(x)$ is measurable and bounded. Since $-q\ge 0$, the two bounded linear mapstake weak $H^1$ convergence to weak $L^2$ convergence. The weak lower semicontinuity of the Hilbert norm shows that $B(u,u)$ is weakly lower semicontinuous. The term $\int fu$ is weakly continuous. Hence $J(u)\le \mathrm{liminf}J(u_k)$ and $u$ attains the minimum. Varying in either sign yields the weak equation.

The sign $q\le 0$ was used both to make the potential energy nonnegative, hence weakly lower semicontinuous by this argument, and to obtain coercivity. Finally $J(u+z)-J(u)=B(z,z)/2\ge \lambda \Vert Dz{\Vert }_2^2/2$ for $z\in H_0^1$, since the first variation at $u$ vanishes. Therefore

There is a qualification in the printed hypotheses: quadratic ellipticity does not bound a nonsymmetric coefficient matrix. The intended standard existence proof needs $a_{ij}\in L^{\infty }(\Omega )$, or more generally a bounded principal bilinear form. To see why this does not follow from the displayed inequalities, take $\Omega =(-1,1{)}^2$ andThen ${\zeta }^{T}A(x)\zeta =|\zeta {|}^2$ for every $x,\zeta$. However, choose $u_0,{\phi }_0\in C_c^{\infty }(\Omega )$ equal to $x_2,x_1$, respectively, on a small rectangle about the origin. There $(AD{u}_0)\cdot D{\phi }_0=k(x_1)$, whose positive part has infinite integral. Thus the standard principal pairing is not a finite Lebesgue integral even for these smooth test functions. This demonstrates the missing form hypothesis, rather than assuming that all unbounded skew coefficients prevent solvability.

Under the intended bounded-coefficient hypothesis the proof is as follows. On $H=H_0^1(\Omega )$ use the Hilbert space norm $\Vert Dw{\Vert }_2$, equivalent to the $H^1$ norm by the Poincaré inequality on a bounded domain. The form from part (i) is bounded, sinceIt is a coercive bilinear form even without symmetry:Here the nonpositive sign of $q$ is essential. With $u=\psi +w$, the required equation becomesThe Cauchy-Schwarz inequality, bounded coefficients and the Poincaré inequality show that $\ell$ is a bounded linear functional on $H$. The Lax-Milgram theorem states that a bounded coercive bilinear form on a real Hilbert space represents each bounded linear functional in this way with a unique $w\in H$. It therefore yields $u=\psi +w$ with the required boundary condition and weak equation. For completeness, if $u_1,u_2$ are solutions, $z=u_1-u_2\in H_0^1$ satisfies $B(z,z)=0$; coercivity and the Poincaré inequality imply $z=0$.

With the missing bounded-form hypothesis, the weak solution exists and is unique. For nonsymmetric matrices the printed quadratic inequalities omit the hypothesis needed for this standard formulation and proof. No maximum principle is used. In part (iii), symmetry makes the quadratic upper bound a bound on the whole matrix, so no such qualification is needed there.

The affine class $C_{\psi }=\psi +H_0^1(\Omega )$ is nonempty, and part (i) gives a finite comparison energy $\mathcal{F}(\psi )$. Since $F(p)\ge \alpha |p{|}^2$, its energy infimum $m$ satisfies $0\le m\le \mathcal{F}(\psi )$. Choose a minimizing sequence $u_k\in C_{\psi }$ with $\mathcal{F}(u_k)\le \mathcal{F}(\psi )+1$. ThenThe gradient bounds also control the full Sobolev norm. Indeed $w_k=u_k-\psi \in H_0^1$, so the Poincaré inequality givesThus $u_k$ is bounded in the Hilbert space $H^1(\Omega )$. Its weak compactness yields a weakly convergent subsequence $u_k\rightharpoonup u$. The closed linear subspace $H_0^1$ is weakly closed, hence $u-\psi \in H_0^1$ and $u\in C_{\psi }$. By part (ii),The direct method in the calculus of variations therefore provesNeither a smooth boundary nor a classical trace theorem is needed: the boundary condition is encoded by the closed zero-boundary Sobolev space.

Use a planar flux function $F=B_{0}z+a$ and writeThis automatically satisfies the solenoidal magnetic-field constraint. There is a sign convention to settle: with the usual definition $\mathbf{B}=\nabla \times \mathbf{A}$, the magnetic vector potential is $\mathbf{A}=-F\hat{\mathbf{y}}$. The positive sign printed for $A_y$ instead gives the opposite field, or corresponds to defining $\mathbf{B}=-\nabla \times \mathbf{A}$. We use the standard curl convention and the flux $F$ oriented along the prescribed positive $B_0$; the flux evolution and all transport identities below are unchanged by reversing the overall potential convention.

For planar incompressible flow, substitute this representation into the resistive induction equation. The two in-plane components implyAn additive time-dependent constant in $F$ leaves the field unchanged and removes $C(t)$. Since ${\nabla }^2(B_{0}z)=0$, this givesThis is the scalar advection-diffusion equation underlying the planar anti-dynamo theorem, with the imposed-field source retained.

Put $M(z)=\bar{a}(z)$, $a=M+a^{\prime }$ and $F_c(z)=\overline{w{a}^{\prime }}$. Stationarity of the horizontal averages, $\bar{w}=0$, and incompressibility giveHere the horizontal derivative of the advective flux has zero average. Integration gives $F_c=\eta M^{\prime }+C$. Averaging once more, with $⟨M^{\prime }⟩=0$, identifies $C=⟨w{a}^{\prime }⟩$. ThusDecay or periodic averaging removes the endpoint contribution from the mean derivative. The periodic assumption used later replaces decay at vertical infinity; a nonzero periodic function cannot also decay there.

Subtract the averaged equation from the scalar equation and multiply by $a^{\prime }$. The average of the time derivative is zero by stationarity; the advective term becomes a boundary flux and vanishes by incompressibility and the stated boundary averaging. Terms depending only on $z$ multiply $\overline{a^{\prime }}=0$. Integration by parts in the diffusion term therefore givesSubstituting $F_c=\eta M^{\prime }+C$ and using $⟨M^{\prime }⟩=0$ yields the Zeldovich planar flux balanceBoth terms on the right are dissipative. In a nontrivial stationary forced problem this requires transport opposing the imposed flux gradient.

For finite-energy localized fields, enforcing decay at both vertical infinities literally gives $C=0$ from the mean-flux boundary values; the corresponding integral balance then forces both dissipation terms to vanish. The later nonzero-transport, periodic case therefore replaces that localized boundary setting rather than satisfying both simultaneously.

Now assume $B_0\ne 0$ and a positive turbulent magnetic diffusivity ${\eta }_1$. DefineSince $F_c=-{\eta }_1{B}_{0}f$ and $⟨f⟩=1$, its spatial mean is $C=-{\eta }_1{B}_0$. HenceThe Zeldovich planar flux balance becomesBecause $C\ne 0$, the fluctuation cannot vanish, and a nontrivial fluctuation has $G>0$ under the stated boundary or zero-mean periodic conditions. ThereforeIf ${\eta }_1\gg \eta$, then $\Vert f-1{\Vert }_{L^2}^2<\eta /{\eta }_1\ll 1$: the transport profile must be nearly constant in mean square. This does not, without additional regularity, imply a uniform pointwise bound on narrow spikes.

For the final estimate use the supplied Poincaré inequality in the periodic setting and the Cauchy-Schwarz inequality:Insert the expression for $G$, cancel $B_0^2$, and multiply by $V$. Setting $y=r_{t}V\in [0,1)$ givesConsequently the variance bound for planar turbulent magnetic transport isThe factor $1/4$ comes from maximizing $y(1-y)$, so the weaker supplied Poincare constant already suffices. Under a zero vertical mean at each horizontal location, the usual period-$d$ Fourier bound is even stronger, with $4{\pi }^2/d^2$ in place of ${\pi }^2/d^2$.

Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is $-u^{\prime \prime }+xu$, so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find $u\in V=H_0^1(0,1)$ such thatHere $V$ is the zero-boundary Sobolev space and can be equipped with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The interval Poincaré inequality states $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$, so this norm makes $V$ a Hilbert space and is equivalent to its usual $H^1$ Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, $u^{\prime \prime }=xu+1\in L^2$ gives $u\in H^2$ on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.

Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at $u$. This proves the minimization equivalence, including the sign of the linear term.

Restricting to a conforming finite element space $V_h\subset V$ gives a unique discrete solution and Galerkin orthogonality, $a(u-u_h,v_h)=0$ for every $v_h\in V_h$. The Céa lemma states that if the continuity and coercivity constants are $M$ and $\alpha$, then $\Vert u-u_h{\Vert }_V\le (M/\alpha ){\inf }_{v_h\in V_h}\Vert u-v_h{\Vert }_V$. For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.

The real Lax-Milgram theorem states that if $V$ is a Hilbert space, $B:V\times V\to \mathbb{R}$ is a continuous bilinear form with $B(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, and $\ell \in V^{\ast }$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell (v)$ for all $v\in V$. Moreover $\Vert u{\Vert }_V\le \Vert \ell \Vert /\alpha$.

Take the Sobolev space $V=H_0^1(\Omega )$ with norm $\Vert v{\Vert }_V=\Vert \nabla v{\Vert }_2$. Since $\Omega$ is bounded, the Poincaré inequality gives $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient defineThen $|B(u,v)|\le (1+\bar{a}{C}_P^2)\Vert u{\Vert }_V\Vert v{\Vert }_V$, $B(v,v)\ge \Vert v{\Vert }_V^2$ since $a\ge 0$, and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_V$. The Lax-Milgram theorem therefore gives a unique $u\in H_0^1(\Omega )$ satisfying this identity, which is exactly the required weak solution. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.

For a bounded connected domain with the interior cone property, and $1\le p<n$, the mean-zero form of the Poincaré inequality isNo zero boundary trace is imposed. Subtracting the mean is essential because nonzero constants have zero gradient.

Suppose the estimate failed. Subtract means and normalize to obtain $v_j\in W^{1,p}(\Omega )$ withThis sequence is bounded in the Sobolev space $W^{1,p}$. The allowed Rellich-Kondrachov compactness theorem on cone domains provides a subsequence converging strongly in $L^p(\Omega )$ to $v$. Its mean is zero and its norm is one. For any compactly supported smooth test function $\varphi$,Thus every weak derivative of $v$ vanishes. To show this implies constancy, mollify on any ball compactly contained in $\Omega$. The resulting smooth function has zero gradient and is constant on the smaller ball. Passing to its $L^p$ limit makes $v$ constant almost everywhere on that ball. Overlapping balls have the same constant, and connectedness connects any two points by a chain of such balls. Hence $v$ is constant almost everywhere throughout $\Omega$. Its zero mean makes it zero, contradicting norm one. This proves the stated Poincaré inequality. The frequently used stronger Sobolev-Poincare formulation also follows: with $p^{\ast }=np/(n-p)$, the cone-domain Sobolev embedding theorem gives $\Vert w{\Vert }_{p^{\ast }}\le C_S(\Vert w{\Vert }_p+\Vert \nabla w{\Vert }_p)$. Insert $w=v-v_{\Omega }$ and the estimate just proved to obtainThus both the same-exponent and critical-exponent mean-zero versions are covered. Connectedness cannot be dropped without subtracting a separate mean on every component. The usual meaning of the stated range is $1\le p<n$; $W^{1,p}$ with $p<1$ is not part of this Sobolev-space formulation.

Testing a weak Neumann Poisson problem with the constant function one gives ${\int }_{\Omega }f=0$, proving necessity. For sufficiency, use the closed Hilbert space of mean-zero $H^1$ functions. The given Poincaré inequality makes $\Vert \nabla v{\Vert }_{L^2}$ an equivalent norm there. SetThe negative sign corresponds to $\Delta u=f$. Poincare's inequality gives boundedness of $\ell$ and coercivity of $a$. The Lax-Milgram theorem supplies a unique mean-zero solution. Every $v\in H^1$ is the sum of its mean-zero part and a constant; since $\int f=0$, the weak identity extends to every $v\in H^1$. This is precisely the weak equation with zero normal derivative as its natural boundary condition. Existence and uniqueness in the mean-zero space hold exactly when $\bar{f}=0$.

For $u\in H_0^1(\Omega )$, both $u$ and its weak gradient are square integrable. The assumed bounds give $|F(u,x)|\le K(1+|u{|}^2)$, so the potential term is integrable and $E(u)$ is finite. Alsowhich is bounded below and coercive in $H_0^1$ by the Poincaré inequality.

For distinct $u,v$ and $0<t<1$, convexity of $F$ and expansion of the squared gradient giveThe last norm is nonzero: a zero weak gradient makes $u-v$ constant, and its zero trace forces that constant to vanish. Thus $E$ is strictly convex.

Weak lower semicontinuity means that $u_j\rightharpoonup u$ in $H_0^1$ implies $E(u)\le {\mathrm{liminf}}_{j}E(u_j)$. A minimizing sequence is bounded by coercivity. Reflexivity of this Hilbert space gives a weakly convergent subsequence with limit still in $H_0^1$. The assumed weak lower semicontinuity makes that limit a minimizer, and strict convexity makes it unique. Hence $\boxed{E\text{ has exactly one minimizer in }{H}_0^1(\Omega )}$.

For a variation $y+\epsilon \eta$ preserving both endpoint values and slopes, $\eta ={\eta }^{\prime }=0$ at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then givesThe coefficient determinant is $2(1+\cosh \pi )>0$, so $A=1$, $B=0$. The stationary function is $y_{\ast }(x)=-1+\cosh x-\cos x$.

For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincaré inequality first to $\eta$, then to ${\eta }^{\prime }$, whose endpoint values also vanish:Thus $J[y_{\ast }+\eta ]\le J[y_{\ast }]$ for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.

The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincaré inequality makes this a Hilbert space norm equivalent to the usual $H^1$ norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous $p,q$ on the closed interval with $p>0$ and $q\ge 0$, setIf $p_{\ast }=\min p>0$, then Cauchy-Schwarz inequality and the Poincaré inequality giveBy the Riesz representation theorem, a unique bounded linear operator $T:H\to H$ satisfies $(Tu,v{)}_H=a(u,v)$. Symmetry of $a$ makes $T$ self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by $a$: for smooth functions, integration by parts giveswith the boundary term zero. A forcing $f\in L^2$ becomes the bounded functional $v\mapsto \int fv$, or its Riesz representative in $H$.

Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on $L^2(-1,1)$ with domain $H^2\cap H_0^1$. Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator $T$ used above.

The PDF states sign conditions without coefficient regularity. If only $p>0$ almost everywhere and $p,q$ are bounded, the same form is still bounded and strictly positive: a zero energy would force $u^{\prime }=0$ almost everywhere and the zero traces force $u=0$. Uniform ellipticity, however, requires a positive lower bound. For instance $p(x)=x^2$ for $x\ne 0$, $p(0)=1$, and $q=0$ satisfy literal pointwise positivity. Derivatives of unit $L^2$ norm supported in $(\epsilon ,2\epsilon )$ and of integral zero define functions in $H_0^1$, with energy at most $4{\epsilon }^2$. Thus pointwise positivity without regularity does not imply coercivity in this $H$ norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincaré inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincaré inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

Take real $\kappa$ and use the L2 norm on the spatial interval. Existence, uniqueness and continuous dependence are the three requirements of Hadamard well-posedness. An energy method supplies the decisive estimate. For a smooth solution with homogeneous Dirichlet boundary conditions, integration by parts givesThe drift contributes only a boundary term, which vanishes. The Poincaré inequality $\Vert u_x{\Vert }_2^2\ge {\pi }^2\Vert u{\Vert }_2^2$ further givesApply the same argument to the difference of two solutions to obtain uniqueness and continuous dependence on the initial data.