Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 (Analysis of Partial Differential Equations)

The Taylor series definition says that $f\in C^{\infty }(\mathbb{R})$ and, for every $a$, there is $r>0$ such thatThe equivalent factorial derivative criterion for real analyticity says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such thatThe uniformity over $I$ matters: bounds only at $a$ do not exclude a flat function.

Assume the factorial derivative criterion for real analyticity. The Taylor theorem with Lagrange remainder giveswhen the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the Taylor remainder tends to zero, proving the Taylor series definition.

Conversely, write the convergent power series at $a$ as $\sum b_k{h}^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\le M{\rho }^{-k}$ for some $M$. Termwise differentiation on $|h|\le \rho /2$ givesHere the sum is $(1-1/2{)}^{-n-1}$, obtained by differentiating the geometric series. This is the required locally uniform bound. The two definitions of a real analytic function are equivalent.

Consider the flat functionAway from zero every derivative has the form $P_n(1/x)e^{-1/x^2}$ for a polynomial $P_n$: differentiating preserves this form. For every $m\ge 0$,because an exponential function decays faster than any power. Inductively, extend each displayed derivative by zero at zero. It is continuous there, and its difference quotient at zero also tends to zero by the same estimate with one extra power of $|x{|}^{-1}$. Thus each extension is the derivative of the preceding extension. This proves $f\in C^{\infty }(\mathbb{R})$ and $f^{(n)}(0)=0$ for all $n$.

Its Taylor series at zero is identically zero, whereas $f(x)>0$ for every $x\ne 0$. It is smooth everywhere but not real analytic at zero.

The complex Liouville theorem states that a bounded entire function is constant. Indeed, if $|f|\le M$, the Cauchy estimate on any disc of radius $R$ centered at $z$ gives $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ gives $f^{\prime }(z)=0$ everywhere.

The analogous conclusion for bounded real analytic functions on $\mathbb{R}$ is false. For example, $\sin x$ is bounded, nonconstant, and real analytic on the whole real line. Boundedness only on that line does not bound its holomorphic extension on the complex plane.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

The total differential order of ${\partial }_t-{\partial }_x^2$ is two, so its principal symbol is $p(\tau ,\xi )=-{\xi }^2$. The conormal to $t=0$ is $(1,0)$, on which $p$ vanishes. The initial line is a characteristic hypersurface for this total-order symbol; the first-order time derivative does not enter it.

Suppose a real analytic solution existed near $(0,0)$. Repeated use of the heat equation gives ${\partial }_t^{k}u={\partial }_x^{2k}u$. The initial power series is ${\sum }_{j\ge 0}(-1{)}^j{x}^{2j}$ near zero, henceThe time Taylor series at $x=0$ would therefore have coefficients $(-1{)}^k(2k)!/k!$. The ratio of successive absolute coefficients is $2(2k+1)\to \infty$, giving radius of convergence zero. This contradicts the assumed real analytic regularity. No such analytic local solution exists, although the initial function itself is real analytic.

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

The Cauchy problem for a partial differential equation here prescribes both the value $u{|}_{\Gamma }=g$ and the normal derivative ${\partial }_{n}u{|}_{\Gamma }=h$, together with $\Delta u=0$. Only one of these traces would be boundary data for a usual elliptic boundary problem, rather than full Cauchy data.

Every real hypersurface is a non-characteristic hypersurface for the Laplace equation. In local real analytic coordinates flattening the real analytic hypersurface $\Gamma$, the coefficient of the second transverse derivative is nonzero: its principal coefficient is the squared length of the conormal. The equation can therefore be solved for that second derivative. The normal derivative data determine the transverse first derivative, because the coefficient relating them is nonzero and the tangential first derivatives are already determined by $g$.

The coefficients, flattened Cauchy data, and coordinate change are all real analytic. The Cauchy-Kovalevskaya theorem applies, giving a unique local real analytic solution around each point of $\Gamma$. This is a local existence assertion, not a claim of stable dependence in arbitrary Sobolev space norms.

The Cauchy-Kovalevskaya theorem cannot be applied to merely $C^2$, non-$C^3$ Cauchy data. It requires real analytic data.

There is also no $C^2$ solution of the Laplace equation on a neighborhood of a point where one of these prescribed traces fails to be $C^3$. By interior elliptic regularity, any such solution would be smooth and real analytic. On the real analytic hypersurface retained from the preceding part, both its restriction and its normal derivative would then be real analytic, hence $C^3$. This contradicts the prescribed trace. There is no solution on a neighborhood of all of $\Gamma$ with the stated non-$C^3$ data. This does not exclude solutions near other points where the data happen to be real analytic.

Use unit speed and write the Cauchy data as $u(0)=f$, $u_t(0)=g$. For finite-energy data define the wave energy estimate quantityMultiply $u_{tt}-{\Delta }_{x}u=0$ by $u_t$ and use integration by parts. With compact support or sufficient decay the boundary flux is zero, soFor general finite-energy solutions, cutoff or approximation arguments justify this identity; equivalently the local estimate below, applied in both time directions and with radii tending to infinity, gives the same equality. Arbitrary smooth data need not have finite global energy; then the global bound with an infinite right side is uninformative, while the local estimate remains useful.

If $f\in H^1$ and $g\in L^2$, the fundamental theorem of calculus and the wave energy estimate further giveTogether these yield an a priori bound for $\Vert u(t){\Vert }_{H^1}+\Vert u_t(t){\Vert }_2$ on each bounded time interval. No existence assumption is proved by the estimate itself; it controls any sufficiently regular solution.

The local wave energy estimate is, for $T\ge 0$, $r>0$, and any center $x_0$,To prove it, let $R=r+T$ and integrate the local energy estimate identity ${\partial }_{t}e=\mathrm{div}(u_t\nabla u)$, where $e=(u_t^2+|\nabla u{|}^2)/2$, over the shrinking ball $B_{R-t}(x_0)$. Differentiation of this moving-domain integral yieldsIntegrating in time proves the local wave energy estimate, without assumptions at spatial infinity.

Let the union of the initial supports be a compact set $K$. If $\mathrm{dist}(x_0,K)>T$, choose $r>0$ with $r+T<\mathrm{dist}(x_0,K)$. The initial energy on $B_{r+T}(x_0)$ vanishes. Applying the shrinking-ball identity up to every intermediate time shows $u_t$ and $\nabla u$ vanish throughout that cone. In particular, along the vertical segment through $x_0$, $u_t=0$; its initial value is also zero, so $u(T,x_0)=0$. This last value check removes the constant ambiguity invisible to gradient energy.

Consequently the finite propagation speed conclusion isThis set is compact for each finite $t$. The speed is at most one in these units, or $c$ for $u_{tt}-c^2\Delta u=0$. Time reversal gives the same conclusion for negative time.

For this Neumann Poisson problem, interpret the forcing in $L^2(U)$, as is automatic if it is smooth up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior smooth but even $\int f\cdot 1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.

Suppose the weak solution is smooth on $\overline{U}$. Testing against compactly supported test functions gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and Green's first identity implyfor every smooth $v$ on $\overline{U}$. Every smooth boundary function has such an extension, so ${\partial }_{n}u=0$ on $\partial U$. This proves both the interior equation and the boundary condition.

Conversely, for $u\in C^2(\overline{U})$ satisfying the equation and zero normal derivative, Green's first identity gives the weak identity for all smooth $v$ on $\overline{U}$. The density of smooth functions in a Sobolev space and the Cauchy-Schwarz inequality extend it continuously to every $v\in H^1(U)$. Thus the classical solution is a weak solution, and the smooth weak solution is classical.

Subtract the two weak solution identities and test with their difference $w\in H^1(U)$. ThenA Sobolev function with zero weak gradient is constant on each connected component. One justification is to mollify locally: each mollification has zero gradient and is constant on its ball, and overlaps identify the constants; taking limits gives the original assertion. Since $U$ is connected, $w$ is one constant on $U$.

Conversely, adding a constant changes neither the weak derivative nor the weak identity. The solution is unique up to an additive constant.

There is a normalization error in the PDF. With the printed unnormalized integral, testing a constant function equal to one would give a left side $|U|(1-|U|{)}^2$ and a right side zero. Thus that formulation fails whenever $|U|\ne 1$.

Use instead the average $v_U=|U{|}^{-1}{\int }_{U}v$. The Poincare-Wirtinger inequality, also called the Neumann-Poincare inequality, isIf no $C_P$ exists, subtract the average and normalize a violating sequence to obtain $w_j\in H^1(U)$ with ${\int }_U{w}_j=0$, $\Vert w_j{\Vert }_2=1$, and $\Vert \nabla w_j{\Vert }_2\to 0$. This sequence is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrachov compactness theorem supplies a subsequence converging strongly in $L^2(U)$ to $w$.

For every compactly supported test function $\phi$, integration by parts and these convergences give $\int w{\partial }_i\phi =0$. Thus $w$ has zero weak gradient. The Sobolev function with zero weak gradient result and connectedness make $w$ constant. Strong $L^2$ convergence preserves its zero integral, so $w=0$. It also preserves its norm one, a contradiction. This proves the correctly normalized Neumann-Poincare inequality.

Work in the mean-zero Sobolev spaceThe integral is a continuous functional on $H^1(U)$, so $H$ is closed. The Neumann-Poincare inequality shows that $\Vert v{\Vert }_a=\Vert \nabla v{\Vert }_2$ is equivalent to the usual $H^1$ norm on $H$, making $a$ a complete inner product there.

For $f\in L^2(U)$ the functional $L(v)={\int }_{U}fv$ satisfiesThe Riesz representation theorem, or the Lax-Milgram theorem, gives a unique $u\in H$ with $a(u,v)=L(v)$ for all $v\in H$. To recover every $H^1(U)$ test, write $v=(v-v_U)+v_U$. The constant contributes zero to $a$ and contributes $v_U{\int }_{U}f=0$ to $L$. Thus the same equality holds for all $v\in H^1(U)$.

A weak solution exists whenever ${\int }_{U}f=0$; fixing its average to zero makes it unique. Moreover $\Vert \nabla u{\Vert }_2\le \sqrt{C_P}\Vert f{\Vert }_2$ and the Neumann-Poincare inequality also controls $\Vert u{\Vert }_2$.

The constant function $v=1$ is an admissible $H^1(U)$ test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately givesThis is necessary, and the preceding Hilbert-space construction proves sufficiency for $f\in L^2(U)$. It is the balance condition corresponding to zero total boundary flux.

The clamped second-order Sobolev space is $H_0^2(U)={\overline{C_c^{\infty }(U)}}^{H^2}$. On a smooth bounded domain the Sobolev trace theorem characterizes it by zero value and zero normal derivative on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.

For a smooth weak solution, compactly supported test functions and two integrations by parts giveso ${\Delta }^{2}u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u={\partial }_{n}u=0$ on $\partial U$. Thus it is a classical solution of the clamped biharmonic problem.

Conversely, a $C^4(\overline{U})$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported test function, two integrations by parts give ${\int }_U\Delta u\Delta v={\int }_{U}fv$. Both sides are continuous for the $H^2$ norm, so the defining density of $C_c^{\infty }(U)$ in $H_0^2(U)$ extends this equality to every required test. The two notions agree under the stated smoothness.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincare inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincare inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.

Let $M=\Vert F^{\prime }{\Vert }_{\infty }$. The sharper energy estimate uses the divergence structure of the viscous scalar conservation law. Multiply by $u$ and integrate over the line. With $G(s)={\int }_0^{s}r{F}^{\prime }(r)dr$,because the decay makes $G(u)$ tend to zero at both ends. Integration by parts in the diffusion term therefore givesOne may take $C_0=0$, uniformly in $\epsilon$. This does not require $F(0)=0$.

If a bound explicitly involving $M$ and $\epsilon$ is desired, retaining the transport term and using the Cauchy-Schwarz inequality and the elementary inequality $ab\le (a^2+b^2)/2$ givesThe Gronwall inequality then gives the valid but weaker choice $C_0=M^2/\epsilon$. The exact cancellation explains why its divergence is unnecessary.

Multiply the viscous scalar conservation law by $-u_{xx}$, rather than estimating $F^{\prime \prime }$ after differentiating. Integration by parts yieldsThus $d\Vert u_x{\Vert }_2^2/dt\le (M^2/\epsilon )\Vert u_x{\Vert }_2^2$. Applying the Gronwall inequality givesOnly the assumed bound on $F^{\prime }$ is used; a global bound on $F^{\prime \prime }$ is not needed for this energy estimate.

For the sharp energy estimate, $C_0=0$ stays uniform while the available bound $C_1=M^2/\epsilon$ diverges as $\epsilon \downarrow 0$ when $M>0$. If the coarser estimate is used for the first part, both displayed bounds $M^2/\epsilon$ diverge, but the first divergence is only an artifact of discarding an exact cancellation.

The method of characteristics explains why uniform control of the gradient cannot generally persist for the inviscid scalar conservation law. Before characteristic crossing, with initial data $u_0$,If $F^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )<0$ somewhere, the denominator reaches zero in finite positive time. The solution steepens and the smooth description breaks down; an entropy solution can subsequently contain shocks. Positive viscosity replaces such a discontinuity by a thin smooth layer, which can have large gradient even while its $L^2$ norm remains controlled. The divergent bound does not assert that every flux and every initial datum form a shock; a linear flux, for example, has no such steepening.

Insert the travelling wave into the viscous scalar conservation law. With $s=x-\sigma t$ the equation becomesand one integration givesFor a nonconstant profile, $Q(v(s))$ never vanishes. Indeed, the autonomous ordinary differential equation $v^{\prime }=Q(v)/\epsilon$ has unique local solutions since $Q$ is $C^1$; reaching an equilibrium would force the whole solution to be constant. Separation and $c=v(0)$ therefore giveA different choice of reference point gives $s-s_0$ on the left, expressing the translation freedom of the travelling wave.

The printed formula needs a nonconstant-profile qualification. Constant profiles also solve the PDE, but their denominator vanishes at their constant value, so the separated integral is not defined. They must be included separately as equilibrium solutions of the integrated ordinary differential equation.

The integrated travelling wave equation is $v^{\prime }=Q(v)/\epsilon$. A finite limiting value $a$ at either end must satisfy $Q(a)=0$. Otherwise continuity of $Q$ makes $v^{\prime }$ eventually have a fixed sign and an absolute value bounded below, which is incompatible with convergence to $a$. ThereforeSubtracting yields the Rankine-Hugoniot conditionFor distinct end states this givesDistinctness is needed for the printed quotient. If $u_l=u_r$, the identity is $0=0$. A nonconstant global profile is strictly monotone by the scalar ordinary differential equation, so cannot have equal finite end states. The equal-state profiles here are constant and their representation allows any $\sigma$.

Fix $u_l>u_r$ and the Rankine-Hugoniot condition speed from the preceding part. The new hypothesis is a uniformly convex scalar flux, $F^{\prime \prime }\ge \kappa >0$; it replaces the globally bounded-$F^{\prime }$ hypothesis of part (a). Choose $b=\sigma u_r-F(u_r)$. Thensince a strictly convex function lies below the chord between its two endpoint values. Also $Q(u_l)=Q(u_r)=0$ andThe zeros at the endpoints are simple, so the separated integral diverges logarithmically there. Thus the travelling wave is a decreasing connection defined for all $s$, unique up to translation.

To specify a limit, fix a number $c\in (u_r,u_l)$ independently of $\epsilon$ and normalize $v_{\epsilon }(0)=c$. If $V^{\prime }=Q(V)$ and $V(0)=c$, uniqueness gives $v_{\epsilon }(s)=V(s/\epsilon )$. ConsequentlyAt $x=\sigma t$ the normalized profile equals $c$ for every $\epsilon$. This single-line value is immaterial to the weak solution. Convergence holds pointwise off the line and in $L_{\mathrm{loc}}^1$ by bounded convergence; it cannot be uniform across a nonzero jump. The transition has thickness of order $\epsilon$.

This is the vanishing viscosity approximation to a compressive entropy shock. The Rankine-Hugoniot condition makes the step a weak solution of the inviscid scalar conservation law, while $F^{\prime }(u_r)<\sigma <F^{\prime }(u_l)$ means characteristic curves enter the shock from both sides. For every smooth convex function $\eta$ used as an entropy, with entropy flux for a scalar conservation law $q^{\prime }={\eta }^{\prime }{F}^{\prime }$, the viscous equation givesAgainst compactly supported tests the right side tends to zero, since $u_{\epsilon }$ stays in $[u_r,u_l]$. Passing to the $L_{\mathrm{loc}}^1$ limit yields the entropy inequality, explaining the direction selected by positive viscosity.