Book 6A

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Course 6Book 6A: The Heat Equation and Its RelativesChapter 1

What a PDE Is

Well-posedness, classification, parabolic scaling, and why the backward heat equation is ill-posed.

26 min read · Updated Oct 2, 2026

Read with Evans, Partial Differential Equations, chapter 1 (the classification of equations and the meaning of well-posedness) and section 2.1 (the transport equation, a first equation solved by hand). Evans's chapter 5 on Sobolev spaces was read alongside Book 4A and can be skipped here.

In this chapter · 8 sections
  1. 1.1Diffusion takes the square of the distance
  2. 1.2What a PDE is
  3. 1.3Elliptic, parabolic, hyperbolic
  4. 1.4Well-posed problems
  5. 1.5Parabolic scaling
  6. 1.6The backward heat equation is ill-posed
  7. 1.7History
  8. 1.8Exercises

This book is about one equation and its relatives. The heat equation

ut=Δuu_t = \Delta u

says that the rate of change of a temperature at a point is proportional to how much the temperature there falls short of the average around it (2B.7 Fourier Series and the First Heat Equation). It is the simplest equation that smooths, the model for every diffusion in physics, chemistry, biology and finance, and, through Hamilton's insight, the model for the Ricci flow: ∂tg=−2Ric⁡(g)\partial_tg = -2\operatorname{Ric}(g) is a nonlinear heat equation for a Riemannian metric (11A.1 The Equation and Its First Solutions). Every analytic tool used on the Ricci flow is first a tool for the heat equation, and this book builds them in their simplest setting.

Before the tools, the vocabulary. This chapter says what a partial differential equation is, how equations are sorted into types, and what it means for one to be well posed. Then it proves two facts about the heat equation that hold for every parabolic equation and shape the whole subject. Parabolic scaling: diffusion over a distance LL takes a time proportional to L2L^2. Irreversibility: the heat equation can be solved forwards in time from any reasonable initial state, but not backwards.

By the end of this chapter you will be able to:

  • say what the order of a PDE is, and whether it is linear, semilinear, quasilinear or fully nonlinear;
  • classify second-order equations as elliptic, parabolic or hyperbolic from their symbol;
  • state Hadamard's three conditions for well-posedness, and recognise an ill-posed problem;
  • use parabolic scaling to estimate diffusion times, and find self-similar solutions of the heat equation;
  • explain, with Fourier modes, why the backward heat equation is ill-posed and why deblurring an image is unstable.

Diffusion takes the square of the distance

In the world Model Why bodies need blood

Oxygen reaches a cell by diffusion: molecules wander at random, and their concentration cc obeys the diffusion equation ct=DΔcc_t = D\Delta c, where DD is the diffusion coefficient. For oxygen in water at room temperature, DD is about 2×10−92 \times 10^{-9} m²/s (measured values lie between about 1.91.9 and 2.3×10−92.3 \times 10^{-9}), and in tissue at body temperature it is of the same order. A diffusing quantity typically travels a distance LL in a time of order

t≈L22Dt \approx \frac{L^2}{2D}

(the precise statement, ⟨x2⟩=2Dt\langle x^2\rangle = 2Dt in each coordinate, is proved in 6A.3 The Heat Equation on ℝⁿ). With D=2×10−9D = 2 \times 10^{-9} m²/s:

distance time
1010 µm, across a cell 0.0250.025 s
11 mm about 44 minutes
11 cm about 77 hours
11 m about 88 years

Diffusion is fast over the size of a cell and hopeless over the size of a body. So animals larger than a few millimetres need a circulation that carries oxygen by flow to within a few cell widths of where it is used, and then lets diffusion finish the job. The square in L2L^2 is the whole story: multiply the distance by 10510^5, and the time is multiplied by 101010^{10}.

The same law governs cooking. Heat conducts into food by the heat equation, with the thermal diffusivity in place of DD (about 1.4×10−71.4 \times 10^{-7} m²/s for water, and of the same order for most foods, which are mostly water). So, other things being equal, a steak twice as thick takes about four times as long to cook through. Rowat and colleagues use exactly this example to teach diffusion in "The kitchen as a physics classroom" (Physics Education, 2014). Other things are rarely exactly equal (evaporation, fat, bone, and the shape of the food all matter), so the square law is a scaling law, not a recipe.

What a PDE is

A partial differential equation is an equation relating an unknown function of several variables to its partial derivatives. Its order is the highest order of derivative that appears. Some of the equations this guide uses, with u=u(x,t)u = u(x, t) or u=u(x)u = u(x) for x∈Rnx \in \mathbb{R}^n:

equation name type
ut+b⋅∇u=0u_t + b\cdot\nabla u = 0 transport linear, first order
Δu=f\Delta u = f Poisson (Laplace if f=0f = 0) linear, elliptic
ut=Δuu_t = \Delta u heat linear, parabolic
utt=Δuu_{tt} = \Delta u wave linear, hyperbolic
ut=Δu+u(1−u)u_t = \Delta u + u(1 - u) Fisher–KPP (6A.4 Maximum Principles) semilinear, parabolic
ut=Δ(um)u_t = \Delta(u^m) porous medium (6A.7 Nonlinear Parabolic Equations) quasilinear, parabolic (degenerate)
det⁡D2u=f\det D^2u = f Monge–Ampère (5A.6 Where This Track Leads) fully nonlinear, elliptic
∂tg=−2Ric⁡(g)\partial_tg = -2\operatorname{Ric}(g) Ricci flow (11A.1 The Equation and Its First Solutions) quasilinear system, weakly parabolic

The words in the last column measure how nonlinear an equation is, according to where the nonlinearity sits. A second-order equation is

  • linear if it has the form ∑aij(x)∂i∂ju+∑bi(x)∂iu+c(x)u=f(x)\sum a^{ij}(x)\partial_i\partial_ju + \sum b^i(x)\partial_iu + c(x)u = f(x);
  • semilinear if the coefficients of the highest derivatives depend only on xx, and the nonlinearity is in the lower-order terms, as in Fisher–KPP;
  • quasilinear if it is linear in the highest derivatives, with coefficients that may depend on uu and its lower derivatives: ∑aij(x,u,∇u)∂i∂ju+…\sum a^{ij}(x, u, \nabla u)\partial_i\partial_ju + \dots;
  • fully nonlinear if the highest derivatives enter nonlinearly, as in det⁡D2u\det D^2u.

The more nonlinear, the harder. Semilinear equations behave much like linear ones on short time scales. Quasilinear equations can be handled by freezing the coefficients and solving a linear problem, which is the strategy of 6A.7 Nonlinear Parabolic Equations. The Ricci flow is quasilinear: written in coordinates, Ric⁡\operatorname{Ric} involves second derivatives of gg linearly, with coefficients depending on gg itself.

Elliptic, parabolic, hyperbolic

For a linear second-order operator Lu=∑i,jaij∂i∂ju+lower orderL u = \sum_{i,j}a^{ij}\partial_i\partial_ju + \text{lower order}, with (aij)(a^{ij}) symmetric, the principal symbol is the quadratic form

σ(x,ξ)=∑i,jaij(x)ξiξj.\sigma(x, \xi) = \sum_{i,j}a^{ij}(x)\xi_i\xi_j.

It records what the operator does to a fast oscillation: if u=eiλx⋅ξu = e^{i\lambda x\cdot\xi} with λ\lambda large, then Lu≈−λ2σ(x,ξ)uLu \approx -\lambda^2\sigma(x, \xi)u, since the second derivatives dominate. The operator is elliptic if σ(x,ξ)>0\sigma(x, \xi) > 0 for every ξ≠0\xi \neq 0, that is, if the matrix (aij)(a^{ij}) is positive definite (2B.8 Calculus in Several Variables). The Laplacian is the model: σ(ξ)=∣ξ∣2\sigma(\xi) = |\xi|^2.

An evolution equation ut=Luu_t = Lu with LL elliptic is parabolic: the heat equation is the model. An equation utt=Luu_{tt} = Lu with LL elliptic is hyperbolic: the wave equation. The types behave completely differently. Elliptic equations describe equilibria and have smooth solutions; parabolic equations describe diffusion and smooth their data instantly; hyperbolic equations describe waves and carry singularities along at finite speed. This book is about the first two.

The symbol explains why. For a Fourier mode u=u^(t)e2πix⋅ξu = \hat u(t)e^{2\pi ix\cdot\xi}, the heat equation becomes the ODE u^′=−4π2∣ξ∣2u^\hat u' = -4\pi^2|\xi|^2\hat u (4A.5 The Fourier Transform), so the mode decays at the rate 4π2∣ξ∣24\pi^2|\xi|^2: the more it oscillates, the faster it dies. The positivity of the symbol is exactly what makes every oscillating mode decay. If σ\sigma were negative in some direction, the corresponding modes would grow, and if σ\sigma vanished in some direction, those modes would be left untouched.

Where this goes The symbol of the Ricci flow

The Ricci flow's symbol is not positive definite: it vanishes on the directions corresponding to changing the metric by a diffeomorphism. Since Ric⁡(ϕ∗g)=ϕ∗Ric⁡(g)\operatorname{Ric}(\phi^*g) = \phi^*\operatorname{Ric}(g) for every diffeomorphism ϕ\phi, the equation cannot tell a metric from its reparametrisations, and the linearised operator has a large kernel. The flow is only weakly parabolic, and the standard existence theory of 6A.7 Nonlinear Parabolic Equations does not apply directly. DeTurck's trick (11A.3 Short-Time Existence and Uniqueness) adds a term that breaks the symmetry and makes the symbol positive definite. Computing a symbol is the first thing to do with any new geometric equation.

Well-posed problems

What does it mean to have solved a PDE? Jacques Hadamard proposed in 1902 that a problem for a PDE (an equation together with initial or boundary conditions) is well posed if

  1. a solution exists;
  2. it is unique;
  3. it depends continuously on the data: small changes in the initial or boundary conditions produce small changes in the solution.

The third condition is the one that matters for applications. Data are measured, so they are never exact; a problem in which arbitrarily small errors in the data can produce large changes in the answer is useless for prediction, even if solutions exist and are unique. Which norms "small" refers to is part of the statement, and choosing them is part of the art.

Hadamard's own example of an ill-posed problem is the Laplace equation with initial data, as if it were an evolution equation. On R×(0,∞)\mathbb{R}\times(0, \infty) consider uxx+uyy=0u_{xx} + u_{yy} = 0 with u(x,0)=0u(x, 0) = 0 and uy(x,0)=1nsin⁡(nx)u_y(x, 0) = \frac1n\sin(nx). The solution is

un(x,y)=1n2sin⁡(nx)sinh⁡(ny).u_n(x, y) = \frac{1}{n^2}\sin(nx)\sinh(ny).

As n→∞n \to \infty the data tend to 00 uniformly, but at y=1y = 1 the solution has size sinh⁡nn2→∞\frac{\sinh n}{n^2} \to \infty. The zero solution has data arbitrarily close to that of a solution which is enormous a unit distance away. Elliptic equations are well posed with boundary conditions (the Dirichlet problem, 5A.4 Harmonic Functions and Conformal Mapping, 6A.2 Harmonic Functions), not with initial conditions.

Parabolic scaling

Proposition 1.1 Parabolic scaling

If u(x,t)u(x, t) solves ut=Δuu_t = \Delta u, then for every λ>0\lambda > 0 so does

uλ(x,t)=u(λx,λ2t).u_\lambda(x, t) = u(\lambda x, \lambda^2t).

Proof. By the chain rule, ∂tuλ=λ2ut(λx,λ2t)\partial_tu_\lambda = \lambda^2u_t(\lambda x, \lambda^2t) and Δuλ=λ2(Δu)(λx,λ2t)\Delta u_\lambda = \lambda^2(\Delta u)(\lambda x, \lambda^2t).

Space and time scale differently: shrinking lengths by λ\lambda shrinks times by λ2\lambda^2. So the natural way to measure a region of space-time is by a length rr in space and r2r^2 in time, and the basic regions are parabolic cylinders Br(x)×(t−r2,t]B_r(x) \times (t - r^2, t] (6A.6 Parabolic Regularity). The diffusion time estimate t≈L2/2Dt \approx L^2/2D is this proposition in physical units (Figure 1.1).

Figure 1.1. Parabolic scaling. The curves ∣x∣=ct|x| = c\sqrt t (for several cc) are the level curves of x2/tx^2/t, along which the heat kernel's shape is unchanged; heat released at the origin spreads like t\sqrt t. Doubling the width of a space-time box quadruples its duration.

Scaling also finds solutions. Look for solutions that are unchanged by parabolic scaling up to a factor, of the form

u(x,t)=t−aF(xt).u(x, t) = t^{-a}F\Big(\frac{x}{\sqrt t}\Big).

In one dimension, substituting and writing y=x/ty = x/\sqrt t gives the ODE F′′+y2F′+aF=0F'' + \frac y2F' + aF = 0 (Exercise 1.6). For a=12a = \frac12 this is (F′+y2F)′=0\big(F' + \frac y2F\big)' = 0, solved by F(y)=e−y2/4F(y) = e^{-y^2/4}: we recover, up to a constant, the heat kernel 14πte−x2/4t\frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}, the subject of 6A.3 The Heat Equation on ℝⁿ. The exponent a=12a = \frac12 is forced by requiring the total heat ∫u dx\int u\,dx to be conserved. For a=0a = 0, the solution F(y)=∫0ye−s2/4dsF(y) = \int_0^ye^{-s^2/4}ds gives the error function profile: the temperature near the surface of a half-space suddenly heated, which Kelvin used to estimate the age of the Earth (6A.3 The Heat Equation on ℝⁿ).

Where this goes Scaling the Ricci flow

The Ricci flow has the same scaling law. Since Ric⁡(cg)=Ric⁡(g)\operatorname{Ric}(cg) = \operatorname{Ric}(g) for a constant c>0c > 0, if g(t)g(t) is a Ricci flow then so is gλ(t)=λg(t/λ)g_\lambda(t) = \lambda g(t/\lambda): metrics scale like length squared, and time scales like the metric, that is like length squared (Exercise 1.9, 11A.1 The Equation and Its First Solutions). Every blow-up argument in the subject, from Hamilton's singularity analysis (11B.4 Singularities) to Perelman's κ-solutions (12B.1 κ-Solutions), rescales a flow near a high-curvature point by this law and takes a limit.

The backward heat equation is ill-posed

Forward in time, the heat equation is well posed in every reasonable sense (6A.3 The Heat Equation on ℝⁿ, 6A.4 Maximum Principles). Backward, it is not. On the circle R/Z\mathbb{R}/\mathbb{Z}, a solution is u(x,t)=∑cne−4π2n2te2πinxu(x, t) = \sum c_ne^{-4\pi^2n^2t}e^{2\pi inx} (2B.7 Fourier Series and the First Heat Equation). Given the state at time TT, recovering the state at time 00 means multiplying the nn-th Fourier coefficient by e+4π2n2Te^{+4\pi^2n^2T}.

Proposition 1.2 The backward heat equation is ill-posed

There are solutions unu_n of ut=uxxu_t = u_{xx} on the circle, 0≤t≤T0 \leq t \leq T, whose final values un(⋅,T)u_n(\cdot, T) tend to 00 uniformly while their initial values un(⋅,0)u_n(\cdot, 0) do not stay bounded.

Proof. Take un(x,t)=e−4π2n2(t−T)sin⁡(2πnx)nu_n(x, t) = e^{-4\pi^2n^2(t - T)}\frac{\sin(2\pi nx)}{n}. At t=Tt = T it has size 1n\frac1n. At t=0t = 0 it has size e4π2n2Tn→∞\frac{e^{4\pi^2n^2T}}{n} \to \infty.

A measurement of the present state, however accurate, contains errors at high frequency, and running the heat equation backwards multiplies them by factors that grow like ecn2e^{cn^2}. Worse, most states have no past at all: since every forward solution is smooth (indeed analytic, 5A.3 Residues and Fourier Transforms) at positive times, a final state that is not smooth cannot be the result of heat flow. The heat equation destroys information, at a rate that grows with frequency. That is the mathematical form of irreversibility, and it reappears as the monotonicity of entropy in 6A.10 Entropy, Information and Diffusion.

In the world In use Why deblurring is unstable

Blurring a photograph by an out-of-focus lens, or by atmospheric turbulence, acts approximately as convolution with a smooth kernel; for a Gaussian blur it is exactly running the heat equation forward for some time (3A.8 Convolution and Mollifiers). Deblurring is running it backward. A naive deblurring divides each Fourier coefficient of the image by the Fourier transform of the blur, which for a Gaussian blur is the factor e+4π2∣ξ∣2te^{+4\pi^2|\xi|^2t} above. Fine detail is recovered in principle, but noise in the image, which is present at all frequencies, is amplified enormously (Figure 1.2). Practical image restoration therefore regularises: it gives up the highest frequencies, or penalises solutions that are too rough (Tikhonov regularisation, Wiener filtering, total-variation methods). Every such method is a deliberate compromise forced by the ill-posedness of the backward heat equation.

Figure 1.2. Forward and backward heat flow on a circle (computed with 40 Fourier modes). Top: a step signal. Middle: after heat flow for time T=2×10−4T = 2 \times 10^{-4}, with and without added noise of size 10−410^{-4} in each Fourier coefficient (the two curves are indistinguishable). Bottom: running backward for the same time. Without noise the steps come back, up to the ripples caused by keeping only 40 modes; with the noise, the highest modes are multiplied by up to e4π2⋅402T≈3×105e^{4\pi^2\cdot40^2T} \approx 3 \times 10^5 and the reconstruction is destroyed (the plot is clipped).
Where this goes A heat equation that must run backward

Perelman's monotonicity formulas use a conjugate heat equation, ∂tu=−Δu+Ru\partial_tu = -\Delta u + Ru along a Ricci flow (12A.2 Ricci Flow as a Gradient Flow). Its sign is reversed, so it is ill-posed forward in time. It is solved backward, from a final condition at a later time, where it is well posed: in the backward time variable τ=T−t\tau = T - t it is an ordinary heat equation. The Gaussian (4πτ)−n/2e−∣x∣2/4τ(4\pi\tau)^{-n/2}e^{-|x|^2/4\tau} that appears in Perelman's W\mathcal W-entropy (12A.3 The 𝓦-Entropy) is a heat kernel run this way. Knowing which direction a parabolic equation can be solved in is not a technicality; it decides how the argument is built.

History

Joseph Fourier's Théorie analytique de la chaleur (1822) derived the heat equation and solved it with series. Adolf Fick's law of diffusion (1855) put diffusing substances under the same equation. The classification of second-order equations into elliptic, parabolic and hyperbolic was settled in the nineteenth century, and the names are by analogy with conic sections: replacing derivatives by variables turns the model operators into polynomials whose level sets are an ellipse, a parabola and a hyperbola (Exercise 1.8). Jacques Hadamard formulated well-posedness in 1902 and gave his example of the ill-posed Cauchy problem for the Laplace equation; his book Lectures on Cauchy's Problem in Linear Partial Differential Equations (1923) developed it. The study of ill-posed problems and their regularisation became a field of its own in the 1960s, with Andrey Tikhonov's work.

Recall Where we stand

A PDE is classified by its order, by how nonlinear it is (linear, semilinear, quasilinear, fully nonlinear), and, for second-order equations, by its symbol: elliptic when the symbol is positive definite, parabolic for ut=Luu_t = Lu with LL elliptic. A problem is well posed when solutions exist, are unique and depend continuously on the data. The heat equation is invariant under u(x,t)↦u(λx,λ2t)u(x, t) \mapsto u(\lambda x, \lambda^2t), so diffusion over a distance LL takes a time of order L2/DL^2/D, and self-similar solutions lead to the heat kernel. Backward in time the heat equation is ill-posed: high frequencies grow like e4π2∣ξ∣2te^{4\pi^2|\xi|^2t}. 6A.2 Harmonic Functions studies the equilibria of the heat equation, the harmonic functions.

Exercises

Exercise 1.3 Classifying equations

For each equation, give the order and say whether it is linear, semilinear, quasilinear or fully nonlinear; for the second-order ones say whether they are elliptic, parabolic or hyperbolic. (a) ut=uxx+sin⁡uu_t = u_{xx} + \sin u. (b) uxx+xuyy=0u_{xx} + xu_{yy} = 0, in the regions x>0x > 0 and x<0x < 0. (c) ut=div⁡(∇u1+∣∇u∣2)u_t = \operatorname{div}\big(\frac{\nabla u}{\sqrt{1 + |\nabla u|^2}}\big). (d) uxxuyy−uxy2=1u_{xx}u_{yy} - u_{xy}^2 = 1. (e) ut+uux=0u_t + uu_x = 0.

Solution

(a) Second order, semilinear, parabolic. (b) Linear; elliptic for x>0x > 0, hyperbolic for x<0x < 0 (an equation of Tricomi type, which changes type across x=0x = 0). (c) Second order, quasilinear: expanding, the coefficients of ∂i∂ju\partial_i\partial_ju depend on ∇u\nabla u; parabolic, since the coefficient matrix 11+∣p∣2(I−p⊗p1+∣p∣2)\frac{1}{\sqrt{1 + |p|^2}}\big(I - \frac{p\otimes p}{1 + |p|^2}\big) is positive definite. It is a close relative of the mean curvature flow of a graph (6A.8 Curve Shortening and the First Geometric Flows), which has an extra factor 1+∣∇u∣2\sqrt{1 + |\nabla u|^2} on the right. (d) Second order, fully nonlinear (a Monge–Ampère equation), elliptic at convex solutions. (e) First order, quasilinear (Burgers' equation).

Exercise 1.4 Diffusion times

(a) Using t≈L2/2Dt \approx L^2/2D with D=2×10−9D = 2 \times 10^{-9} m²/s, check the table in the oxygen example. (b) A sugar molecule in water has D≈5×10−10D \approx 5 \times 10^{-10} m²/s. Roughly how long does it take a lump of sugar's molecules to spread 11 cm through a still cup of tea by diffusion alone? What does this tell you about why we stir? (c) If a 22 cm steak takes 88 minutes to cook through, estimate the time for a 33 cm steak from the square law, and list two reasons the estimate might be off.

Solution

(a) L2/2DL^2/2D: 10−10/4×10−9=0.02510^{-10}/4\times10^{-9} = 0.025 s; 10−6/4×10−9=25010^{-6}/4\times10^{-9} = 250 s; 10−4/4×10−9=2.5×10410^{-4}/4\times10^{-9} = 2.5 \times 10^4 s, about 77 hours; 1/4×10−9=2.5×1081/4\times10^{-9} = 2.5 \times 10^8 s, about 88 years. (b) 10−4/10−9=10510^{-4}/10^{-9} = 10^5 s, a little over a day: convection from stirring (or from temperature differences) does the mixing, not diffusion. (c) 8×(3/2)2=188 \times (3/2)^2 = 18 minutes. The pan heats one side only, so the geometry is not symmetric; heat transfer at the surface is not instantaneous; meat contracts and loses water as it cooks.

Exercise 1.5 Hadamard's example

(a) Check that un=1n2sin⁡(nx)sinh⁡(ny)u_n = \frac{1}{n^2}\sin(nx)\sinh(ny) solves Laplace's equation with un(x,0)=0u_n(x, 0) = 0 and ∂yun(x,0)=1nsin⁡(nx)\partial_yu_n(x, 0) = \frac1n\sin(nx). (b) Which of Hadamard's three conditions fails, and in which norms? (c) Why does the same construction not work for the heat equation forward in time?

Exercise 1.6 Self-similar solutions

(a) Substitute u=t−aF(x/t)u = t^{-a}F(x/\sqrt t) into ut=uxxu_t = u_{xx} and derive F′′+y2F′+aF=0F'' + \frac y2F' + aF = 0. (b) Show that ∫u(x,t) dx\int u(x, t)\,dx is independent of tt only if a=12a = \frac12 (when the integral is finite and non-zero). (c) Solve the ODE for a=12a = \frac12 and for a=0a = 0. (d) In Rn\mathbb{R}^n, show that u=t−n/2F(∣x∣/t)u = t^{-n/2}F(|x|/\sqrt t) conserves ∫u\int u, and find FF.

Solution

(a) ut=−at−a−1F−12t−a−1yF′u_t = -at^{-a-1}F - \frac12t^{-a-1}yF' and uxx=t−a−1F′′u_{xx} = t^{-a-1}F''. (b) ∫u dx=t−a∫F(x/t) dx=t12−a∫F(y) dy\int u\,dx = t^{-a}\int F(x/\sqrt t)\,dx = t^{\frac12 - a}\int F(y)\,dy. (c) a=12a = \frac12: (F′+y2F)′=0(F' + \frac y2F)' = 0; the solution decaying at infinity has F′+y2F=0F' + \frac y2F = 0, so F=Ce−y2/4F = Ce^{-y^2/4}. a=0a = 0: F′′=−y2F′F'' = -\frac y2F', so F′=Ce−y2/4F' = Ce^{-y^2/4} and F=A+C∫0ye−s2/4 dsF = A + C\int_0^ye^{-s^2/4}\,ds. (d) F=Ce−r2/4F = Ce^{-r^2/4}, the nn-dimensional heat kernel (4πt)−n/2e−∣x∣2/4t(4\pi t)^{-n/2}e^{-|x|^2/4t} when C=(4π)−n/2C = (4\pi)^{-n/2}.

Exercise 1.7 How bad is backward heat flow?

On the circle R/Z\mathbb{R}/\mathbb{Z}, suppose the state at time T=10−3T = 10^{-3} is known to within an error of 10−610^{-6} in every Fourier coefficient. Up to which frequency nn can the initial state be recovered with error at most 10−210^{-2} in each coefficient? Explain why this cut-off is a form of regularisation.

Solution

We need 10−6e4π2n2⋅10−3≤10−210^{-6}e^{4\pi^2n^2\cdot10^{-3}} \leq 10^{-2}, i.e. 4π2n2×10−3≤ln⁡104≈9.214\pi^2n^2 \times 10^{-3} \leq \ln10^4 \approx 9.21, so n2≤233n^2 \leq 233 and n≤15n \leq 15. Discarding all frequencies above 1515 trades the lost detail for stability: that is the simplest regularisation, spectral truncation.

Exercise 1.8 Where the names come from

In each of the operators ut−uxxu_t - u_{xx}, utt+uxxu_{tt} + u_{xx} and utt−uxxu_{tt} - u_{xx} (in the two variables xx, tt), replace each ∂x\partial_x by a variable ξ\xi and each ∂t\partial_t by τ\tau, to get the polynomials τ−ξ2\tau - \xi^2, τ2+ξ2\tau^2 + \xi^2 and τ2−ξ2\tau^2 - \xi^2. Show that setting each equal to a non-zero constant gives a parabola, an ellipse (here a circle) and a hyperbola.

Exercise 1.9 Rehearsal: scaling the Ricci flow

Assume the fact, proved in 9A.4 Curvature and What It Means, that Ric⁡(cg)=Ric⁡(g)\operatorname{Ric}(cg) = \operatorname{Ric}(g) for every constant c>0c > 0. (a) Show that if g(t)g(t) solves ∂tg=−2Ric⁡(g)\partial_tg = -2\operatorname{Ric}(g), then so does gλ(t)=λg(t/λ)g_\lambda(t) = \lambda g(t/\lambda). (b) The round sphere of radius rr in dimension nn has Ric⁡=n−1r2g\operatorname{Ric} = \frac{n - 1}{r^2}g. Show that g(t)=(1−2(n−1)t)g0g(t) = (1 - 2(n - 1)t)g_0 is the Ricci flow from the unit sphere, so it disappears at T=12(n−1)T = \frac{1}{2(n - 1)}. (c) Use (a) to deduce, without computing, that the sphere of radius rr disappears at time r22(n−1)\frac{r^2}{2(n - 1)}: time scales like length squared, as in the heat equation.

Solution

(a) ∂tgλ(t)=λ⋅1λ(∂tg)(t/λ)=−2Ric⁡(g(t/λ))=−2Ric⁡(λg(t/λ))\partial_tg_\lambda(t) = \lambda\cdot\frac1\lambda(\partial_tg)(t/\lambda) = -2\operatorname{Ric}(g(t/\lambda)) = -2\operatorname{Ric}(\lambda g(t/\lambda)). (b) ∂tg=−2(n−1)g0\partial_tg = -2(n - 1)g_0, and Ric⁡(g(t))=Ric⁡(g0)=(n−1)g0\operatorname{Ric}(g(t)) = \operatorname{Ric}(g_0) = (n - 1)g_0. (c) The sphere of radius rr is gλ(0)g_\lambda(0) with λ=r2\lambda = r^2, so its flow is r2g(t/r2)r^2g(t/r^2), which ends at t=r2Tt = r^2T.

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