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Let $I_C[F]=(2\pi {)}^{-1}{\int }_C{e}^{ikx-(k^2-i\alpha k)t}F(k)dk$. The drift-diffusion half-line global relation and the half-line drift null contour identity give the formula in the secondary title. Its real-space kernel is the difference of two drifted Gaussian heat kernels, and its boundary-forcing kernel is $x(2\sqrt{\pi }{t}^{3/2}{)}^{-1}\exp [-(x+\alpha t{)}^2/(4t)]$. This kernel is an approximate identity in time as $x\downarrow 0$, which verifies the Dirichlet boundary condition.

For diffusion with a constant diffusion coefficient $D>0$ and no sources, the model flux is $J=-D\nabla u$. A continuity equation gives $u_t=-\nabla \cdot J=D\Delta u$. This proportionality is a constitutive assumption about transport; the continuity equation expresses conservation. If $D$ varies with position, it must remain inside the divergence, giving $u_t=\nabla \cdot (D\nabla u)$ instead.^[ref]

An exact constructed heat equation example on the whole plane uses dimensionless coordinates with $D=1$ and the smooth initial Gaussian function $u(x,y,0)=e^{-r^2}$. Set $s=1+4t$. For $t\ge 0$,Direct differentiation verifies both the initial condition andThe center decreases, while the region $r^2>s$ increases at that instant. The sign-change circle expands with radius $\sqrt{s}$. A fixed outer location can initially increase and later decrease as this circle passes it.

At time $t$, the peak is $1/s$, and the radius where the profile is $1/e$ times its peak is $\sqrt{s}$. The width grows as the height falls. Integration in polar coordinates givesThus a decreasing peak is compatible with a conserved spatial integral. Dividing $u$ by $\pi$ makes a probability density function with coordinate variances $s/2=1/2+2t$, connecting the same spreading to Brownian motion with diffusion coefficient one. Equivalently the profile is evolved by the Gaussian heat kernel with time parameter $2t$, since that article uses generator $\frac{1}{2}\Delta$.

A bounded entire harmonic function satisfies $u(x)={\mathbb{E}}_{x}u(B_t)$ because $u(B)$ is a bounded local martingale. The Gaussian heat kernel has a directional derivative with $L^1$ norm $\sqrt{2/\pi }/\sqrt{t}$. Consequently $|u(x)-u(y)|\le \Vert u{\Vert }_{\infty }\sqrt{2/\pi }|x-y|/\sqrt{t}$; letting $t\to \infty$ proves the harmonic Liouville theorem in every dimension.

Write $\mathbf{c}=(U,V{)}^T$ and $\Gamma =\left(\begin{array}{cc}{\gamma }_{11}& {\gamma }_{12}\\ {\gamma }_{12}& {\gamma }_{22}\end{array}\right)$. A well-posed anisotropic diffusion problem requires $\Gamma$ to be symmetric positive definite: ${\gamma }_{11}>0$ and $\Delta ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}^2>0$. For the normal mode in the source the dispersion relation isThe group velocity vanishes when $\mathbf{c}-2i\Gamma \mathbf{k}=0$. ConsequentlyThe requested anisotropic absolute-instability threshold for positive diffusion isFor positive definite diffusion this is also sufficient: the anisotropic Gaussian heat kernel has exponential factor $\exp [\mu t-(\mathbf{x}-\mathbf{c}t{)}^T{\Gamma }^{-1}(\mathbf{x}-\mathbf{c}t)/(4t)]$ and a $t^{-1}$ prefactor, so its fixed-position growth rate is precisely $\mu -f$.

Let the prescribed speed be $q$ and let $d_{\min }\le d_{\max }$ be the positive eigenvalues of $\Gamma$. The Rayleigh quotient for ${\Gamma }^{-1}$ givesAbsence of absolute hydrodynamic instability for every direction therefore requiresThis is an upper bound on $\mu$, not a minimum required value. The source's last sentence reverses this monotonicity: reducing the growth parameter always helps, so there is no finite lower threshold guaranteeing absence. If it had asked for instability in every direction, the corresponding condition would be $\mu >q^2/(4d_{\min })$. Without the positive-definite assumption, negative diffusion directions make the forward problem ill posed; a singular diffusion tensor requires a separate transport analysis and does not support the displayed inverse-matrix formula.

Select the causal Green function, zero for $t<0$; otherwise the impulse equation permits arbitrary homogeneous additions. Use the spatial Fourier transform convention $e^{-ikx}$ and the temporal Laplace transform boundary convention $e^{i\omega t}$. The transformed impulse equation isIf $\mu >0$, a literal real-frequency Fourier transform of the growing response need not exist. Its causal inverse instead starts on $\mathrm{Im}\omega =c>\max (\mu ,0)$ and is continued from there. For $t>0$ close the frequency contour below; its clockwise orientation and the pole residue $1/(-i)=i$ give $e^{-i\omega (k)t}$. For $t<0$ the upper closure contains no pole. The remaining inverse spatial transform is a Gaussian integral:The unit-mass Gaussian heat kernel tends to $\delta (x)$ at $t\downarrow 0$, giving the required impulse normalization.

At the advected center $x=Ut$, its amplitude is $e^{\mu t}/\sqrt{4\pi \gamma t}$. At a fixed spatial point,Thus $\mu >0$ permits growth of the moving wave packet, but it is convective hydrodynamic instability if the fixed-position rate is negative. A positive fixed-position rate is absolute hydrodynamic instability. At its equality threshold the $t^{-1/2}$ factor decays, explaining the marginal exponential classification. This directly verifies the physically selected Briggs-Bers criterion saddle rather than assuming its accessibility.

To keep the normalization visible, denote the integral defining the printed constant by $A_d$. Substituting $r=|y{|}^2/(2s)$ gives, for $y\ne 0$,This is the Newtonian potential of the Brownian heat kernel. The singularity at zero is locally integrable in dimension $d$, since its radial integral behaves like ${\int }_0^1{r}^{2-d}{r}^{d-1}dr=1/2$. All integrands are nonnegative, so Tonelli's theorem permits both orders of integration.

Integrating in $s$ first givesIntegrating in $y$ first uses the Gaussian heat kernel convolution. Completing the square in $|y-x{|}^2/t+|y{|}^2/s$ givesConsequently $I={\int }_t^{\infty }p(r,0,x)dr$, and the actual identity isThe original PDF uses $A_d$ rather than $A_d^{-1}$ as the multiplier while defining its constant to equal $A_d$. That formula is false. For example $A_3=1/(2\pi )$, so the required multiplier in dimension three is $2\pi$. Equivalently, as $t\downarrow 0$ at $x\ne 0$, the left side tends to $|x{|}^{2-d}$, whereas the printed right side tends to $A_d^2|x{|}^{2-d}$. One can retain the printed form only by defining its multiplier as $c_d=A_d^{-1}$.

Continue with $A_d$ equal to the integral defined in the PDF and $u(z)=|z{|}^{2-d}$. Until the Brownian motion hits the radius-$\epsilon$ ball, $u(B_s)$ is a bounded local martingale, hence a true martingale. At the finite stopping time $t\wedge T_{\epsilon }$, optional stopping givesRearranging,The fixed random variable $u(B_t)$ is integrable by the Gaussian heat kernel and the local-integrability calculation above. Also ${\mathbb{P}}_x(T_{\epsilon }\le t)\le (\epsilon /|x|{)}^{d-2}\to 0$. For any $R>0$, the final expectation is at mostFirst send $\epsilon \downarrow 0$, then $R\to \infty$, proving that it vanishes. The identities from part (b) giveSubtracting yields the small-ball Brownian hitting asymptoticThus the constant in this printed limit requires exactly the same reciprocal correction as in part (b). For $d=3$ the multiplier is $2\pi$, not $1/(2\pi )$. The argument proves the corrected finite-time limit directly, rather than inferring it from the eventual hitting probability alone.

Far from the inner edge of a steady Keplerian accretion disk, the inner angular-momentum flux is negligible compared with $\dot{M}j$. Since $\Omega \propto R^{-3/2}$, the viscous torque in an accretion disk becomes $\mathcal{G}=3\pi \nu \Sigma j$. The steady balance $\mathcal{G}\simeq \dot{M}j$ therefore gives $\nu \Sigma =\dot{M}/(3\pi )$. With the stated surface density of a disk,Steady conservation of mass gives $\dot{M}=-2\pi R\Sigma v_R$, and consequentlyBoth relations use the far-inner-edge approximation, rather than extrapolating the stress-free boundary profile to the inner edge.

For the trace passive scalar, put $\sigma =\Sigma C$. Substitute this into the contaminant conservation equation and expand its advective flux:The braces multiplying $C$ vanish by background conservation of mass. Hence the concentration satisfiesThe small contaminant fraction ensures it does not alter the ambient disk dynamics. The surface-density model already averages vertically; an axisymmetric background and ring-like injection give $C=C(R,t)$. More generally, rapid differential orbital motion winds azimuthal structure and facilitates mixing, but radial-only dependence remains a symmetry or mixing assumption, not a property of every initial injection.

For the axisymmetric advection-diffusion equation, cylindrical geometry givesHere $k=\zeta \nu$ and $k\Sigma =\zeta {\nu }_0{\Sigma }_0$ are independent of radius. ThusLet $Y=\ln (R/R_{\ast })$, with any fixed reference length $R_{\ast }$. Since $C_Y=R{C}_R$ and $C_{YY}=R^2{C}_{RR}+R{C}_R$, this is the tracer diffusion in a power-law accretion disk equationAssume $\zeta >0$, ${\nu }_0>0$, and write $v=3{\nu }_0/2$, $D=\zeta {\nu }_0$. The change $X=Y+vt$ turns this into the heat equation $C_t=D{C}_{XX}$. The radial Dirac delta function must be transformed with its Jacobian:Convolution with the Gaussian heat kernel givesThis full-line logarithmic solution extends the far-edge power-law model to $0<R<\infty$. An actual finite inner boundary would require its own boundary condition and modified kernel.

The fraction must be computed with the contaminant mass measure, rather than with $CdR$. IndeedThe total contaminant mass is the conserved quantity $M_c=2\pi {\Sigma }_0{C}_0/R_0$. Its normalized logarithmic radius is therefore a normal distribution with mean $Y_0-vt$ and variance $2Dt$. Integrating this tail gives, for $r>0$,The complementary error function accounts for the one-time mass fraction outside the specified radius.

For an outward target $r_1>1$, put $\ell =\ln r_1>0$. Since $\mathrm{erfc}$ is decreasing, maximize $F$ by minimizing $z(t)=(\ell +vt)/\sqrt{4Dt}$. Its derivative has the sign of $vt-\ell$, so the unique maximum occurs at $t_{\ast }=\ell /v=2\ln r_1/(3{\nu }_0)$. At this time $z(t_{\ast })=\sqrt{v\ell /D}$. Thus the printed expression is correctly derived as the maximum instantaneous exterior fraction:

It is not, however, the probability of ever reaching the radius. A tracer can pass through the target and later return; its history is not counted by a one-time exterior fraction. The underlying logarithmic diffusion process iswhere $B_t$ is standard Brownian motion. To compute its actual hitting probability, first put absorbing levels at $-A$ and $\ell$ relative to $Y_0$. Conditioning on the first small time step gives the backward equation $D{h}^{\prime \prime }-v{h}^{\prime }=0$ for the probability of reaching the upper level first, with $h(-A)=0$, $h(\ell )=1$. Solving givesLetting the lower level tend to $-\infty$ exhausts all finite-time upward-hitting paths. Therefore the correct eventual probability in this idealized model isFor example, if $\ln r_1=2\zeta /3$, the maximum exterior fraction is $\frac{1}{2}\mathrm{erfc}(1)\simeq 0.07865$, whereas the actual hitting probability is $e^{-1}\simeq 0.36788$. This is an explicit counterexample to the last inference in the question and an instance of maximum occupation is not a hitting probability.

For $r_1=1$, the target is the initial radius and the hitting probability is one, while ${\sup }_{t>0}F(1,t)=1/2$ with no attained maximum. For $0<r_1<1$, the inward drift ensures eventual hitting with probability one, and ${\sup }_{t>0}F(r_1,t)=1$ as $t\downarrow 0$. The square-root expression involving $\ln r_1$ is therefore only meaningful for the outward-target case, and even there it describes occupancy rather than first passage.

Write $h=\log \psi$ and $q=2ϵh_{\theta }$, with $ϵ>0$. Direct differentiation givesThus the Cole-Hopf transformation with this negative-flux convention maps the heat equation to the stated viscous Burgers equation. Conversely its residual is independent of $\theta$ and can be removed by multiplying $\psi$ by a function of $Z$.

Convolving each initial exponential with the Gaussian heat kernel givesThe resulting finite-exponential Cole-Hopf solution isIt is a softmax function average of the constant states, so $q_1\le q\le q_N$ and $q_{\theta }={\mathrm{Var}}_w(q_j)/(2ϵ)\ge 0$.

For $N=2$, put $\Delta =q_2-q_1>0$ and $\overline{q}=(q_1+q_2)/2$. Equality of the two exponents occurs atThereforeThis is an increasing viscous shock from $q_1$ to $q_2$, moving at speed $-\overline{q}$ and with transition width of order $ϵ/\Delta$. That speed agrees with the Rankine-Hugoniot condition for flux $-q^2/2$; its concavity makes the increasing jump the entropy shock.

For general $N$, use the shock frame $\theta =-(q_1+q_N)Z/2+\xi$. The leading $Z$ coefficient in $A_j$ is $q_j(q_j-q_1-q_N)/(4ϵ)$. The endpoint coefficients coincide, and every interior coefficient is smaller by $(q_j-q_1)(q_N-q_j)/(4ϵ)>0$. The extremal-state asymptotics of exponential Burgers solutions therefore reduce the large-$Z$ profile to the same tanh shock with states $q_1,q_N$ and centreAt a fixed $\theta$, rather than in the moving frame, the endpoint of largest $|q_j|$ dominates. If $q_N=-q_1$, both endpoint rates coincide and the limiting shock is stationary. Intermediate plateaus disappear in the moving-frame limit.

Substitute $C=t^{\alpha }F(\xi )$, $\xi =r{t}^{-\beta }$. The derivatives giveFor a nontrivial spreading similarity solution, equality at all times requires $\boxed{\beta =1/2}$. The conserved total number of molecules isA finite nonzero constant $N$ therefore requires $\alpha +3\beta =0$, giving $\boxed{\alpha =-3/2}$. In one dimension the same diffusion length exponent is $1/2$, but conservation gives amplitude exponent $-1/2$; generally it is $-d/2$ in dimension $d$.

The profile equation is consequentlyMultiplication by ${\xi }^2$ puts it in the integrated form $[D{\xi }^2{F}^{\prime }+{\xi }^{3}F/2{]}^{\prime }=0$. For the regular localized Gaussian heat kernel profile, the constant is zero at the origin, so $D{F}^{\prime }+\xi F/2=0$. Integrating givesThe second expression chooses the optional normalization $A=N/(4\pi D{)}^{3/2}$. Its mass is $N$, and its width tends to zero with $t^{1/2}$ while its mass stays fixed, so it approaches $N$ times a Dirac delta distribution at the origin as $t\downarrow 0$.

Analytically continuing the free-particle propagator to imaginary time gives the Gaussian heat kernelThe two image endpoints have displacements $2rL$ and $2rL-2q$ from the starting point. Inserting their kernels into the thermal trace of an interval image kernel givesThe prefactor comes from the normalization of the free-particle propagator; it cannot be dropped from a thermal trace.

For a harmonic function $u$ on $\mathbb{C}\cong {\mathbb{R}}^2$, the Itô formula givesStop on leaving increasing discs and at increasing deterministic times. On each such interval the gradient is bounded, so the stochastic integral is a martingale. This proves that $u(B)$ is a continuous local martingale.

If $|u|\le K$, this is a bounded local martingale and hence a true martingale. Start Brownian motion at an arbitrary $z$. Its Gaussian heat kernel givesFor a unit vector $e$, ${\partial }_e{p}_t(v)=-(e\cdot v)p_t(v)/t$, so $\int |{\partial }_e{p}_t(v)|dv=\sqrt{2/\pi }/\sqrt{t}$. Integrating the directional derivative along the segment from $z$ to $w$ yieldsLetting $t\to \infty$ gives the harmonic Liouville theorem:This Gaussian heat-kernel proof of the harmonic Liouville theorem uses the bounded martingale identity established above.

Work with $t>0$ and write $u=t^{-1/2}{e}^{-x^2/(4t)}$. The chain rule givesThus $u_{xx}=u_t$: it is a constant multiple of the one-dimensional Gaussian heat kernel.

For the modified equation, put $f=g(t)u$. Substitution gives $g{u}_{xx}=g^{\prime }u+g{u}_t-gu/t$. The heat-equation terms cancel, and $u>0$ permits division to obtain $\boxed{g^{\prime }=g/t}$. Solving this separable differential equation givesThe zero solution is included by $C=0$. This is the scalar integrating factor for a reaction-diffusion equation with reaction coefficient $1/t$, because the modified equation is $f_t=f_{xx}+f/t$. Its sign produces growth. Indeed ${\int }_{\mathbb{R}}udx=2\sqrt{\pi }$ is constant, while ${\int }_{\mathbb{R}}fdx=2C\sqrt{\pi }t$ increases linearly for $C>0$; neither profile is being assigned unit normalization here.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

In a steady Keplerian accretion disk with $\Sigma ={\Sigma }_0{R}^{-2}$, $\nu ={\nu }_0{R}^2$, and tracer diffusivity $k=\zeta \nu$, the trace passive scalar concentration obeys the displayed constant-coefficient advection-diffusion equation in logarithmic radius. Contaminant mass is proportional to $CdY$, because $2\pi R\sigma dR=2\pi {\Sigma }_{0}CdY$. A radial impulse $C_0\delta (R-R_0)$ therefore has logarithmic amplitude $C_0/R_0$ and evolves by the Gaussian heat kernel centered at $\ln R_0-vt$. The full-line solution idealizes radii far from the disk boundaries.