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Taking the spatial Fourier transform of the diffusion equation gives ${\tilde{\theta }}_t=-D{k}^2\tilde{\theta }$ and $\tilde{\theta }(k,0)=\tilde{g}(k)$. Hence $\tilde{\theta }=e^{-D{k}^{2}t}\tilde{g}$. Fourier inversion and the Gaussian transform from part (a), or direct evaluation of the inverse Gaussian integral, give the heat kernelSubstituting the integral for $\tilde{g}$ and interchanging absolutely integrable terms produces the convolution solutionFor the stated nonnegative compactly supported data, understood to be nonzero on a set of positive measure, the kernel is strictly positive everywhere. Therefore $\theta (x,t)>0$ for every real $x$ and every $t>0$. Property P holds: diffusion has infinite propagation speed. The distant value can be extremely small, but is not zero. For continuous initial data, any nontrivial nonnegative $g$ automatically has the required positive-measure set.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

The insulated boundary imposes the Neumann boundary condition $T_{x_1}(0,x_2,t)=0$. Use the method of images by reflecting the initial data evenly across $x_1=0$. At the origin, the original sector and its reflection contribute equally. Using the given two-dimensional heat kernel,Differentiation givesThis vanishes exactly when $e^{3/(4Kt)}=4$. Its sign is positive before that time and negative afterwards. Moreover $T\to 0$ as $t\downarrow 0$ and as $t\to \infty$. Therefore this is the unique global maximum:The factor two from the insulated-boundary reflection is essential; applying the full-plane heat kernel only to the unreflected sector would give half the temperature.

For a deterministic localized initial peak in an unbounded $d$-dimensional sample, the mean obeys ${\partial }_t⟨\varphi ⟩=\Gamma \kappa {\nabla }^2⟨\varphi ⟩-\Gamma a⟨\varphi ⟩$. The Gaussian Model A impulse response is therefore the heat kernel multiplied by the local relaxation factor:The peak spreads over a distance of order $\sqrt{\Gamma \kappa t}$ and its central height falls both by spreading and by exponential relaxation. Its integrated mean is $A{e}^{-\Gamma at}$, rather than $A$: the equation describes nonconserved order-parameter dynamics, despite the use of the word “density” in this part.

An individual realization also develops thermal fluctuations; it does not tend pointwise to zero. For an initially deterministic field, the connected Fourier mode variance grows as $[k_{B}T/(a+\kappa q^2)](1-e^{-2r(q)t})$, while the deterministic mean decays. It approaches the zero-mean Gaussian equilibrium field at the coarse-graining cutoff. On a periodic or bounded sample use the corresponding boundary-adapted heat kernel; at $\kappa =0$ there is no spatial spreading. The Dirac delta distribution is an ideal initial profile, recovered as a distributional limit; a finite-energy physical peak has a microscopic width.

Taking the Laplace transform in $t$ givesThe bounded resolvent solution supplied in the hint can be writtenPart (a) inverts the kernel, yielding the heat kernel

Either essay option suffices; both are developed here to make the two mathematical constructions available.

Diffusion for images. Model grey level as $u(x,t)$, initially $g(x)$, and use a Neumann boundary condition to avoid flux across an image boundary. The linear heat equation $u_t=\Delta u$ smooths the data. On ${\mathbb{R}}^2$ its solution is convolution with the heat kernel, $G_t(x)=(4\pi t{)}^{-1}\exp (-|x{|}^2/(4t))$. Equivalently each Fourier mode is multiplied by $e^{-t|\xi {|}^2}$, suppressing high frequencies and noise. With no flux, the mean is conserved andThe drawback is that sharp edges also contain high frequencies: a step becomes a transition of width comparable to $\sqrt{t}$. Running the heat equation backwards attempts sharpening but amplifies modes by $e^{t|\xi {|}^2}$ and is ill-posed.

The Perona-Malik equation instead uses a decreasing diffusivity,Small gradients are smoothed strongly and large ones less strongly. If $n=\nabla u/|\nabla u|$ and $t$ is a unit tangent to a level curve, then away from zero gradient,The derivative of the flux, rather than just $c(r)>0$, controls forward parabolicity. Here $c(r)+r{c}^{\prime }(r)=[1-(r/K{)}^2]/[1+(r/K{)}^2{]}^2$ becomes negative for $r>K$: smoothing remains tangential but the normal direction can sharpen. This formal edge enhancement comes with forward-backward ill-posedness, so existence and stability of the unregularized continuum equation must not be presumed. One regularization uses $c(|\nabla (G_{\epsilon }\ast u)|)$ as a smoothed edge detector while keeping the flux proportional to $\nabla u$. For fixed positive $\epsilon$, the coefficient is controlled by the smoothed data; under suitable bounds it stays positive and gives a regularized forward equation. It preserves edges through reduced cross-edge transport, without the same local backward-diffusion calculation.

Diffusion from a gradient energy connects these equations with variational regularization. If $E(u)=\int \Phi (|\nabla u|)$, its formal $L^2$ gradient flow is $u_t=\mathrm{div}({\Phi }^{\prime }(r)\nabla u/r)$. Linear diffusion corresponds to $\Phi (r)=r^2/2$. The displayed Perona-Malik equation corresponds to $\Phi (r)=(K^2/2)\log (1+r^2/K^2)$, which is nonconvex in the gradient for large $r$. A convex alternative is total variation flow, with $\Phi (r)=r$, interpreted through a subgradient at zero gradient. Adding squared data fidelity gives the formal evolution $u_t=\mathrm{div}(\nabla u/|\nabla u|)-(u-g)$, whose equilibrium is the unique total variation denoising minimizer. Convex TV preserves sharp interfaces more effectively than the heat energy, but it may produce piecewise constant plateaux, known as staircasing in total variation denoising. A stopping time or fidelity weight controls the smoothing scale. Linear diffusion is stable but blurs edges; nonlinear diffusion must balance edge selectivity with parabolicity and regularization.

Wavelets on an interval. Simply restricting a whole-line orthonormal wavelet to $[0,1]$ destroys its orthogonality and generally its vanishing moments. Periodizing the basis restores orthogonality on the circle, but treats the two endpoints as neighbors. This is suitable for periodic data and can create an artificial endpoint jump for nonperiodic data.

A localized interval-adapted wavelet basis instead uses unchanged interior functions and finitely many special boundary wavelets at each endpoint. Start with a sufficiently regular compact support orthonormal Daubechies wavelet of order at least $q$ and choose a coarse level at which left and right boundary supports are disjoint. At each boundary, take appropriate finite combinations of the scaling functions that meet the endpoint, restricted to the interval. Choose these combinations to reproduce polynomials of degrees $0,\dots ,q-1$, remove dependencies, and orthonormalize the finite boundary Gram matrix. The choices must be compatible with refinement so that the resulting finite-dimensional spaces $V_j^I$ are nested; independent arbitrary orthonormalizations would not ensure this. Interior functions retain their whole-line filters, while the boundary functions use finite boundary refinement matrices.

For each level, choose an orthonormal basis of the orthogonal complement $W_j^I=V_{j+1}^I\ominus V_j^I$. It consists of interior wavelets and a bounded number of left and right boundary wavelets. Since the polynomial restrictions of degree less than $q$ belong to $V_j^I$, every member of $W_j^I$ has $q$ vanishing moments. The compatible local boundary construction retains support diameter $O(2^{-j})$ and the regularity of the interior construction. Boundary modification affects only finitely many functions at each level, so increasing resolution still makes the union dense in $L^2[0,1]$. Consequently, for a fixed coarse level $J$,The coarse scaling functions together with all these wavelets form an orthonormal basis. This construction is the Cohen-Daubechies-Vial interval wavelet construction. It preserves localization, polynomial cancellation and stable coefficient extraction without imposing periodic or zero boundary data. The finite boundary refinement matrices also permit a fast transform; a mere restriction followed by one unrelated Gram-Schmidt process at each scale does not establish all these properties.

Use the weighted Neumann heat kernel with constant drift. The spatial operator has the self-adjoint formso its homogeneous Neumann boundary conditions give a regular Sturm-Liouville problem in the weighted inner productThe constant eigenfunction is ${\varphi }_0=1$, with eigenvalue ${\lambda }_0=0$ and squared norm $N_0=(e^{\beta L}-1)/\beta$. For $k\ge 1$, put $q_k=k\pi /L$. Substitution of $\varphi =e^{-\beta x/2}v$ in $\mathcal{L}\varphi =-\lambda \varphi$ gives $v^{\prime \prime }+(\lambda -{\beta }^2/4)v=0$ and the Robin boundary conditions $v^{\prime }=\beta v/2$ at both endpoints. HenceIndeed,The Sturm-Liouville eigenfunction expansion is complete and orthogonal. Its heat kernel, relative to the measure $e^{\beta y}dy$, isFor a coefficient $a_k(t)=⟨{\varphi }_k,u(\cdot ,t){⟩}_{\beta }/N_k$, two applications of integration by parts yieldSolving these scalar linear differential equations and summing gives the Neumann boundary-forcing formula:This depends only on the given data. In particular, the zero eigenvalue retains the changing weighted mean:The endpoint derivatives of the boundary integral are understood as interior limits. Differentiating each homogeneous Neumann eigenfunction at the endpoint before summing would incorrectly discard the prescribed boundary forcing: the short-time heat kernel makes that interchange invalid. The stated compatibility conditions give the classical boundary traces.

For the heat kerneldirect partial differentiation givesThus $K$ solves the heat equation. For bounded continuous $f$, differentiation under the integral sign givesWith $Y=(x-y)/\sqrt{4t}$,The Gaussian integral makes the weight have total mass one, and the dominated convergence theorem gives $u(x,t)\to f(x)$. This is the Gaussian approximate identity.

For the viscous Burgers equation with unit viscosity, the Cole-Hopf transformation $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose(up to an irrelevant positive constant), and set $u=K_t\ast f$. ThereforeEquivalently,and the approximate-identity limit gives $w(x,t)\to g(x)$.

Removing the death term with $u(x,t)=e^{\mu t}n(x,t)$ gives the heat equation $u_t=D{u}_{xx}$ with initial data $N\delta (x)$. The heat kernel therefore yieldsFor each fixed $x$, the exponential mortality factor and diffusive prefactor show that $n(x,t)\to 0$ as $t\to \infty$. Its spatial integral is $N{e}^{-\mu t}$, the expected number of surviving larvae.

Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.

LetConvolution with the initial Dirac delta function translates the heat kernel, while part (b) propagates the impulsive source from time one. HenceAt $(x,t)=(0,2)$ this becomesIt vanishes exactly whenThis is the cancellation of two heat-kernel impulses.

For each $x$, the bounded local martingale $u(x+B_t)$ is a true martingale. Taking expectations givesfor every $t>0$. Thus $u$ equals its convolution with every heat kernel. The convolution is smooth, so the originally Borel function $u$ is smooth. Differentiating the heat semigroup identity at $t=0$ gives $\Delta u=0$, so $u$ is a bounded harmonic function on the plane. The harmonic Liouville theorem now implies that $u$ is constant.

For $a=0$ and $\kappa >0$, the heat kernel givesAs $t\downarrow 0$, $K_t$ is a Gaussian of total mass one whose width is of order $\sqrt{\kappa t}$; it becomes a narrow spike at zero and converges to the delta distribution. Thus the convolution tends to $u_0(x)$.

Let $p_t(z)=(2\pi t{)}^{-1/2}{e}^{-z^2/(2t)}$ be the heat kernel for $u_t=\frac{1}{2}{u}_{xx}$. The symmetry of the Gaussian distribution gives the method-of-images formulaThis is the Neumann heat kernel on a half-line. Differentiation under the integral sign shows that $v\in C^{1,2}$ for $t,x>0$ and that $v_t=\frac{1}{2}{v}_{xx}$. At $x=0$, the two differentiated kernel terms cancel, so $v_x(t,0)=0$. The Gaussian approximate identity gives $v(t,x)\to f(x)$ as $t\downarrow 0$, while the dominated convergence theorem gives continuity up to $x=0$. Finally $|v(t,x)|\le \Vert f{\Vert }_{\infty }$, which is stronger than the required exponential bound. Thus $v$ satisfies every condition in the displayed boundary-value problem.