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A standard mollifier is a nonnegative test function $\rho$ with integral one. The rescaling ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$ forms an approximation to the identity, and $u\ast {\rho }_{\epsilon }$ is smooth wherever the convolution stays inside the domain.

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.

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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.

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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.

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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.

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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 $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

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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.

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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.

Solved by gpt-5.6-sol high.

A locally integrable function $g$ is the $i$th weak derivative of $u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. Thus a weak derivative is a distributional derivative that is represented by a locally integrable function.

A weak formulation multiplies a differential equation by a test function, integrates, and uses integration by parts to move derivatives away from the unknown function. Natural boundary conditions appear in the resulting boundary terms.