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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.

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.

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.

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.

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.

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.

Put $h=|u-v|$, $q=Q(u,v)$ and $h_0=|u_0-v_0|$. The mean value theorem gives $|q|\le Mh$ on the bounded state range. Fix a final time $t$ and let $L(\tau )=a-M(t-\tau )$ and $R(\tau )=b+M(t-\tau )$. Choose a smooth increasing function $\rho$ equal to zero on $(-\infty ,0]$ and one on $[1,\infty )$. For small $\delta >0$, defineIts two nonnegative derivative contributions imply $(W_{\delta }{)}_{\tau }+M|(W_{\delta }{)}_x|\le 0$. Therefore $h(W_{\delta }{)}_{\tau }+q(W_{\delta }{)}_x\le 0$.

Use the nonnegative test $\phi (\tau ,x)={\eta }_{ϵ}(\tau )W_{\delta }(\tau ,x)$, where ${\eta }_{ϵ}(0)=1$, it is one up to just before $s$, decreases smoothly to zero near $s$, and remains zero afterward. The Kato inequality for scalar conservation laws givesbecause the other interior contribution is nonpositive. At Lebesgue times $s$, the temporal approximate identity tends to $\int h(s)W_{\delta }(s)$; take a countable sequence $\delta \downarrow 0$ and use dominated convergence on the bounded cone. This yieldsfor almost every $s\in [0,t]$. This is local L1 contraction for scalar conservation laws. It uses only local integrability, so the bounded data need not have finite global $L^1$ norm. If $M=0$, the flux difference is zero and the same proof uses a stationary interval.

If $u_0=v_0$, the right-hand side of the local L1 contraction for scalar conservation laws estimate is zero. Applying it on a countable collection of rational intervals and final times proves $u=v$ almost everywhere in space-time. Thus

To deduce nonnegativity one needs a one-sided comparison, rather than merely putting $v=0$ in the absolute-value estimate. The difference $w=u-v$ satisfies the weak equation with flux $f(u)-f(v)$. Subtract its weak integral identity from the Kato inequality for scalar conservation laws and divide by two. The result is the same inequality forWhen $w\ge 0$ both expressions vanish; when $w<0$, $|q_{-}|\le M{w}_{-}$. The shrinking-interval proof therefore bounds the negative part by its initial negative part. This is order preservation for scalar entropy solutions.

The constant $v=0$ is an entropy solution for every $f$, even if $f(0)\ne 0$, because the constant flux has zero spatial derivative. If $u_0\ge 0$, its initial negative part is zero, and comparison givesThis explicitly establishes the additional sign conclusion without incorrectly identifying uniqueness alone with order preservation.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.