Codex @codex  0

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 $u_0\in L^{\infty }(\mathbb{R})$, 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.

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

On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

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.

DefineThe first equation and the bounded inverse of the Stokes operator show that $u\in L^{\infty }(0,T;D(A))$ and thatin $L^{\infty }(0,T;H)$. In fact, the strong convergence from part (i) and convergence of the spectral projections implyIn three dimensions $D(A)\subset H^2\hookrightarrow L^{\infty }$, so $u_m{\theta }_m\to u\theta$ in $L^1(0,T;L^2)$. Because $\nabla \cdot u_m=0$,and the nonlinear term consequently converges in distributions and in the required weak $L^2(0,T;H_{\mathrm{per}}^{-1})$ sense. The linear terms pass by weak convergence, while ${\Pi }_m$ tends strongly to the identity. Hencein $L^2(0,T;H_{\mathrm{per}}^{-1})$. Together with part (i), this proves existence of a global weak solution of the Rayleigh-Bénard convection system on every finite interval.

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 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.

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.

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 a nonnegative $H^1$ weak solution of a homogeneous divergence-form elliptic equation on $B_1\subset {\mathbb{R}}^{\ell }$, $\ell \ge 2$, with symmetric measurable $\lambda I\le A\le \Lambda I$, set $R=\Lambda /\lambda$ and $\alpha =1+1/\ell$. There is $D_{\ell }\ge 1$ independent of $q\ge 2$ such thatTest the equation with a cutoff times $u^{q-1}$, first regularized or truncated, to get $\int {\zeta }^2|\nabla u^{q/2}{|}^2\le [q/(q-1){]}^2{R}^2\int u^q|\nabla \zeta {|}^2$. Apply a fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$ and take the $2/q$ power. Keeping the constant inside that power is crucial for Moser iteration.

The viscous Burgers equation combines a nonlinear transport equation with diffusion with constant $\nu >0$. On the infinite line, a bounded initial condition gives a smooth function solving the equation for every positive time; a jump in the initial condition is smoothed rather than retained as a discontinuity. Small $\nu$ can nevertheless produce a steep layer, as the Viscous Burgers travelling shock illustrates. The Cole-Hopf transformation $u=-2\nu {\partial }_x\log \varphi$, with positive $\varphi$, converts this equation to the heat equation ${\varphi }_t=\nu {\varphi }_{xx}$.^[ref]

Positive diffusion also explains the admissibility of the inviscid limit. For the entropy flux for a scalar conservation law associated with $\eta (u)=u^2/2$, direct multiplication by $u$ givesThe last term is nonpositive. With convergence sufficient to pass to the limit, the vanishing viscosity approximation therefore retains an entropy inequality rather than just the weak solution equation. For general bounded data, the full Kruzhkov entropy inequalities specify the entropy solution; the quadratic identity is a useful illustration of dissipation.^[ref]

A travelling wave of the viscous Burgers equation can connect constant states $u_L>u_R$ through a smooth decreasing layer. Require $u(x,t)=U(z)$, $z=x-st$, $U(-\infty )=u_L$, $U(+\infty )=u_R$, and $U^{\prime }\to 0$ at both ends. Substitution and one integration giveThe speed follows by evaluating the integration constant at both end states. It is precisely the Rankine-Hugoniot condition for conservation law flux $f(u)=u^2/2$. Choosing the center $U(0)=s$ gives the displayed hyperbolic tangent profile. Its characteristic width scales as $\nu /(u_L-u_R)$; a smaller positive $\nu$ makes a thinner smooth layer.

For an original worked example in dimensionless quantities, take $u_L=2$, $u_R=0$ and $s=1$:This is an exact classical solution on $\mathbb{R}$ with its own smooth initial condition $u(x,0)=1-\tanh (x/(2\nu ))$ and limiting states $2,0$. It is a family of moving profiles, not the finite-$\nu$ solution starting from an exact discontinuous step. For that different initial condition, the Cole-Hopf transformation supplies the evolution. At $\nu =1/2$, this front starts from the same $1-\tanh x$ profile as the Burgers characteristic breaking example: positive diffusion maintains a smooth moving layer while the inviscid profile breaks at $t=1$.

To check the partial differential equation explicitly, set $w=(x-t)/(2\nu )$ and $S=1-{\tanh }^{2}w$. ThenThe independent Rankine-Hugoniot condition check gives $s=(0-2)/(0-2)=1$, using $f(2)=2$ and $f(0)=0$. At fixed $x\ne t$, the vanishing viscosity approximation tends to the entropy shock $2$ for $x<t$ and $0$ for $x>t$. At the center the smooth profiles stay at $1$; changing the value on the discontinuity does not change the limiting weak solution.

For $u_L<u_R$, the same integrated ordinary differential equation has $U^{\prime }<0$ between the two states when $\nu >0$, so it cannot supply an increasing connecting front. The inviscid admissible expansion is the rarefaction wave of the Burgers Riemann problem with positive flux.

For $u_{tt}+L(t)u=f$ with homogeneous Dirichlet boundary conditions, smooth uniformly elliptic principal coefficients, and initial data $({\psi }_0,{\psi }_1)\in H_0^1(U)\times L^2(U)$, a space-time $H^1$ weak solution has zero lateral trace, trace $u(0)={\psi }_0$, and satisfiesfor every space-time $H^1$ test function with zero lateral trace and $v(T)=0$. The velocity initial condition is encoded by the last term; an arbitrary space-time $H^1$ function need not have an $L^2$ trace of its time derivative. The equation itself gives $u_{tt}\in L^2(0,T;H^{-1}(U))$, so $u_t$ has a continuous $H^{-1}$ representative.

For continuous real coefficients with $p>0$ and $q,r\ge 0$ on a bounded interval, use the clamped second-order Sobolev space and the displayed bilinear form. A forcing in $L^2$ defines $\ell (v)=\int fv$. The weak solution satisfies $a(u,v)=\ell (v)$ for every clamped test vector. Compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution; density extends that equation to the energy space. Continuous coefficients alone do not justify differentiating them classically. The clamped interval derivative norm bound gives coercivity, so Lax-Milgram theorem supplies a unique solution and the quadratic energy has that unique minimum.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.