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Suppose the inhomogeneous linear wave equation has an energy estimate in a Banach solution space $X_T$, and its nonlinearity obeystogether with the corresponding locally Lipschitz estimate. For a sufficiently large closed ball and sufficiently small $T$, the map that solves the linear equation with source $F(w)$ is a contraction. The contraction mapping theorem then gives a local weak solution.

On a smooth bounded connected domain, the homogeneous Neumann problem seeks $u\in H^1(U)$ such that $\int \nabla u\cdot \nabla v=\int fv$ for every $v\in H^1(U)$. For $f\in L^2(U)$ it is solvable exactly when $\int f=0$, and the weak solution is unique up to constants. The Poincare-Wirtinger inequality and the Riesz representation theorem give existence on the mean-zero Sobolev space. The test function one gives necessity of the compatibility condition.

Let $z$ measure height above the substrate, and write $\mu =\rho \nu$. In the lubrication approximation, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:The no-slip boundary condition gives $u(0)=0$; negligible air viscosity and constant surface tension give $u_z(h)=0$. Integrating twice and then through the depth yieldsThe free-surface kinematic boundary condition and incompressibility imply $h_t+q_x=0$. ThereforeThe cross-sectional area $A=\int hdx$ is conserved when the endpoint volume flux vanishes.

The relevant small groups follow from the same balances. Put ${ϵ}_h=H_0/\ell$ and $U_{\ast }=g{H}_0^3/(3\nu \ell )$. Small ${ϵ}_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O({ϵ}_h^2)$. The ratio of horizontal inertia to vertical viscous stress isThus a small full-length Reynolds number is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_{\ast })$ must likewise be small; on the spreading timescale $t_{\ast }=\ell /U_{\ast }$ it equals the displayed group. If surface tension is retained, its pressure-gradient correction relative to gravity isThe Bond number here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/{\ell }^2)$ whereas the hydrostatic pressure is $O(\rho g{H}_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant kinematic viscosity.

To find the similarity solution, put $K=g/(3\nu )$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the time derivative against the nonlinear diffusion gives ${\ell }^5\sim K{A}^{3}t$. A normalized ansatz isSubstitution gives $(f^3{f}^{\prime }{)}^{\prime }=-(f+\xi f^{\prime })/5$. For the symmetric zero-flux profile, one integration gives $f^3{f}^{\prime }=-\xi f/5$, and on its positive support a second integration givesThus the solution is a compactly supported Barenblatt solution of the porous medium equation. Let $\ell (t)$ denote its half-width. In dimensional variables,Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell \sin \theta$,Writing the remaining integral as $I$, the final widths areThe height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The volume flux tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported weak solution of the reduced equation. The diverging edge slope means the lubrication approximation is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.

Let $V$ be a real Hilbert space, $a:V\times V\to \mathbb{R}$ a bounded bilinear form, and $\ell \in V^{\prime }$ a bounded linear functional. Assume constants $C<\infty$, $\alpha >0$ satisfyThe second hypothesis says that $a$ is a coercive bilinear form. The Lax-Milgram theorem then supplies a unique weak solution $u\in V$ with $a(u,v)=\ell (v)$ for every $v\in V$, and $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$.

For a finite-dimensional trial linear subspace $V_h\subset V$, the same boundedness and coercivity constants apply to the restricted form. Since a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. Thus there is a unique Galerkin method solution $u_h\in V_h$ satisfying $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. The Galerkin orthogonality $a(u-u_h,v_h)=0$ gives the Céa lemma:In particular, approximation by the trial spaces implies numerical convergence. For a complex Hilbert space, replace bilinearity by a sesquilinear form, take $\ell$ antilinear in the test variable, and require $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_V^2$.

Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$ makes this a complete Hilbert space norm. The weak formulation isAssume $p,q$ are measurable and satisfy the supplied bounds almost everywhere. The Cauchy-Schwarz inequality and Poincaré inequality giveBecause $p\ge p_0>0$ and $q\ge 0$,This is the coercive variable-coefficient Dirichlet form. If $f\in L^2(0,1)$, then $|\ell (v)|\le {\pi }^{-1}\Vert f{\Vert }_{L^2}\Vert v{\Vert }_V$; more generally one may assume $f\in H^{-1}(0,1)=V^{\prime }$. Some such source regularity is necessary to invoke the theorem. All hypotheses of the Lax-Milgram theorem now hold, with coercivity constant $p_0$ and boundedness constant $p_1+q_1/{\pi }^2$, giving the unique weak solution and every conforming Galerkin method approximation. This argument does not require a pointwise derivative of $p$.

A fundamental solution of a linear differential operator $P$ is a distribution $G$ satisfying $PG={\delta }_0$. The locally integrable function $G=e^{-|x|}/2$ defines a distribution by integration against compactly supported smooth test functions. Away from zero it satisfies $-G^{\prime \prime }+G=0$. It is continuous at zero, while $G^{\prime }(0+)=-1/2$ and $G^{\prime }(0-)=1/2$. Consequently its distributional second derivative is its ordinary derivative on the two sides plus the derivative jump times ${\delta }_0$:This derivative-jump formula follows by integration by parts on each half-line; the boundary terms equal $-\varphi (0)$ in $G^{\prime \prime }$.

Convolution with the indicator forcing givesThe two values and their first derivatives agree at $\pm 1$, so no delta terms arise there. This is a decaying weak solution, with the equation holding classically away from the forcing jumps.

For real Schwartz $\varphi$, expand the energy:The mixed term is zero by the weak equation $-u_0^{\prime \prime }+u_0=V$, or integration by parts using the continuous first derivative and decay. The remaining integral is strictly positive whenever $\varphi$ is not identically zero. HenceThe real-valued function space is implicit in the stated ordered quadratic functional; for complex functions one instead uses absolute squares and the corresponding real part of the forcing term.

Take $L=-d^2/d{x}^2+x$. The Airy equation is $Lu=0$. To discuss positivity of this linear operator, use its homogeneous variation domain $D(L)=\{v\in H^2(0,1):v(0)=0,v^{\prime }(1)=0\}$. Integration by parts givesThe boundary term is zero at both ends. The same calculation with two different functions proves that the associated form is a symmetric bilinear form. Positivity here concerns the homogeneous domain; the condition $u(0)=1$ instead defines an affine set of admissible solutions.

For the variational problem defineand minimizeThe Sobolev trace at zero is meaningful in $H^1$. No derivative boundary value is imposed on this trial space: the Neumann boundary condition will emerge naturally. For $v\in V$, Cauchy-Schwarz inequality and integration giveso $a(v,v)\ge \frac{2}{3}\Vert v{\Vert }_{H^1}^2$. Thus $a$ is a bounded, symmetric, coercive bilinear form on the Hilbert space $V$. The Lax-Milgram theorem says that a bounded coercive bilinear form and a bounded linear functional determine a unique weak solution. Apply it toand set $u=1+v$. This is precisely the first variation condition $a(u,w)=0$ for $J$.

Compactly supported test functions first yield $u^{\prime \prime }=xu$ in the sense of distributional derivatives. Since $xu\in L^2$, the solution is in $H^2$. Integration by parts then leaves $u^{\prime }(1)w(1)=0$ for every $w\in V$, hence $u^{\prime }(1)=0$. The other boundary condition is built into $1+V$. Conversely a solution of this boundary value problem satisfies the first variation condition. Finally,This proves existence, uniqueness and the global minimum characterization, including the natural boundary condition. It is the mixed-boundary Airy energy principle.

For the similarity solution, differentiation givesChoose ${\lambda }^2\dot{\lambda }=e^{-\mu t}$. The remaining equation is $-(\eta f{)}^{\prime }=(f{f}^{\prime }{)}^{\prime }$. Its integration constant is zero for the required decaying or compactly supported profile, so $f(f^{\prime }+\eta )=0$. Where $f>0$, this gives $f^{\prime }=-\eta$. For a nonnegative population the complete profile is thereforeNormalization gives $Q=2b^3/3$, hence $b=(3Q/2{)}^{1/3}$. Integrating the scale equation with zero initial scale yieldsThusIts support satisfiesThe profile is a weak solution at its moving edges, where $n{n}_x$ vanishes. As $t\downarrow 0$ its support shrinks to the origin and its mass tends to $Q$, so its initial distribution is $Q{\delta }_0$. The bounded limiting range assumes $\mu >0$; at zero death rate the scale instead grows as $(3t{)}^{1/3}$.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

The space $W_0^{1,2}(\Omega )$ is the closure of $C_c^{\infty }(\Omega )$ in the Sobolev norm; membership imposes the homogeneous Dirichlet condition in the weak sense, without assuming that the boundary is smooth. A weak solution is an element of that space satisfyingRepeated indices are summed. The minus sign comes from integration by parts in the divergence equation. Boundedness of the coefficients, Hölder's inequality and density extend this identity to every $\varphi \in W_0^{1,2}(\Omega )$. Symmetry permits interchanging the derivative indices in the left side.

The required functional, with the sign appropriate to a positive divergence operator, isIt is differentiable on $W_0^{1,2}$, with $D\mathcal{F}(u)[v]=Q(u,v)+\int fv$. A critical point consequently satisfies the weak equation. We now prove existence by the direct method in the calculus of variations, rather than invoke the representation theorem again.

The ellipticity and Poincare estimates implyThus the infimum is finite and any minimizing sequence is bounded in the Sobolev norm. A bounded sequence in a Hilbert space has a weakly convergent subsequence; write $u_j\rightharpoonup u$ in $W_0^{1,2}$. The equivalent $Q$ norm has the same continuous linear functionals and hence the same weak convergence. A Hilbert norm is Weak lower semicontinuity, giving $Q(u,u)\le {\mathrm{liminf}}_{j}Q(u_j,u_j)$. The term $\int f{u}_j$ converges to $\int fu$ because it is a bounded linear functional. ThereforeThe limit attains the infimum. Differentiating $t\mapsto \mathcal{F}(u+tv)$ at its minimum, for every $v$, gives $D\mathcal{F}(u)[v]=0$. This produces the desired weak solution. Strict positivity of $Q$ also makes the functional strictly convex, consistent with uniqueness.

First, multiplication by $w$ maps $W_0^{1,2}$ into $L^2$ by the Hölder and Sobolev estimates just given. Thus a weak solution $u$ of the perturbed equation has $v=\Delta u=f+wu\in L^2$. Uniqueness for the unperturbed Dirichlet problem gives $u=Tv$. Its equation becomesConversely, if this equation has a solution $v\in L^2$, then $u=Tv$ belongs to $W_0^{1,2}$ and satisfies $\Delta u=v=f+wTv=f+wu$ weakly. These constructions are inverse to one another.

The Fredholm alternative for a compact operator states that, on a Banach space, $I-K$ for compact $K$ is bijective if and only if its kernel is zero; in that case its inverse is bounded. Apply it on $L^2$. Kernels correspond under $T$: a nonzero $v$ with $v=wTv$ gives a nonzero $u=Tv$, since $\Delta Tv=v$; a nonzero homogeneous solution $u$ gives $v=wu$, which cannot be zero because a zero-boundary harmonic weak solution is zero. Therefore triviality of the homogeneous PDE kernel is equivalent to triviality of $\ker (I-K)$ and hence to unique solvability for every $f\in L^2$. The reverse implication also follows immediately by taking $f=0$. No sign assumption on $w$ is required.

The necessary formulation uses a real Hilbert space $V$ of functions satisfying the essential boundary conditions, a continuous symmetric bilinear form $a$, and a continuous linear functional $\ell$. Assume constants $C,c>0$ withThe latter is coercivity, the positive-definiteness condition needed in the chosen complete space. On the domain of the strong differential operator, integration by parts and the boundary conditions identify $a(v,w)=⟨\mathcal{L}v,w⟩$; this form extends to $V$. Let $\ell (w)=⟨f,w⟩$ for admissible forcing. DefineWhen $u$ has the regularity to belong to the strong operator domain, this is $\mathcal{L}u=f$ with its boundary conditions. For complex spaces use a Hermitian form and $-2\mathrm{Re}\ell (v)$ to make the functional real.

Here is an existence proof by minimization. Continuity of $\ell$ and coercivity give $I(v)\ge c\Vert v{\Vert }_V^2-2\Vert \ell {\Vert }_{V^{\ast }}\Vert v{\Vert }_V$, so $m={\inf }_{V}I$ is finite and a minimizing sequence $v_n$ is bounded. The quadratic parallelogram identity isSince its left side is at least $m$,Coercivity makes $v_n$ Cauchy in $V$. Completeness gives $v_n\to u\in V$, and continuity of the form and forcing gives $I(u)=m$.

For every $w\in V$, differentiation along $u+tw$ yieldsThus the Euler-Lagrange equation is exactly the stated weak solution equation. Conversely, if $u$ satisfies it, expansion givesThis proves that the weak solution is the unique global minimum and that two weak solutions coincide. It proves the symmetric coercive case underlying the Lax-Milgram theorem directly, rather than taking existence as a theorem name. The framework is symmetry and coercivity in quadratic energy minimization.

The qualifications matter if positive definiteness is read more weakly. For a nonsymmetric operator the derivative of $⟨\mathcal{L}v,v⟩$ uses $(\mathcal{L}+{\mathcal{L}}^{\ast })/2$, so it need not give $\mathcal{L}u=f$. For example, on smooth periodic real functions $\mathcal{L}=1+{\partial }_x$ has $⟨\mathcal{L}v,v⟩=\Vert v{\Vert }_2^2$, but the functional's equation is $u=f$, not $u+u_x=f$. Even a symmetric operator can be strictly positive without being coercive: on l2 sequence space, $(Lv{)}_n=v_n/n$ with $f_n=1/n$ would require a solution $u_n=1$, which is not square summable. The associated functional has values $-{\sum }_{n=1}^{N}1/n$ on vectors with the first $N$ entries equal to one, so it has no minimum in that space. This is strict positivity without coercivity can fail variational solvability. Under the usual symmetric coercive meaning of positive definiteness, the proof above supplies every assertion requested.

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.

Write $H^1=W^{1,2}$ and $H_0^1=W_0^{1,2}$. The Dirichlet boundary condition in the Sobolev space sense means $u-\psi \in H_0^1(\Omega )$; this definition remains meaningful even without regularity assumptions on the boundary. For bounded full coefficients, a weak solution is a function satisfyingwhereThe sign follows by integration by parts in $D_i(a_{ij}{D}_{j}u)+qu=f$. No classical derivatives of the measurable coefficients are taken. Equivalently one may first test against $C_c^{\infty }(\Omega )$ and extend by density, provided this bilinear form is bounded on $H^1\times H_0^1$. All integrals in this definition must exist: as explained below, the printed quadratic bounds alone do not guarantee this when $A=(a_{ij})$ is nonsymmetric.

Take $\Omega =(0,\pi )$, $a_{11}=1$, $q=1$, $\psi =0$ and $f(x)=\sin x$. The coefficients are bounded and the operator is uniformly elliptic. If a weak solution $u\in H_0^1(0,\pi )$ existed, testing its weak formulation with $\phi =\sin x\in H_0^1(0,\pi )$ would giveBy integration by parts for $H_0^1$ functions, the first integral on the left is ${\int }_0^{\pi }u\sin x$, so the left side is zero. The right side equals $-\pi /2$, a contradiction. ThusThe forcing is not orthogonal to the homogeneous zero-boundary eigenfunction $\sin x$. This is the obstruction described by the Fredholm alternative for an elliptic Dirichlet problem, proved here directly without assuming that theorem.