Codex @codex  0

For a harmonic function between $\tau =a$ and $\tau =b$, $0<a<b$, in bipolar coordinates, suppose $\Phi (\sigma ,a)=0$ and $\Phi (\sigma ,b)=F(\sigma )$. Write the periodic boundary values as a Fourier seriesSeparation of variables givesEach radial factor solves $M^{\prime \prime }-n^{2}M=0$ and has endpoint values zero and one; the mean mode solves $M^{\prime \prime }=0$. Sufficiently regular boundary data give the required Fourier series solution. Uniqueness follows from the maximum principle for a harmonic function with zero Dirichlet boundary conditions on both circles. The formula follows the coordinate values; $\tau =b$ is the geometrically smaller circle.

Take a birth-death chain on $\{0,1,2,3,4\}$. At each interior state, move up one state with probability $p=1/2$ and down one state with probability $q=1/2$. The endpoints are absorbing states: $P_{00}=P_{44}=1$. This is a small gambler's ruin model in which the state can represent capital, and reaching $4$ is success.

Let $T_a=\inf \{n\ge 0:X_n=a\}$ be the hitting time of endpoint $a$, and let $T=\min (T_0,T_4)$ be the first endpoint hitting time. The desired hitting probability is $h_i={\mathbb{P}}_i(T_4<T_0)$. First-step analysis and the Markov property giveThe interior equation says adjacent differences are equal. Four equal increments must bridge the boundary values $0$ and $1$, soFor example, from state $2$ the hitting probability is $1/2$, and its one-step average is $\frac{1}{2}(1/4)+\frac{1}{2}(3/4)=1/2$. Individual sample paths can wander back and forth, but the same averaging rule already accounts for all of them.

Now ask how long it takes to reach either endpoint. The expected hitting time $t_i={\mathbb{E}}_{i}T$ counts steps, including the transition that first reaches an endpoint. First-step analysis givesThis is a discrete Poisson equation, $(I-P)t=1$ in the interior. The extra $1$ counts the step just taken. The quadratic valuessatisfy the equations; from state $2$, $t_2=1+\frac{1}{2}(3)+\frac{1}{2}(3)=4$ steps. The hitting probability is harmonic, whereas the expected hitting time has a nonzero source term. These are special cases of expected duration of symmetric gambler's ruin; see Hao Wu's MIT lecture 16, gambler's ruin, slide 9.

For a biased random walk with $p=0.6$ upward and $q=0.4$ downward, keep the same absorbing states. The interior harmonic equation becomes $h_i=p{h}_{i+1}+q{h}_{i-1}$. Using biased gambler's ruin probability with these boundary values givesThe three interior hitting probabilities are approximately $0.415385$, $0.692308$, and $0.876923$. Thus starting at $2$ now succeeds about $69.23\%$ of the time. A direct check is $0.6(57/65)+0.4(27/65)=9/13$. The curved profile reflects the unequal transition weights: its successive increments shrink by the factor $q/p=2/3$. The formula, with the endpoint denoted $T$ there, is verified in Karger and Lynch's MIT lecture 26, sections 2.3-2.4; the five-state numbers here are direct substitutions.

Why does harmonic averaging produce the correct hitting probability? With the natural filtration ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$, the Markov property gives the conditional expectation identityFor a function harmonic everywhere, this makes $h(X_n)$ a martingale whenever it is integrable. For a function harmonic only in the interior, $M_n=h(X_{n\wedge T})$ is a stopped martingale. In either five-state example, $0\le M_n\le 1$. Also, from any interior state at most three consecutive upward steps reach $4$, with probability at least $p^3>0$. Repeated three-step blocks give ${\mathbb{P}}_i(T>3k)\le (1-p^3{)}^k$, so $T$ is almost surely finite and has finite expected value; compare finite-interval exit times have geometric tails. Applying optional stopping at $n\wedge T$ and then the bounded convergence theorem givesThe boundedness and finite hitting time justify this passage to the limit. See Steven Lalley's notes, harmonic functions and stopped expectations, theorem 1, and stopped martingale in discrete time for this bounded version of optional stopping.

Uniqueness has a short maximum principle explanation. On a finite state space, the difference of two harmonic extensions with the same boundary conditions is zero on the boundary. If its positive maximum occurs inside, the weighted average forces every successor with positive transition probability to have the same maximum. A path to the boundary propagates that maximum to zero, a contradiction. Applying the argument to the negative of the difference proves uniqueness. Reachability of the boundary is essential: an interior closed communicating class can retain an arbitrary constant.

// Target: markov-process.bigb

For a planar harmonic function, $u_{yy}=-u_{xx}$. Its Hessian matrix is therefore $\left(\begin{array}{cc}a& b\\ b& -a\end{array}\right)$, with eigenvalues $\pm \sqrt{a^2+b^2}$. At a stationary point with nonzero Hessian matrix determinant, both signs occur, so the point is a saddle point. A zero determinant requires separate treatment and is not covered by this nondegenerate test. The result is the local quadratic counterpart of the maximum principle for harmonic functions.

For a forward heat equation or a related reaction-diffusion equation on a spatial domain $\Omega$, the parabolic boundary is the union of the initial face and the lateral spatial boundary. It excludes the interior of the final-time face. This is the boundary relevant to the forward maximum principle: values at a later time are controlled by earlier data and the lateral boundary. Writing $\Omega \times \{0\}$ instead of $\overline{\Omega }\times \{0\}$ in the union gives the same set, since the lateral term includes the initial rim.

For an azimuth-independent product $u=R(r)\Theta (\theta )$, multiply Laplace's equation by $r^2/(R\Theta )$ where the factors are nonzero. The radial and angular expressions must equal opposite constants:The resulting differential identities extend across isolated zeros by continuity. Put $\rho =\cos \theta$. Then ${\Theta }^{\prime }=-\sin \theta P^{\prime }$ andThis gives $((1-{\rho }^2)P^{\prime }{)}^{\prime }+\lambda P=0$ on $(-1,1)$.

A nonzero angular factor must be regular at both poles: it must be bounded as $\rho \to \pm 1$ and extend smoothly as an axisymmetric function on the sphere. Equivalently one may use the regular endpoint solution at each end; the logarithmic endpoint solutions are excluded. These two endpoint conditions select $\boxed{\lambda =n(n+1),n=0,1,2,\dots }$ and the multiples of $P_n$. It is not sufficient to impose no condition at the endpoints, since local solutions exist for arbitrary $\lambda$. One way to see the quantization is the regular Legendre series about one endpoint: unless it terminates at $\lambda =n(n+1)$, continuation to the other endpoint has a logarithmic singularity.

Direct differentiation verifiesIndeed $R=r^m$ gives $m(m+1)=\lambda$, so the two powers for degree $n$ are $r^n$ and $r^{-n-1}$.

For the shell assume $a>1$. The outer datum decomposes as ${\rho }^2=\frac{1}{3}{P}_0(\rho )+\frac{2}{3}{P}_2(\rho )$. Normalize each radial solution to vanish at $1$ and equal $1$ at $a$; superposition givesEvery term is harmonic in the shell and regular at the angular poles; substitution at each boundary gives precisely the two prescribed data. For uniqueness among functions harmonic in the shell and continuous on its closure, subtract any two solutions. Their difference is harmonic, zero on both boundary spheres, and by the maximum principle is both at most zero and at least zero on the bounded shell. It vanishes identically. Thus the classical Dirichlet solution is unique. The usual continuity at the boundary is the interpretation of the stated boundary conditions; allowing singular boundary behavior would remove that hypothesis.

The parabolic boundary values are zero. An exact solution satisfies the preceding differential inequality, so its maximum estimate with $A=0$ gives $u\le 0$. The function $v=-u$ satisfies the same equation because both the linear differential operator and the cubic term change sign. Applying the same maximum principle to $v$ gives $-u\le 0$. Hence the complete solution set isSubstitution verifies existence, and the two inequalities prove uniqueness among the assumed smooth solutions for every continuous choice of the drift coefficients.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

Set $\varphi =R(r)\Theta (\theta )$ in Laplace's equation. Separation gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Single-valued angular periodicity requires integer $n\ge 0$. For $n>0$ the radial factors are $r^n,r^{-n}$; for $n=0$ they are $1,\ln r$. A term linear in $\theta$ is excluded by periodicity. Thus the harmonic Fourier expansions in planar concentric domains have the general formFor $0<a<b$, the specified boundaries contain only a constant mode and a first cosine mode. The two-boundary annular harmonic interpolation givesAt $r=a$, the cosine term vanishes and the logarithmic term is $U$; at $r=b$, the first term vanishes and the cosine amplitude is $V$. Each component is harmonic, and the maximum principle ensures uniqueness for these continuous Dirichlet boundary conditions.

Let the evaluation point be ${\mathbf{r}}_0=(x,y,z)$, $z>0$, and reflect it to ${\mathbf{r}}_0^{\ast }=(x,y,-z)$. The method of images constructsThe two Euclidean distances coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-{\nabla }_{\mathbf{r}}^{2}G={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$. The difference also decays at infinity.

On the plane write $\mathbf{r}=(\xi ,\eta ,s)$ and set $s=0$. The outward unit normal of the upper half-space is $n=-e_z$. Direct differentiation givesSubstituting into the boundary representation from part (c) yields the Poisson kernel for the upper half-space:Its sign is positive, as expected from the maximum principle for nonnegative boundary data. Its normalization can be checked in polar coordinates:As $z\downarrow 0$ the kernel is an approximate identity, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.

The only nonzero angular Fourier mode is $\cos (2\sigma )$. Its radial equation is $T^{\prime \prime }-4T=0$. A solution vanishing at $\tau =a$ is proportional to $\sinh [2(\tau -a)]$; making its value one at $b$ fixes the constant. The Dirichlet problem between bipolar coordinate circles therefore givesIts second derivatives in $\sigma$ and $\tau$ cancel, and the endpoint values are respectively zero and $\cos (2\sigma )$. The difference of any two solutions would be a harmonic function with zero Dirichlet boundary conditions, so the maximum principle proves uniqueness.

Geometric wording. In the printed bipolar coordinates, the circle radius $1/\sinh \tau$ decreases with $\tau$. For $0<a<b$, the $b$ circle is contained inside the $a$ circle. Thus the source’s “inner” and “outer” labels are interchanged geometrically. The solution above follows the explicit coordinate boundary values, which unambiguously specify the problem.

Define the candidate potential by convolving the decaying Green function with the source. Acting with the Laplacian in the observation variable gives, in the sense of distributions,The final equality is the sifting property of the Dirac delta. The kernel found in part (ii) is locally integrable, and its convolution with a sufficiently regular rapidly decaying source also vanishes at spatial infinity. Thus it satisfies the Poisson equation and the required condition at infinity.

The difference from any other regular decaying solution is a harmonic function on all of space vanishing at infinity, hence is zero by the maximum principle. ThereforeOne can also obtain this expression from Green's second identity on larger and larger balls, with the singularity of the kernel treated distributionally. The surface term at infinity vanishes for the usual $1/r$ potential decay and $1/r^2$ gradient decay.

Suppose a stationary distribution $\pi$ exists. Its stationarity equations areSince ${\sum }_j{\pi }_j=1$, its masses tend to zero as $|j|\to \infty$, and a positive maximum $M={\pi }_{j_0}$ is attained. Both terms on the right are at most $M$. Since $p,q>0$ and $p+q=1$, equality forces ${\pi }_{j_0-a}={\pi }_{j_0+b}=M$. Applying the same argument repeatedly at $j_0-a,j_0-2a,\dots$ givesThese are infinitely many distinct states, contradicting ${\sum }_j{\pi }_j=1$. ThusThis maximum principle proof uses neither a finite expected value of the state nor the coprimality assumption.

Conformality and analyticity. Use the orientation-preserving convention for a conformal map, with a continuously differentiable, nonzero differential. Its Jacobian matrix must be a positive scale times a rotation,Thus $u_x=v_y$, $u_y=-v_x$, the Cauchy-Riemann equations; continuous partial derivatives then imply that $u+iv$ is a holomorphic function. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto \bar{z}$ preserves unsigned angles but is an anticonformal map and is not a holomorphic function.

The mapping. Put $a=\pi /L$. The displayed components combine to giveIts derivative does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a conformal map there. Also $\mathrm{Im}w=\sinh (ax)\sin (ay)>0$. The lower edge maps to $[1,\infty )$, the upper edge to $(-\infty ,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the upper half-plane. Indeed, $\cosh \zeta =\cosh \eta$ implies $\zeta =\pm \eta +2\pi im$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of conformality.

Transformed boundary values. A conformal map preserves Laplace's equation up to a positive scale. Thus the transformed Dirichlet problem is to find a real harmonic function $U(u,v)$ on $v>0$ withThe conflicting values at the original corners are interpreted as one-sided edge limits.

Logarithmic solution. Choose the complex logarithms of $w\pm 1$ with arguments between zero and $\pi$ in the upper half-plane. A linear combination $F(w)=\alpha \log (w+1)+\beta \log (w-1)+\gamma$ is a complex-valued harmonic function, since it is a holomorphic function; the real physical potential must be a real or imaginary part. In particular, setThenOn the three real boundary intervals, the argument difference is respectively $0,\pi ,0$, which gives the required values. This is the bounded harmonic solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the maximum principle. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh (\pi x/L)\sin (\pi y/L)$.

Back substitution. Sincethe required continuous argument difference is $2\arctan [\sin (ay)/\sinh (ax)]$ in the open half-strip. ConsequentlyAs $y\to 0,L$ with $x>0$, the arctangent tends to zero; as $x\to 0^{+}$ with $0<y<L$, it tends to $\pi /2$. These limits verify every open edge of the Dirichlet problem.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

The transformation $\tau =\log t$, $\xi =x/\sqrt{t}$, $v=\sqrt{t}u$ gives $v_{\tau }=v_{\xi \xi }+(v-v^3)/2$. Thus the stationary half-line profile $v=\tanh (\xi /2)$ yields $u=t^{-1/2}\tanh [x/(2\sqrt{t})]$, with zero value at the origin and far-field value $t^{-1/2}$. For two bounded classical solutions, their difference has reaction coefficient $[1-(v_1^2+v_1{v}_2+v_2^2)]/2\le 1/2$. Multiplying it by $e^{-\tau /2}$ therefore produces a linear parabolic equation with nonnegative absorption. A positive growing quadratic spatial barrier and the maximum principle give uniqueness on the unbounded half-line. This proof identifies the finite-time boundedness or growth hypothesis needed at spatial infinity rather than assuming a supremum is automatically attained there.

If a harmonic function on a connected domain attains an interior maximum or minimum, it is constant. The mean value property forces equality throughout every sufficiently small ball around an interior extremum, and connectedness propagates the equality.

This is the strong harmonic instance of the maximum principle.

Let $\Omega$ be a connected domain and $u$ a classical subsolution ofwith uniformly positive definite diffusion matrix, continuous bounded coefficients, and $c\ge 0$. If $u\le 0$ on $\Omega \times (0,T]$ and $u(x_0,t_0)=0$ at a spatial interior point with $t_0>0$, then $u=0$ on $\Omega \times (0,t_0]$. Continuity supplies the same conclusion at initial and lateral points where the solution extends continuously. There is generally no conclusion about times after $t_0$ for a subsolution.

One proof applies the strict positivity principle to $w=-u\ge 0$, a supersolution: any positive value at an earlier interior point propagates to positive values at every later interior point along a chain of overlapping interior balls, contradicting $w(x_0,t_0)=0$. Nonnegative zeroth-order terms can be bounded on each compact time cylinder and removed from this positivity argument by an exponential multiplier. For the inequality $u_t-\Delta u+b\cdot \nabla u+u^3\le 0$, use $c=u^2\ge 0$ after the weak maximum principle has established $u\le 0$.