Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 64 / 3

The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le \Vert g{\Vert }_{\infty }$. Its approximate identity property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is continuous. This bounded solution satisfies the required Gaussian-growth bound.

For heat-equation uniqueness under Gaussian growth, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0{e}^{a|x{|}^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le \tau <1/(4b)$. The positive comparison solutionsatisfies $({\Phi }_b{)}_t=\Delta {\Phi }_b$. For any $\epsilon >0$, its faster spatial growth makes $|w|\le \epsilon {\Phi }_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The heat equation maximum principle applied to $\pm w-\epsilon {\Phi }_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\epsilon \downarrow 0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.

At time $T$, this is Gaussian filtering with covariance $2TI$, hence $\sigma =\sqrt{2T}$ per coordinate. With the angular-frequency Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, the Gaussian filtering Fourier multiplier isFor merely bounded $g$, this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency $k$, the multiplier is $e^{-4{\pi }^{2}T|k{|}^2}$.

For the printed Perona-Malik equation, retain the supplied conductance $c(s)=s{e}^{-s^2/(2{\lambda }^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2{\lambda }^2)}$. Then $u_t=F^{\prime }(u_x)u_{xx}$, withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt $|u_x|=0$ and $\sqrt{2}\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2{\lambda }^2)}$ would give threshold $\lambda$, but that is a different equation.

The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf $u(x_0)={\mathrm{mean}}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the limit give $\Delta u(x_0)=0$. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.

For the area median over the disk, the disk-median curvature expansion, at a point where $|\nabla u|\ne 0$, isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.