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The adjoint transport equation moves a transport equation operator from a solution to a test function by integration by parts. For $Lu={\partial }_{t}u+b\cdot \nabla u$, its formal space-time adjoint operator isIndeed $\int (Lu)\phi =\int u{L}^{\ast }\phi$ when both functions have compact support away from the time endpoints. The term $-(\mathrm{div}b)\phi$ is essential for a compressible velocity field. By contrast, the conservative operator $M\rho ={\partial }_t\rho +\mathrm{div}(b\rho )$ has $M^{\ast }\phi =-{\partial }_t\phi -b\cdot \nabla \phi$.

For a smooth complete characteristic flow map with compact supports remaining bounded on $[0,T]$, solve $L^{\ast }\phi =g$ backward with $\phi (T)=0$, for a smooth compactly supported $g$. Integration by parts in time gives ${\int }_0^T\int ug=\int u_0\phi (0)$ for a weak solution of $Lu=0$. Thus zero initial condition forces zero pairing with every such $g$, proving uniqueness. The characteristic curves solve this terminal problem with a divergence weight; omitting that weight would test a different equation.

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 Barenblatt solution is a mass-preserving source-type similarity solution of the porous medium equation. For the one-dimensional normalizationthe nonnegative profile isIts support is $|x|\le R(t)$ with $R(t)=\sqrt{6\beta A}{t}^{1/3}$, and its peak is $A{t}^{-1/3}$. Substituting the quadratic profile verifies the partial differential equation wherever it is positive. At the edge, $u{u}_x=0$, so extending by zero gives a weak solution without a singular interface flux.

The total mass determines $A$:It is independent of time. As $t\downarrow 0$, the support shrinks and the measures $u(x,t)dx$ converge to $M{\delta }_0$, a point source. This is a nonlinear counterpart of the heat kernel, with a finite moving front instead of positive Gaussian tails. The normalization matters: $\beta (u{u}_x{)}_x=(\beta /2)(u^2{)}_{xx}$.

For a smooth bounded domain and a bounded continuous scalar function $f$, this problem has a weak solution. Let $S$ be the zero-Dirichlet inverse of $-\Delta$, defined by the Lax-Milgram theorem. The Poincaré inequality gives $\Vert Sh{\Vert }_2\le C_P^2\Vert h{\Vert }_2$. The Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. The Nemytskii operator $N_f$ is continuous and has bounded image, so $T=S\circ N_f$ is a nonlinear compact operator mapping a sufficiently large closed $L^2$ ball into itself. The Schauder fixed-point theorem gives a fixed point, necessarily in $H_0^1$, and its defining identity is the desired weak formulation. This proves existence; uniqueness requires additional assumptions.

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]

Let $b(t,x)$ be continuous in time and continuously differentiable in space, with a uniform spatial Lipschitz constant and at most linear growth on each compact time interval. Its characteristic flow map exists forward and backward over every bounded time interval. For each initial probability measure with finite first absolute moment, the displayed pushforward measure is the unique weak solution of the continuity equation ${\partial }_t\rho +\mathrm{div}(b\rho )=0$ continuous in the first Wasserstein distance.

The Gronwall inequality bounds the flow by $C_T(1+|x|)$ and therefore preserves finite first absolute moments. The chain rule establishes the weak equation. For uniqueness, transport a compactly supported terminal test function backward along the flow; it solves ${\partial }_t\phi +b\cdot \nabla \phi =0$ and remains compactly supported on a bounded time interval. Inserting it into the weak identity identifies every terminal integral with that of the claimed pushforward measure. A zero divergence makes the corresponding scalar transport equation and the continuity equation coincide.

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.

A symmetric uniformly elliptic divergence-form energy with zeroth-order coefficient at least $\mu >0$ controls the full $H^1$ norm, including constants. Its quadratic energy minus a bounded load has a unique minimizer by the direct method, and vanishing first variation gives the natural Neumann weak equation. Unlike a pure Neumann Laplacian, this problem needs no zero-mean compatibility condition. The exact energy difference from the weak solution is half the bilinear energy of the difference.

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 a bounded open set, $A\in {\mathbb{R}}^d$ a constant vector, and $f\in L^2(U)$. On the real Sobolev space $H_0^1(U)$, consider the bounded bilinear formIts weak solution solveswith zero Dirichlet boundary condition. The minus sign follows from integration by parts: $\int uA\cdot \nabla v=-\int (A\cdot \nabla u)v$ because $A$ is constant. To represent the plus-drift equation, instead use the form with $-uA\cdot \nabla v$.

No smallness condition on $\Vert A\Vert$ is needed. For a smooth compactly supported $u$,Density extends the identity to $H_0^1(U)$, since the drift pairing is continuous in the $H^1$ norm. Therefore $B(u,u)=\Vert u{\Vert }_{H^1}^2$ for every constant $A$. This is a coercive bilinear form, so the Lax-Milgram theorem gives a unique weak solution for arbitrary constant drift. The drift changes the equation but contributes no energy to this homogeneous-Dirichlet estimate.

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.

Suppose $k>0$ and two sufficiently regular steady state temperature profiles solve the same heat equation with the same PDE source term $H$ and the same Dirichlet boundary condition $h$. Their difference $w$ satisfies $\nabla \cdot (k\nabla w)=0$ and $w=0$ on the boundary. The divergence theorem applied to $wk\nabla w$ givesThus $\nabla w=0$, and the zero boundary values force $w=0$ on each connected component. The argument proves uniqueness without asserting existence and without replacing the spatially varying thermal conductivity by a constant. For weak solutions, the same calculation is the energy argument obtained by choosing $w$ as the test function, under the usual boundedness and positive lower bound assumptions on $k$.

An elliptic boundary value problem couples an elliptic differential equation in a domain to specified boundary conditions. An important example is the Poisson equationHere the Dirichlet boundary data fix the boundary value. Prescribing outward flux instead gives a Neumann boundary-value problem, which has different compatibility and normalization requirements. Thus the differential expression alone does not specify the problem.

For zero Dirichlet boundary conditions, multiply the Poisson equation by a smooth compactly supported test function and apply integration by parts:This weak formulation requires only first weak derivatives, so it makes sense for $u\in H_0^1(\Omega )$ and $f\in L^2(\Omega )$. On a bounded domain, the Poincaré inequality makes its bilinear form coercive, and the Lax-Milgram theorem gives a unique weak solution. The right side also extends to suitable forcing in the continuous dual space of $H_0^1(\Omega )$.^[ref]

The energy method explains the value of this formulation: subtracting two solutions and testing with their difference proves uniqueness. Elliptic regularity asks whether additional smoothness of coefficients, forcing and boundary upgrades the weak solution to a classical one. A Schauder estimate quantifies this gain in Hölder spaces; rough coefficients instead lead to results such as the De Giorgi-Nash-Moser theorem.

For the focusing nonlinear Schrödinger equation on ${\mathbb{R}}^d$,the energy functional isFor a sufficiently smooth solution with decay permitting differentiation under the integral sign and integration by parts, it is constant in time. Indeed,The cancellation is the Hamiltonian conservation mechanism: the evolution rotates the energy gradient by multiplication by $i$. The focusing sign makes the nonlinear contribution negative, so conserved energy alone need not bound the L2 norm of the gradient.^[ref]

For less regular solutions the displayed computation is only formal. To inherit the identity, approximate by smooth solutions and justify convergence of both energy terms. Convergence in Sobolev space $H^1$ controls the nonlinear term through a suitable Sobolev embedding only in the applicable exponent range, or with additional integrability. An arbitrary weak solution must not be assumed to satisfy the identity solely because the smooth equation does.

For a scalar conservation law $u_t+q(u{)}_x=0$, an entropy condition restricts its weak solutions to physically admissible ones. For every convex entropy function $\eta$, define an entropy flux by $Q^{\prime }={\eta }^{\prime }{q}^{\prime }$. The displayed inequality is imposed in the distributional sense. Positive diffusion explains this choice: for $u_t+q(u{)}_x=ϵ[D(u)u_x{]}_x$ with $D>0$, the entropy balance is $\eta (u{)}_t+Q(u{)}_x=ϵ[{\eta }^{\prime }(u)D(u)u_x{]}_x-ϵ{\eta }^{\prime \prime }(u)D(u)u_x^2$. The last term is nonpositive, and an appropriate vanishing-diffusion limit yields the entropy inequality. For a strictly concave flux and a single shock, this selects states with incoming characteristics, rather than a discontinuity whose characteristics spread outward.