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Introduce the Cauchy approximate identityIt is an approximate identity converging to the Dirac delta function $\delta$, and $g_{\epsilon }={\delta }_{\epsilon }^{\prime }$. For a smooth rapidly decaying test function $\varphi$, integration by parts givesBy the definition of the derivative of the Dirac delta, $⟨{\delta }^{\prime },\varphi ⟩=-{\varphi }^{\prime }(0)$. Therefore the limit as a distribution is

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

Use the interior-domain definition of total variation seminorm on a domain, so there is no boundary-extension term at $\pm 1$. For any admissible scalar test function $\phi$,Thus $\mathrm{TV}(u_R)\le 2$.

Choose $0<\epsilon <\min \{R,1-R\}$. From the supplied smooth bump function, defineThe two supports are disjoint and inside $(-1,1)$. Each bump has maximum $e^{-1}$, so $\Vert \phi {\Vert }_{\infty }\le 1$, while $\phi (R)=1$ and $\phi (-R)=-1$. This attains the upper bound, provingEquivalently, its distributional derivative is $D{u}_R={\delta }_{-R}-{\delta }_R$, with total variation two. This is total variation of an interval indicator.

If a test function $\phi$ has integral zero, thenis another compactly supported smooth test function with ${\psi }^{\prime }=\phi$. Hence $v^{\prime }=0$ implies $⟨v,\phi ⟩=-⟨v^{\prime },\psi ⟩=0$ for every such $\phi$.

Choose a test function $\eta$ with integral one. For arbitrary $\phi$, the function $\phi -\eta \int \phi$ has zero integral, soThus $v$ is the constant distribution $⟨v,\eta ⟩$. This proves a distribution with zero derivative is constant.

Choose $h_j\to 0$. The assumed bound and the weak subsequence of a bounded Hilbert-space sequence result give a subsequence for whichFor every test function $\phi \in C_c^{\infty }(V)$, a change of variables yields the difference-quotient integration-by-parts identityThe right side converges to $-{\int }_{V}u{D}_i\phi$, while the left side converges to ${\int }_V{g}_i\phi$. Thus $g_i$ is the $i$th weak derivative of $u$. This holds for every $i$, so

By multiplication of a distribution by a smooth function, for every test function $\psi$,Hence $\rho {\delta }_0={\delta }_0$.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear partial differential operator $P(D)$ on ${\mathbb{R}}^n$ has a fundamental solution of a linear differential operator: there is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ such that $P(D)E={\delta }_0$.

Write $D=-i\partial$. After an orthogonal change of coordinates and multiplication by a nonzero constant, its polynomial symbol may be written as a monic polynomial in the last frequency,For each real ${\mu }^{\prime }$, this polynomial has $M$ complex roots counted with multiplicity. Among a fixed finite collection of horizontal lines at bounded heights, one can choose a line that stays a positive distance from all those roots. Continuity of the roots preserves the choice on a neighborhood $N({\mu }^{\prime })$. Take a countable locally finite cover by such neighborhoods, refine it to a measurable disjoint partition ${\mathbb{R}}^{n-1}={⨆}_j{\Delta }_j$, and let $c_j$ be the chosen height on ${\Delta }_j$. The resulting Hörmander staircasehas bounded heights and may be chosen so that $|P({\xi }^{\prime },s+i{c}_j)|\ge 1$ on each step.

For a test function $\phi$, defineThe Paley–Wiener–Schwartz theorem gives rapid decay in the real frequency directions and at most a fixed exponential factor in the bounded imaginary direction. Together with $|P|\ge 1$, this proves that the integral defines a continuous distribution. Applying $P(D)$ cancels the denominator. The remaining integrand is entire in $z$, so the Cauchy integral theorem shifts every horizontal contour to the real axis; the partition then recombines into ${\mathbb{R}}^{n-1}$. The Fourier inversion theorem giveswhich proves the theorem.

Choose a cutoff function $\eta$ supported in $B_{\rho }(x)$, equal to one on $B_{{\rho }^{\prime }}(x)$, and satisfying $|D\eta |\le (\rho -{\rho }^{\prime }{)}^{-1}$. Use ${\eta }^{2}u$ as a test function in the weak formulation:The Cauchy-Schwarz inequality followed by Young inequality givesand henceThis is the Caccioppoli inequality. The numerical constant is convention-dependent and is normally absorbed into the displayed estimate; replacing the radii by fixed intermediate radii gives the stated form with one universal constant.

For $u(x)=\sin |x|$, the ordinary derivative away from zero isThis bounded function is locally integrable. For a test function $\phi$, integration by parts on $(-\pi ,0)$ and $(0,\pi )$ produces boundary terms at zero which cancel because $u$ is continuous there and $u(0)=0$. HenceTherefore $u$ has the weak derivativeThe jump in the ordinary derivative creates no Dirac delta function; such a term would arise from a jump in the function itself.

For $u\in L_{\mathrm{loc}}^1(\mathbb{R})$, a function $g\in L_{\mathrm{loc}}^1(\mathbb{R})$ is its weak derivative whenfor every test function $\phi \in C_c^{\infty }(\mathbb{R})$. Equivalently, the distributional derivative of $u$ is represented by the locally integrable function $g$.

A phase function is a real smooth function $\Phi$ on $X\times ({\mathbb{R}}^k\setminus \{0\})$ that is positively homogeneous of degree one in $\theta$ and has nonzero total differential $d_{x,\theta }\Phi$. The symbol classconsists of smooth amplitudes for which, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,

Choose a smooth cutoff function $\chi$ equal to one near zero. The associated oscillatory integral is defined on a test function $f$ byOn the compact $x$-support of $f$, use an integration-by-parts operator $L$ satisfying $L{e}^{i\Phi }=e^{i\Phi }$. Repeated application of its formal adjoint lowers the effective symbol order until the integral is absolutely convergent. The resulting bounds involve only finitely many derivatives of $f$, prove that the limit is independent of $\chi$, and give the seminorm estimate required for

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

For $u_0\in L^{\infty }(\mathbb{R})$, call $u\in L^{\infty }({\mathbb{R}}_{+}\times \mathbb{R})$ a weak solution when, for every $\phi \in C_c^1([0,\infty )\times \mathbb{R})$,This follows by multiplying the equation by a test function and applying integration by parts in time and space; $(F\phi {)}_x$ includes the term $F_x\phi$ because the original transport operator is not in divergence form.

If $u$ is $C^1$, test functions supported away from $t=0$ show that $u_t+F{u}_x=g$ as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give $u(0,x)=u_0(x)$. Thus a $C^1$ weak solution is the unique classical solution from part a.

Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

Across a straight discontinuity $x=\sigma t$, integration by parts on its two sides shows that the boundary terms cancel exactly when the Rankine-Hugoniot condition holds:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; its initial datum is therefore zero in the weak identity.

The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an entropy condition would exclude.

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

For any distribution $E\in {\mathcal{D}}^{\prime }$ and any test function $f$, the convolution $(E\ast f)(x)=⟨E_y,f(x-y)⟩$ is a smooth function. On a compact set of $x$ values, the translated test functions and all their derivatives have supports in one fixed compact set, so distributional continuity permits every derivative:For constant-coefficient operators, $P(D)(E\ast f)=(P(D)E)\ast f$. Hence a fundamental solution of a linear differential operator supplies a smooth particular solution for every compactly supported smooth datum. No temperedness of $E$ is needed.