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For a weak solution of a uniformly elliptic homogeneous divergence-form elliptic operator, testing with ${\eta }^2(u-c)$ and choosing the gradient of the cutoff function in $B_{2R}\setminus B_R$ gives ${\int }_{B_R}|\nabla u{|}^2\le C{R}^{-2}{\int }_{B_{2R}\setminus B_R}|u-c{|}^2$. The arbitrary constant $c$ can be chosen as an annular average, which makes the estimate suitable for a hole-filling argument.

For a $C^1$ flux $F$ and a weak solution of $\mathrm{div}F(Du)=0$, discrete integration by parts and the fundamental theorem of calculus along a line segment show that $w_h={\delta }_{\ell ,-h}u$ solves $\mathrm{div}(A_{h}D{w}_h)=0$, with $p_{-}=Du(x-h{e}_{\ell })$ and $p_{+}=Du(x)$. Coercive quadratic-form bounds and operator norm bounds for $DF$ on all segments between these gradients pass to $A_h$. This produces a divergence-form elliptic operator without initially assuming second partial derivatives of $u$.

A Bessel process of dimension $d$ is the radial part of $d$-dimensional Brownian motion. Away from zero it solves the stochastic differential equationThis equation also defines noninteger dimensions through its weak solutions.

Consider the Riemann problem on $\mathbb{R}$ with initial condition $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$. No endpoint boundary conditions are imposed. When $u_L>u_R$, its entropy solution is a shock wave with constant speed $s=(u_L+u_R)/2$. In the original example $u_L=2$, $u_R=0$, the Rankine-Hugoniot condition givesThe left characteristic speed is $2$ and the right characteristic speed is $0$, so $2>1>0$: both sides enter the entropy shock. The Viscous Burgers travelling shock gives an exact smooth family tending to this solution.

Reversing the states to $u_L=0$, $u_R=2$ still gives a Rankine-Hugoniot condition speed of $1$, but now characteristic curves move away from the discontinuity. That expansion jump is a weak solution and fails entropy admissibility. For general $u_L<u_R$, the selected entropy solution is the continuous self-similar solutionThis rarefaction wave fills the expanding gap between the two exterior characteristic curves. Inside its fan, $u_t=-x/t^2$ and $u{u}_x=x/t^2$, verifying the Inviscid Burgers equation; its values agree continuously with both exterior states. For $0$ to $2$, the fan occupies $0<x<2t$. If $u_L=u_R$, the solution is constant.^[ref]

The clamped problem uses the weak solution identity $\int \Delta u\Delta v=\int fv$ for every $v$ in the clamped second-order Sobolev space. For $f\in L^2(U)$, the clamped Hessian identity and the Poincaré inequality make this a bounded coercive bilinear form. The Lax-Milgram theorem gives a unique weak solution without a mean-zero condition on $f$.

A classical solution has enough ordinary derivatives for the differential equation and its data to hold pointwise. For the heat equation, $u_t$ and $u_{xx}$ should be continuous in the interior, with the specified initial and boundary traces. A weak solution instead imposes an integrated identity, allowing less regular data; uniqueness must always be stated for the solution class used.

On the zero-boundary Sobolev space $H_0^1(0,1)$ with norm $\Vert u^{\prime }{\Vert }_{L^2}$, assume measurable coefficients $0<p_0\le p\le p_1$ and $0\le q\le q_1$. The Poincaré inequality and Cauchy-Schwarz inequality give $|a(u,v)|\le (p_1+q_1/{\pi }^2)\Vert u^{\prime }{\Vert }_2\Vert v^{\prime }{\Vert }_2$, while nonnegativity of $q$ gives $a(u,u)\ge p_0\Vert u^{\prime }{\Vert }_2^2$. Thus the form is a bounded bilinear form and a coercive bilinear form. For a source in $H^{-1}$, the Lax-Milgram theorem yields a unique weak solution and every conforming Galerkin method approximation. No derivative of the bounded coefficient $p$ is needed for this weak formulation.

For $H=H_0^1(U)\times H^1(U)$, the coupled formsatisfiesThe Poincaré inequality for $u$ then controls both $\Vert u{\Vert }_2$ and $\Vert w{\Vert }_2$, so the form is coercive on $H$. The Lax-Milgram theorem therefore gives a unique weak solution. The first component has a Dirichlet boundary condition, while the second component's Neumann boundary condition is the natural boundary condition in the weak formulation.

On a bounded open set, let $Tv\in W_0^{1,2}$ be the weak solution of $\Delta Tv=v$. The energy estimate bounds $T:L^2\to W_0^{1,2}$. For $p>n$, put $q=2p/(p-2)<2n/(n-2)$. The zero-boundary Rellich-Kondrachov compactness theorem gives compact embedding into $L^q$, and Hölder's inequality gives $\Vert wz{\Vert }_2\le \Vert w{\Vert }_p\Vert z{\Vert }_q$. Hence $Kv=wTv$ is compact on $L^2$, without a regularity assumption on the boundary. Writing $v=\Delta u$ makes $\Delta u-wu=f$ equivalent to $(I-K)v=f$. The Fredholm alternative for a compact operator then identifies unique solvability with triviality of the homogeneous kernel.

Let $U\subset {\mathbb{R}}^d$ be bounded, $A\in {\mathbb{R}}^d$ satisfy $\Vert A{\Vert }_2<2$, and $f\in L^2(U)$. The bilinear formon $H_0^1(U)$ is bounded. The Cauchy-Schwarz inequality and $2ab\le a^2+b^2$ giveso it is a coercive bilinear form. The Lax-Milgram theorem therefore gives a unique weak solution of $-\Delta v+v+A\cdot \nabla v=f$ with zero Dirichlet boundary condition.

If $Lu=\mathrm{div}F+g$ and $L$ contains $\mathrm{div}(bu)+du$, then $L(u-k)=\mathrm{div}(F-kb)+(g-kd)$. Thus shifting a weak solution to make it nonnegative changes the forcing unless the operator annihilates constants. This bookkeeping is essential in applying the Weak Harnack inequality to upper and lower oscillations.

For the wave equation $u_{tt}=c^2{u}_{xx}$ with zero initial displacement and initial velocity $g$, D'Alembert's formula becomesFor $c=1$ this is the original speed-one form $u(t,x)=\frac{1}{2}{\int }_{x-t}^{x+t}f(s)ds$ when the initial velocity is named $f$. The integral records the total initial kick inside the domain of dependence.

For a computed example, set $g(x)=V$ on $[-a,a]$ and zero elsewhere, with $V>0$. This discontinuous datum gives a weak solution:At the centre, $u(0,t)=V\min (t,a/c)$. For $a=1\mathrm{m}$, $c=2\mathrm{m}{\mathrm{s}}^{-1}$ and $V=0.4\mathrm{m}{\mathrm{s}}^{-1}$, this is $0.1\mathrm{m}$ at $t=0.25\mathrm{s}$ and $0.2\mathrm{m}$ at $t=1\mathrm{s}$. Once $ct>a$, a central plateau $u=Va/c$ occupies $|x|\le ct-a$. In contrast, central zero region for odd compactly supported initial velocity follows when the full integral of $g$ vanishes. Finite propagation speed permits an interior tail; it does not imply the Strong Huygens principle.

An entropy solution is a weak solution of a nonlinear scalar conservation law that also satisfies entropy inequalities selecting physically admissible shocks. For convex fluxes, this excludes expansion shocks and restores uniqueness for bounded initial data.

The Galerkin method seeks an approximate solution in a finite-dimensional subspace and requires the equation's residual to be orthogonal to that subspace. Uniform energy estimates and compactness can then produce a weak solution as the subspaces become dense.

For a bounded measurable coefficient matrix $A$, an $H^1$ weak solution of $\mathrm{div}(A\nabla u)=0$ satisfies $\int A\nabla u\cdot \nabla \phi =0$ for every compactly supported smooth test function. By density this identity extends to compactly supported $H^1$ tests. If $A$ is symmetric and $\lambda I\le A\le \Lambda I$, Caccioppoli inequalities follow without differentiating $A$.

For a transport equation on a domain, prescribe boundary data only where the velocity points into the domain. On $x>0$ with speed $a>0$, a bounded weak solution with initial data $u_0$ and inflow data $f$ satisfiesThe integrals are over the quadrant and its two boundary rays. Backtracking characteristic curves gives $u_0(x-at)$ when $x\ge at$ and $f(t-x/a)$ otherwise. Corner values need not match for a bounded weak solution.

Let $L=a^{ij}(x)D_{ij}$ be uniformly elliptic with continuous coefficients. The coefficient-freezing interior second-derivative estimate extends from smooth functions to the graph norm closure of $L$ on $H^1(U)$: if $u_k\to u$ in $H^1(U)$ and $L{u}_k\to f$ in $L^2(U)$, then $(u_k)$ is Cauchy in $H^2(W)$ for every $W⋐U$. Hence $u\in H_{\mathrm{loc}}^2(U)$ and $Lu=f$. Equivalently, the standard local regularization argument gives the same conclusion for weak solutions.

The Inviscid Burgers equation is the scalar conservation lawFor a classical solution, the chain rule makes this equivalent to $u_t+u{u}_x=0$, a nonlinear transport equation with characteristic speed $u$. The conservative form remains meaningful for a discontinuous weak solution, whereas the product $u{u}_x$ cannot simply be evaluated pointwise at a jump. Its discontinuities obey the Rankine-Hugoniot condition $s=(u_L+u_R)/2$; this condition alone permits inadmissible expansion jumps. For bounded initial conditions on the infinite line, imposing the Kruzhkov entropy inequalities and the initial trace selects a unique entropy solution. For step data, incoming characteristic curves select $u_L>s>u_R$, while $u_L<u_R$ produces a rarefaction wave.^[ref]