Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 105 (Analysis of Partial Differential Equations)

The general-order Cauchy-Kovalevskaya theorem applies to analytic data on a non-characteristic hypersurface. In local coordinates $(t,z)$, a real analytic system of order $m$ in an unknown vector $U$ is in normal form whenwith $F$ real analytic near the initial jet. Analytic data ${\partial }_t^{j}U(0,z)=h_j(z)$ for $0\le j<m$ determine a unique real analytic solution near each initial point. More general systems, including different orders for the components, have the same conclusion after solving for the highest normal derivatives: the corresponding coefficient matrix, or highest-derivative Jacobian for a nonlinear system, must be invertible. This is the system's non-characteristic hypersurface condition. The uniqueness asserted by the Cauchy-Kovalevskaya theorem is initially uniqueness in the real analytic class.

A real analytic coordinate change flattens a real analytic initial hypersurface. Non-characteristicity permits solving for its highest normal derivatives by the real analytic implicit function theorem. Introduce all derivatives through order $m-1$ as additional unknowns. Their normal derivatives are either another jet variable, a tangential first derivative of a jet variable, or the right-hand side of the original equation. This gives a first-order real analytic system. The derivative-compatibility identities have zero initial data and are preserved by the system; equivalently, its real analytic coefficient recurrence reproduces the derivatives of $U$. This explains both reductions, rather than assuming arbitrary real analytic hypersurfaces are already flat.

For the first example, the principal symbol of the Laplace equation at the conormal $d{x}_1$ is $1$, so the plane is a non-characteristic hypersurface. The theorem therefore applies to real analytic $u{|}_{x_1=0}$ and ${\partial }_{1}u{|}_{x_1=0}$:This does not give Hadamard well-posedness in smooth or Sobolev space norms. For $n\ge 2$, the harmonic functionshave zero value data and normal derivative data $e^{-\sqrt{N}}\cos (N{x}_2)$ tending to zero with every fixed tangential derivative. At any fixed $x_1=\delta >0$, however, $u_N(\delta ,0)$ diverges. Thus the elliptic Cauchy problem for a partial differential equation is unstable despite real analytic solvability. The exceptional case $n=1$ is the elementary affine ordinary differential equation and has no such tangential high-frequency instability.

For the wave equation, the principal symbol is $-{\tau }^2+|\xi {|}^2$, so its value on the conormal $dt$ is $-1$. Consequently the Cauchy-Kovalevskaya theorem applies to real analytic displacement and velocity data on this non-characteristic hypersurface:In addition, this Cauchy problem for a partial differential equation has Hadamard well-posedness for nonanalytic data in the usual energy spaces. For example, data in $H^1({\mathbb{R}}^n)\times L^2({\mathbb{R}}^n)$ give a unique solution in $C_t{H}^1\cap C_t^1{L}^2$ on bounded time intervals, with continuous dependence; conservation of $\int (|u_t{|}^2+|\nabla u{|}^2)$ is its basic estimate. Thus real analytic existence and energy-space stability are different assertions, and in this example both are available.

For the heat equation, viewed as a second-order equation in all $(t,x)$ variables, the principal symbol is $|\xi {|}^2$. It vanishes on $dt$, so $t=0$ is a characteristic hypersurface andWriting $u_t={\Delta }_{x}u$ does not repair its hypotheses: the right side has second spatial derivatives, exceeding the first-order normal-form allowance. Indeed real analytic initial data need not produce a jointly real analytic solution through time zero. In one dimension, $u_0(x)=(1+x^2{)}^{-1}$ would forceThe time Taylor coefficients $(-1{)}^k(2k)!/k!$ have zero radius of convergence.

Nevertheless the forward heat equation has Hadamard well-posedness in standard $L^2$ or Sobolev space classes for $t\ge 0$; its Fourier multiplier $e^{-t|\xi {|}^2}$ gives stable smoothing. Backward evolution has the growing multiplier $e^{|t||\xi {|}^2}$ and is unstable. Neither forward stability nor backward instability is decided simply by whether the real analytic theorem applies.

Write the order-$m$ principal symbol as $p_m(x,y;\xi ,\eta )={\sum }_{|\alpha |=m}{a}_{\alpha }(x,y){\xi }^{{\alpha }_1}{\eta }^{{\alpha }_2}$. The conormal of the plane is $(1,0)$. Therefore the non-characteristic hypersurface hypothesis is exactlyIn particular its value at the origin is nonzero, and continuity provides a uniform positive lower bound for its modulus in a sufficiently small neighborhood. No lower-order coefficient enters this condition.

Repeated integration by parts against a compactly supported test function gives the bilinear formal adjointThere is no boundary remainder because $v$ has compact support. The product rule makes this an order-at-most-$m$ linear differential operator with real analytic coefficients. The displayed integral pairing has no complex conjugation, so this is the formal transpose; for a sesquilinear pairing the coefficients would be conjugated.

For uniqueness, suppose $Q$ is the difference of two candidates. Then $\int uQv=0$ for every smooth compactly supported $u$ and $v$, so $Qv=0$ for every such $v$. At an interior point, a polynomial times a cutoff can prescribe every derivative of $v$ through order $m$ independently. Hence each coefficient of $Q$ is zero. This proves uniqueness as a differential operator, not only uniqueness of its integrated action.

Taken literally, the printed condition is insufficient: the choice of which factor vanishes may depend on the multi-index. For a counterexample already at $m=2$, let $P={\partial }_x^2$, let $\Omega =(0,a)\times (-b,b)$ be a small rectangle compactly contained in the unit ball, and choose $\chi \in C_c^{\infty }((-b,b))$ nonnegative with positive integral. PutOn the vertical edges $u=0$, and on the horizontal edges $u$ and its derivatives vanish. Every first derivative of $v$ is zero. Thus for every $|\beta |\le 1$ the same-index alternative in the PDF holds at every boundary point. But

The valid complete boundary-jet condition for formal adjoints is that, at each boundary point, either all derivatives of $u$ through order $m-1$ vanish or all such derivatives of $v$ vanish. On a bounded piecewise smooth domain with smooth coefficients up to its closure, repeated integration by parts produces boundary terms of the form $({\partial }^{\gamma }u){\partial }^{\delta }(a_{\alpha }v)$, multiplied by a normal component, with $|\gamma |,|\delta |\le m-1$. Expanding the second factor shows that every summand vanishes under this stronger whole-jet alternative. Hence the desired identity holds under that condition. The later cap argument satisfies this stronger condition face by face, so it remains valid despite the literal error in this subpart. Boundary regularity and finite integrals are also needed for a general noncompact-support integration formula; the caps used below have them.

The order-$m$ principal symbol of the formal adjoint is $p_m^{\ast }=(-1{)}^m{p}_m$. Derivatives falling on coefficients contribute only lower-order terms. Thus the original plane remains a non-characteristic hypersurface.

For the curved face use the defining function ${\rho }_{\epsilon }=x+y^2-\epsilon$, whose conormal is $(1,2y)$. On the face with $x>0$, $|y|<\sqrt{\epsilon }$ and $0<x\le \epsilon$. Therefore $(x,y,1,2y)$ tends uniformly to $(0,0,1,0)$ as $\epsilon \downarrow 0$. Since $p_m^{\ast }(0,0;1,0)\ne 0$, continuity givesThe same argument supplies a common neighborhood where the transformed coefficient of the highest normal derivative is nonzero, permitting the real analytic normal form used below.

The coordinate change has inverse $x=\tilde{x}-{\tilde{y}}^2+\epsilon$, $y=\tilde{y}$. Thus the coefficient function needed after changing variables isThe PDF's displayed formula has an undefined subscript $f_{\alpha }$ and writes the forward change in the argument. It is not the coefficient pullback by the stated coordinates; the inverse expression above is the one used in transforming the equation. The forward polynomial composition, if intended instead, has the same type of uniform bounds, but is not that pullback.

For $|(\tilde{x},\tilde{y})|\le 1/2$ and $0\le \epsilon \le 1/32$, the inverse image lies in the fixed compact subset specified by $|x|\le 25/32$, $|y|\le 1/2$, whose Euclidean radius is less than one. A real analytic function extends holomorphically to a complex neighborhood of this compact set: local extensions agree on overlaps by uniqueness, and a finite covering gives a common smaller neighborhood. The polynomial inverse maps a fixed small complex neighborhood of the closed half-ball into that neighborhood, uniformly in $\epsilon$. The extensions are bounded there by a common constant $A$. Applying the multivariable Cauchy estimate on equal-radius polydiscs givesThis is uniform analyticity under polynomial coordinate changes. The argument gives a compact margin and a common complex radius, not merely separate analyticity for each parameter.

In the flattened coordinates, ${\partial }_x={\partial }_{\tilde{x}}$ and ${\partial }_y=2\tilde{y}{\partial }_{\tilde{x}}+{\partial }_{\tilde{y}}$. The transformed formal adjoint consequently has coefficients obtained from finitely many real analytic coefficient pullbacks and polynomial factors. They have uniform analyticity bounds on a fixed small ball. By the preceding non-characteristic argument, its coefficient of ${\partial }_{\tilde{x}}^m$ has modulus bounded away from zero there, uniformly for small $\epsilon$. Division by it preserves uniform real analytic bounds.

The transformed forcing ${\tilde{g}}_{\epsilon }$ has analogous bounds, with constants that may depend on $g$. Apply the Cauchy-Kovalevskaya theorem in normal form with zero normal data through order $m-1$ at $\tilde{x}=0$. The supplied majorant series radius result yields a common solution radius $r_g>0$ for all sufficiently small parameters. Tangential differentiation of zero normal data shows that the entire boundary jet through order $m-1$ is zero; changing coordinates preserves this vanishing.

For $(x,y)$ in the closed cap, $|\tilde{x}|\le \epsilon$ and $|\tilde{y}|\le \sqrt{\epsilon }$. Choose $\epsilon$ so small that $\sqrt{{\epsilon }^2+\epsilon }<r_g$, with a strict margin. The whole closed cap then lies in the common real analytic solution neighborhood. Pulling the solution back provesIn fact $v$ is real analytic on a neighborhood of its closure, which is useful for the boundary integration below.

Fix the coefficient normal form and a positive analyticity exponent ${\zeta }_0$ once and for all. If $g$ is a polynomial, its pullback ${\tilde{g}}_{\epsilon }=g(\tilde{x}-{\tilde{y}}^2+\epsilon ,\tilde{y})$ is again a polynomial, with degree bounded independently of $\epsilon$. On a fixed closed coordinate ball, its finitely many nonzero derivatives are uniformly bounded as the parameter ranges over a fixed compact small interval. Hence there is a finite $A_g\ge 1$ such thatAll higher derivatives vanish, so arbitrarily high polynomial degree causes no failure of this bound; it only changes $A_g$.

The normalized forcing belongs to one common majorant class. The Cauchy-Kovalevskaya theorem therefore gives a radius $r_{\ast }>0$ independent of $g$. The problem is linear with zero data, so multiplying the normalized solution by $A_g$ solves the original forcing without changing its radius. Choose one $\epsilon$ with the cap inside this radius. ThusThis is the uniform Cauchy radius for polynomial forcing. It asserts a common domain, not a uniform solution-amplitude bound over all polynomials.

For each polynomial $g$, take the real analytic adjoint solution on the common cap from the preceding part. On the flat boundary face the full boundary jet of $u$ vanishes; on the curved face that of $v$ vanishes. The complete boundary-jet condition for formal adjoints, proved under its corrected hypothesis, therefore applies. The two corner points have zero boundary measure, and the cap is a bounded piecewise smooth domain. It follows thatBy the Weierstrass approximation theorem, polynomials uniformly approximate $\bar{u}$ on the compact cap closure. Taking the limit gives ${\int }_{{\omega }_{\epsilon }}|u{|}^2=0$. Continuity then gives $u=0$ pointwise in the cap. For real $u$, simply approximate $u$ itself.

A positive cap alone is not a neighborhood of the origin. Apply the same argument after reflecting $x\mapsto -x$; the reflected operator remains real analytic and non-characteristic. Shrinking the two parameters to their minimum proves vanishing onThis set is open and contains the origin, including the flat face where the value data vanish. The argument is the parabolic-cap proof of Holmgren uniqueness, and it establishes Holmgren uniqueness theorem for nonanalytic $C^m$ solutions here.

Use the interior $\Omega =\{1/2<|x|<2\}$ for the Sobolev space notation; the closed shell in the PDF specifies its boundary. Put $A(x)=\mathrm{diag}(1,1,g(|x|))$. A weak solution has zero boundary trace in the sense of the Sobolev trace theorem, equivalently $u\in H_0^1(\Omega )$, and satisfiesThe minus sign is necessary because the operator in the paper is positive divergence, not its negative. One can first test with $C_c^{\infty }$ functions and then use density in the zero-boundary Sobolev space. Smoothness of $g$ makes its restriction to the compact radial interval bounded, so both sides define continuous functionals on this space. Merely belonging to $H^1$ without the zero trace would omit the boundary condition.

Positivity and continuity of $g$ give $\mu ={\min }_{[1/2,2]}g>0$ and a finite maximum $G$. Hence, for every $x$ in the shell and every $\xi \in {\mathbb{R}}^3$,This is the uniformly elliptic operator condition. Positivity at separate points without a compactness/continuity lower bound would not alone supply a common ellipticity constant.

Test the weak formulation with $u$ itself. Uniform ellipticity and the Cauchy-Schwarz inequality giveThe Poincaré inequality for the bounded shell with zero trace gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$. If the gradient is nonzero, divide to obtain $\Vert \nabla u{\Vert }_2\le C_P{\lambda }^{-1}\Vert f{\Vert }_2$; if it is zero the estimate is immediate and the zero trace forces $u=0$. ThusThe constant depends on the shell and $g$, not on $u$ or $f$. Subtracting two solutions with the same forcing gives a zero-forcing solution and hence proves uniqueness. This is a coercive energy argument, not an appeal to real analytic Cauchy solvability.

A fixed smooth choice $g\equiv 0$ already gives a counterexample. Select nonzero smooth functions $\varphi (x_1,x_2)$ and $\chi (x_3)$ with product support compactly contained in a small box inside the shell. LetThen $u_k$ has zero boundary trace and is a classical, hence weak, solution. The $L^2$ norms of $f_k$ are bounded independently of $k$. On the other hand,The squared norm of $\chi \cos (k{x}_3)$ tends to $\frac{1}{2}\Vert \chi {\Vert }_2^2$, by the oscillatory integral identity for ${\cos }^2$. The other term has bounded norm, so $\Vert u_k{\Vert }_{H^1}$ grows at least proportionally to $k$. ThereforeThis failure of a coercive Dirichlet estimate under degeneracy loses control in the third direction; it does not rely on varying $g$ with the sequence.

For the spherical coordinates in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin \theta$. Their metric tensor is therefore $\mathrm{diag}(1,r^2,r^2{\sin }^2\theta )$ and the Jacobian determinant is $r^2\sin \theta$. Applying the divergence formula to the gradient givesFor example, the formula follows directly from $\Delta u=|\det h{|}^{-1/2}{\partial }_i(|\det h{|}^{1/2}{h}^{ij}{\partial }_{j}u)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the Laplacian is a smooth geometric operator there as well, described by other charts. The coordinate volume form integrates over $1/2<r<2$, $0<\theta <\pi$, $0\le \varphi <2\pi$, with the usual periodic identification.

Here is a direct use of the spherical representation that also avoids estimating separate coordinate derivatives at the poles. Put $A={\Delta }_{S^2}$, $T={\partial }_r^2+2r^{-1}{\partial }_r$, and write $d\omega$ for unit-sphere area. Then $f=Tu+r^{-2}Au$. Expanding its squared norm with the physical measure givesOn each boundary sphere $u=0$, so every angular derivative, including $Au$ and ${\nabla }_{S}u$, is zero there. Radial and angular integration by parts consequently giveIndeed the $u_{rr}$ term gives $\int |{\nabla }_S{u}_r{|}^2$ with boundary remainder $[\int u_{r}Au]=0$, while the $2u_r/r$ term is $-\int r^{-1}{\partial }_r|{\nabla }_{S}u{|}^2=-\int r^{-2}|{\nabla }_{S}u{|}^2$, its boundary remainder also being zero. Thuswhere the last step uses the preceding $H^1$ energy estimate and $r\ge 1/2$. All unmarked integrals in these two identities are with $drd\omega$.

To control every angular second derivative, the spherical Hessian identity isIt follows by integrating the derivative-commutation identity ${\nabla }^a{\nabla }_a{\nabla }_{b}w={\nabla }_{b}Aw+{\mathrm{Ric}}_b{}^c{\nabla }_{c}w$ and using $\mathrm{Ric}=h$ on the unit sphere. There is no sphere boundary. Hence the preceding bound controls ${\nabla }_S^{2}u$, ${\nabla }_S{u}_r$, and $u_{rr}=Tu-2u_r/r$ in their appropriate weighted norms.

In an orthonormal polar frame the Cartesian Hessian matrix has componentsand $D^{2}u(e_A,e_B)=r^{-2}({\nabla }_S^{2}u{)}_{AB}+r^{-1}{u}_r{\delta }_{AB}$. Since $r$ is bounded above and below, these bounds plus the first-order estimate proveThis is the direct spherical-shell H2 estimate. Intrinsic angular integration combines the suggested angular tests, with no spurious boundary at the polar coordinate singularities.

The coefficients in the spherical Laplacian are independent of $\varphi$, so $[\Delta ,{\partial }_{\varphi }]=0$. Also ${\partial }_{\varphi }$ is tangent to both boundary spheres; differentiating the zero boundary trace gives another zero trace. Therefore $w={\partial }_{\varphi }u$ satisfies $\Delta w={\partial }_{\varphi }f$ and the same homogeneous Dirichlet boundary conditions.

In Cartesian coordinates ${\partial }_{\varphi }=x_1{\partial }_2-x_2{\partial }_1$, a globally smooth rotation field even where angular coordinates are singular. Permuting coordinate axes gives $R_{12}$, $R_{23}$ and $R_{31}$. For each, the rotational commutator for the Laplacian follows directly from $\Delta (x_i{\partial }_{j}u)=x_i{\partial }_j\Delta u+2{\partial }_i{\partial }_{j}u$; antisymmetrizing cancels the extra terms. ThusIf $f$ is radial, all these right-hand sides vanish. Uniqueness from the energy estimate makes every $R_{ij}u$ zero. At each point of a sphere the rotation fields span its tangent plane, so all tangential derivatives of $u$ vanish. Each sphere is connected, giving $\boxed{u(x)=U(|x|)}$. This proves radiality without assuming it as a separation-of-variables ansatz.

The Sobolev embedding theorem $H^2(\Omega )\hookrightarrow L^{\infty }(\Omega )$ in dimension three, combined with the preceding estimate, givesOne direct justification of this embedding is a bounded Sobolev extension operator into $H^2({\mathbb{R}}^3)$, followed by Fourier inversion and the Cauchy-Schwarz inequality: ${\int }_{{\mathbb{R}}^3}(1+|\xi {|}^2{)}^{-2}d\xi <\infty$ bounds the inverse Fourier integral by the $H^2$ norm.

For radial forcing, the preceding part makes $u=U(r)$ and $f=F(r)$. Set $a=1/2$, $b=2$ and $H(r)={\int }_a^r{s}^{2}F(s)ds$. The radial equation gives $r^2{U}^{\prime }(r)=C_0+H(r)$. Both zero boundary values implyMoreover $\Vert H{\Vert }_{\infty }\le ({\int }_a^b{s}^{2}ds{)}^{1/2}({\int }_a^b{s}^2|F{|}^{2}ds{)}^{1/2}\le C\Vert f{\Vert }_2$. Since $r\ge a$, this proves the radial Poisson gradient estimate:

For a nonradial counterexample choose $x_0$ in the shell and a nonconstant $\phi \in C_c^{\infty }(B(0,1))$. For sufficiently small $\epsilon$, set $u_{\epsilon }(x)={\epsilon }^{1/2}\phi ((x-x_0)/\epsilon )$ and $f_{\epsilon }=\Delta u_{\epsilon }$. These are smooth with support away from the boundary. Scaling givesThus the failure of L2-to-Linfinity Poisson gradient bounds in three dimensions is explicit, despite the valid $H^2$ and value-supremum estimates.

The growth bound is $|F(t,x)|\le K(1+|x|)$. A characteristic curve through $(s,a)$ solvesThe $C^1$ field is locally Lipschitz in space, giving local existence and uniqueness. On either direction of a bounded time interval the Gronwall inequality gives $1+|X(t;s,a)|\le (1+|a|)e^{K|t-s|}$. Thus no trajectory can escape to infinity at finite time. On the resulting compact space-time region, local ordinary differential equation existence continues it, proving global definition for all $t\in \mathbb{R}$.

Uniqueness gives the flow composition law and inverse $X(s;t,\cdot )$. Differentiation with respect to $a$ solves the variational equation, making each map a $C^1$ diffeomorphism. Its positive Jacobian determinant isThese facts constitute the global characteristic flow under linear growth. A uniform global bound on $D_{x}F$ is not required; its bounds on each compact characteristic tube suffice.

Write $X(t,a)=X(t;0,a)$ and $A(t,x)=X(0;t,x)$. If $u_0\in C^1({\mathbb{R}}^d)$, the unique global classical solution isIt is $C^1$, and the chain rule makes its value constant along every characteristic curve. The same calculation proves classical uniqueness. No boundedness of $u_0$ is needed for this classical statement.

For $u_0\in L^{\infty }$, a bounded weak solution on the positive half-line is defined byfor every compactly supported $C^1$ test function on $[0,\infty )\times {\mathbb{R}}^d$. The analogous backward-time identity defines the solution on all real time. The $u\mathrm{div}F$ term is essential: this is transport, not the conservative equation $u_t+\mathrm{div}(Fu)=0$.

The same inverse-flow formula defines the unique weak solution, with $\Vert u(t){\Vert }_{\infty }=\Vert u_0{\Vert }_{\infty }$, equality almost everywhere, and a representative continuous into $L^1$ on each compact spatial set. For a direct uniqueness argument, change variables $x=X(t,a)$ in the weak identity and put $w(t,a)=u(t,X(t,a))$. Using $J_t=(\mathrm{div}F)(t,X)J$ givesinitially for $C^1$ compactly supported $\zeta$. Since $J>0$ and $J,J_t$ are continuous on compact sets, approximation in the spatial parameter permits $\zeta =\eta /J$: both $\zeta$ and ${\zeta }_t$ are uniformly approximable there, which are the only derivatives left in this identity. Hence $\int w{\eta }_t+\int u_0\eta (0)=0$ and $w=u_0$ almost everywhere. This also verifies existence by reversing the argument, and avoids assuming second derivatives of the $C^1$ field. Local $L^1$ continuity follows by approximating $u_0$ by continuous functions on compact flow tubes.

This is the weak transport solution under characteristic coordinates. The weak-strong uniqueness principle says that a weak solution in the specified class agrees with a classical solution having the same data for as long as that classical solution exists. Here it holds globally in the bounded class whenever the classical initial data are also bounded.

When the velocity depends on $u$, different initial values travel at different speeds. The resulting characteristic crossing can destroy a classical solution in finite time, even though the initial function is smooth and bounded. For the Inviscid Burgers equation, take $u_0(a)=-\tanh a$. Before crossing,andAt $a=0$ this tends to $-\infty$ as $t\uparrow 1$, while $|u|\le 1$. Thus $\boxed{\text{bounded states can undergo finite-time gradient blow-up and shock formation}.}$ The bounded-derivative Burgers flux agrees with this equation on the whole state range, if a global bounded-speed flux is desired. After crossing, arbitrary weak solutions need not be unique; an entropy solution condition is needed to select the admissible continuation.

A bounded weak solution satisfiesfor every compactly supported $C^1$ test function, of either sign, allowed to meet $t=0$. Across a moving jump $x=s(t)$ with distinct left/right constant states, the distributional delta coefficient vanishes exactly when the Rankine-Hugoniot condition holds:It is required on every interface of a piecewise constant solution; the initial trace must also agree with the data.

For nonuniqueness respecting the paper's global bound on $f^{\prime }$, use the bounded-derivative Burgers flux $f(z)=z^2/2$ on $|z|\le 1$ and $f(z)=|z|-1/2$ outside. It is $C^1$ and $|f^{\prime }|\le 1$. Take data $-1$ for $x<0$, $1$ for $x>0$. A stationary jump obeys the condition because $f(-1)=f(1)$. Another weak solution is the rarefaction waveThe fan solves the equation inside each smooth region, is continuous at its edges, and has the same local $L^1$ initial trace. Thus weak solutions are not unique. The stationary expansion shock violates the entropy inequality, whereas the fan is admissible; using the unmodified quadratic flux alone would miss the printed global bounded-derivative hypothesis.

For a bounded $C^1$ classical solution and a smooth convex entropy $\eta$, let its entropy flux for a scalar conservation law satisfy ${\psi }^{\prime }=f^{\prime }{\eta }^{\prime }$. The chain rule givesIntegration against a nonnegative test function, including its initial boundary contribution, gives equality in the entropy inequality. Approximate a convex piecewise $C^1$ entropy by smooth convex functions on the bounded state range. Their values and flux antiderivatives converge uniformly, so the integrated equality passes to the limit. Therefore every bounded classical solution is an entropy solution. The bounded qualifier is required by the definition's $L^{\infty }$ membership; an arbitrary unbounded $C^1$ solution would not satisfy that membership condition.

Conversely, the affine entropies ${\eta }_{+}(z)=z$ and ${\eta }_{-}(z)=-z$ are both convex, with fluxes $f$ and $-f$. Their inequalities force the same linear weak integral to be both nonnegative and nonpositive. It is therefore zero for every nonnegative test, hence for every test by subtraction of nonnegative compactly supported tests. Thus $\boxed{\text{entropy solution}⟹\text{weak solution}.}$ No assumption that the flux itself is convex is used.

Since the derivative of $|z-k|$ is $\mathrm{sgn}(z-k)$ away from $k$, choose the Kruzhkov entropy fluxOn either side of $k$ its derivative is $f^{\prime }(z){\eta }_k^{\prime }(z)$. The factor $f(z)-f(k)$ vanishes at the change of sign, so the function is continuous and there is no extra delta contribution to its weak derivative. Adding a constant to the flux changes none of the test integrals. Substitution into the entropy inequality pairs $|u-k|$ with this flux, not with an arbitrary antiderivative unrelated to the chosen entropy.

Set $H(a,b)=|a-b|$ and $Q(a,b)=\mathrm{sgn}(a-b)(f(a)-f(b))$. Both are symmetric in $a,b$. For fixed $(s,y)$ apply the entropy inequality for $u$ at the constant level $k=v(s,y)$ with test $\Phi (t,x,s,y)$, then integrate over $(s,y)$. This producesHere the first integral is over both time-space pairs and the second over the three remaining variables. Repeat with $v$, level $u(t,x)$, integrating over $(t,x)$, to obtain the analogous terms with ${\Phi }_s,{\Phi }_y$ and initial value $v_0$.

Adding and using symmetry giveswhere $B_u=\int H(u_0(x),v(s,y))\Phi (0,x,s,y)dsdxdy$ and $B_v=\int H(u(t,x),v_0(y))\Phi (t,x,0,y)dtdxdy$. All integrals are finite on the test's compact support because the states are bounded. The levels being inserted are constants relative to the variables in each entropy inequality; there is no differentiation of the merely measurable other solution. Fubini's theorem and continuity of $H,Q$ justify the integration. This is doubling of variables for scalar conservation laws.

For the stated mollifier, $({\partial }_t+{\partial }_s){\chi }_{\epsilon }=({\partial }_x+{\partial }_y){\chi }_{\epsilon }=0$. Thus the interior doubled integral isOn a bounded state range, $H$ is Lipschitz in each argument, and so is $Q$: away from the diagonal each partial derivative has modulus at most $\sup |f^{\prime }|$, and the function is continuous across the diagonal. Local $L^1$ translation continuity therefore makes this converge to the corresponding single-time, single-space integral. The compact time support is chosen below $T$, with a margin for $s\le t+2\epsilon$.

The initial terms need a separate argument. Since the time kernel is supported at $t-s\in [-2\epsilon ,-\epsilon ]$, $B_v$ is identically zero for $t\ge 0$, whereas $B_u$ samples $v(s,\cdot )$ at $\epsilon \le s\le 2\epsilon$.

The averaged initial trace of an entropy solution follows directly from its inequalities. Test the constant-level entropy for $v$ with $(1-s/\delta {)}_{+}\rho (y)$, using smooth approximations of that temporal cutoff. Boundedness of $v$ and its entropy flux givesApproximate $v_0$ locally in $L^1$ by finitely many constants with a smooth nonnegative partition of unity. Applying this estimate on each partition member and using the triangle inequality makes the limsup of ${\delta }^{-1}{\int }_0^{\delta }\Vert v(s)-v_0{\Vert }_{L^1(K)}ds$ arbitrarily small. This proves its convergence to zero for each compact $K$.

The time kernel has size $O({\epsilon }^{-1})$ and the spatial kernel is a normalized approximate identity. The averaged trace, followed by spatial $L^1$ translation continuity of $v_0$, therefore gives $B_u\to \int |u_0-v_0|\phi (0)$. Its time mass is one, not one-half, because its whole support lies on the positive $s$ side. ConsequentlyThis is the Kato inequality for scalar conservation laws with its initial contribution. Assuming an initial trace solely from interior translation continuity would leave a gap; the entropy argument above supplies it.