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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.

The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.

Lévy characterization of Brownian motion says that a continuous local martingale $M$ with $M_0=0$ is standard Brownian motion with respect to its filtration if its quadratic variation is $⟨M{⟩}_t=t$. Equivalently one may require $M_t^2-t$ to be a local martingale.

Any weak solution has the integral form $Z_t={\int }_0^t\mathrm{sgn}(Z_s-2)d{W}_s$. Its bounded integrand makes it a continuous local martingale starting at zero, andThe specified value of the sign at zero has square one as well. Lévy characterization of Brownian motion therefore proves that every such weak solution is Brownian motion. This is a statement about its law and filtration; it does not assert $Z=W$ or provide a strong solution as a function of a specified driver.

A weak solution is a field $\mathbf{u}$ for which the composed densities $\mathbf{P}(x,t,\mathbf{u})$ and fluxes $\mathbf{Q}(x,t,\mathbf{u})$ are locally integrable andfor every smooth compactly supported vector test function $\mathbf{\phi }$ in the spacetime domain. This is the distributional form of the conservation law, obtained by moving derivatives onto the test function. If tests are allowed to meet an initial boundary $t=0$, the specified initial density contributes the additional term $\int \mathbf{P}(x,0,{\mathbf{u}}_0)\cdot \mathbf{\phi }(x,0)dx$ in the usual initial-value formulation.

Suppose $D=a{\partial }_x+b{\partial }_y$, $DA=f$, $Dq=0$, and $A,q$ are smooth functions. Then $u=A+g(q)$ solves $Du=f$ wherever $g$ is differentiable. If $g$ is continuous and piecewise continuously differentiable with a jump in $g^{\prime }$ at $q=q_0$, the one-sided gradients of $u$ differ by $[g^{\prime }]\nabla q$ across that characteristic curve. Thus a nonzero derivative jump in the data persists along the characteristic unless the relevant component of $\nabla q$ vanishes. A locally Lipschitz continuous $g$ also gives a weak solution across the curve.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

For the planar porous medium equation $n_t=[D_0/(3n_0^2)]\Delta (n^3)$, the mass-$Q$ point-source similarity solution is $n=n_0{\lambda }^{-2}\sqrt{[1-r^2/(r_0^2{\lambda }^2){]}_{+}}$, with $\lambda =(6D_{0}t/r_0^2{)}^{1/6}$ and $r_0^2=3Q/(2\pi n_0)$. Its front grows as $t^{1/6}$ and its central density falls as $t^{-1/3}$. The mass conservation integral uses the planar area element $2\pi rdr$. Although the density slope diverges at the front, its diffusion flux vanishes there, giving a compactly supported weak solution with initial measure $Q{\delta }^{(2)}$.

The Rankine-Hugoniot condition expresses conservation laws across a moving discontinuity of a weak solution. Across a discontinuity $x=\sigma t$ in $u_t+f(u{)}_x=0$, with left and right states $u_{-}$ and $u_{+}$, conservation requiresWriting $[u]=u_{+}-u_{-}$ and $[f(u)]=f(u_{+})-f(u_{-})$ gives $\sigma [u]=[f(u)]$. The same relation holds componentwise for a system of conservation laws; in compressible flow, conservation of mass, momentum and energy supplies the jump conditions for a shock wave. Integrating the weak solution identity over a thin space-time region crossing the discontinuity yields this balance between the swept-up conserved quantity and the difference of fluxes.

A rarefaction wave is a continuous expanding fan of characteristic curves joining constant states of a hyperbolic system or scalar conservation law. For $u_t+f(u{)}_x=0$ with $f^{\prime \prime }>0$ between $u_L<u_R$, the Riemann problem has the self-similar solutionTo derive the fan, substitute $u(t,x)=V(\xi )$, $\xi =x/t$, into the smooth equation. It gives $(f^{\prime }(V)-\xi )V^{\prime }=0$, so the nonconstant branch has $f^{\prime }(V)=\xi$. Strict convexity makes $f^{\prime }$ invertible on the state interval, and the two fan edges match the exterior states continuously. Derivatives can jump at those edges.

For the Inviscid Burgers equation with $u_L=0$ and $u_R=1$, the solution is $0$ for $x\le 0$, $x/t$ for $0<x<t$, and $1$ for $x\ge t$. The characteristic speed equals $u$, so the gap between the outgoing exterior characteristic curves fills with all intermediate speeds. This is the selected entropy solution, whereas the corresponding expansion discontinuity is only a weak solution.^[ref]

A rectangular initial displacement with zero initial velocity splits into two equal half-height copies moving in opposite directions. The D'Alembert formula reads $u(x,t)=[u_0(x+ct)+u_0(x-ct)]/2$. Overlap adds the heights; separated copies each retain the original width. For discontinuous fronts the formula defines a weak solution of the wave equation.

A one-dimensional scalar conservation law describes the balance of a scalar density $u$ and its conservation law flux $f(u)$:The second identity follows from the first for smooth $u$: the quantity inside a fixed interval changes only through its boundary fluxes. For a classical solution, the chain rule also gives the quasilinear partial differential equation $u_t+f^{\prime }(u)u_x=0$, with characteristic speed $f^{\prime }(u)$.

The conservative form still defines a discontinuous weak solution. With initial condition $u_0$, it requiresfor every smooth compactly supported test function on $[0,\infty )\times \mathbb{R}$. For $f(u)=u^2/2$, a jump from $1$ to $0$ travels at speed $1/2$ by the Rankine-Hugoniot condition, between incoming characteristic speeds $1$ and $0$. Reversing those states still gives a conservative jump of speed $1/2$, but entropy solution admissibility selects a rarefaction wave. Thus conservation determines jump balance while an entropy condition selects the continuation after characteristic crossing.^[ref]

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.

For a globally Lipschitz nonlinearity of constant $ϵ$, the map taking $v$ to the zero-Dirichlet Poisson solution with load $f(v)$ is a contraction in the gradient norm if the displayed inequality holds. Poincaré gives the factor $C_P$ from the Sobolev input to $L^2$, and testing the linear Poisson equation gives another factor $C_P$ back to the gradient norm. The condition yields a unique weak solution without requiring $f(0)=0$. With the optimal Poincaré constant it reads $ϵ<{\lambda }_1$.

A shock wave is a moving discontinuity of a weak solution to a conservation law. Its speed is fixed by the Rankine-Hugoniot condition, while an entropy condition selects the physically admissible direction of compression.

A classical solution is a sufficiently differentiable function that satisfies a differential equation and any prescribed initial or boundary conditions. A weak solution instead satisfies an appropriate integral or distributional identity in a specified function space. For a smooth locally Lipschitz continuous vector field, an ordinary differential equation has a unique local solution for each initial state; this does not by itself ensure global existence or boundedness.

For the nonlinear diffusion equation $u_t=(u{u}_x{)}_x+u/(2t)$ on $x>0,t>0$, the fixed boundary value $u(0,t)=1$ and zero slope at $x=0$ admit the displayed compact-support similarity solution, with front $x_F=2\sqrt{t}$. Substitution $u=f(x/\sqrt{t})$ reduces the equation to $(f{f}^{\prime }+\eta f/2{)}^{\prime }=0$. The boundary data make the integration constant zero. While $f>0$, division gives $f^{\prime }=-\eta /2$, hence $f=1-{\eta }^2/4$. At its first zero the diffusive flux $u{u}_x$ vanishes, so extension by zero creates no singular flux term. The solution is smooth inside its positive region and is a weak solution at the moving front.

In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar{v}(s)=(4\pi {)}^{-1}{\int }_{S^2}v(s\omega )d\omega$. Outside $B_a$, the weak solution is a harmonic function. The weak formulation against radial test functions gives $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ for $a<s<R$. The zero Sobolev trace gives $\bar{v}(R)=0$; integration proves the displayed formula, without assuming the source is a radial function.

Cauchy-Schwarz inequality on spheres gives $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding balls cannot remain bounded in the full $H^1({\mathbb{R}}^3)$ norm after zero extension of H01.