Book 6B

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Course 6Book 6B: Linear PDE: Characteristics, Waves and RegularityChapter 5

Power Series and the Cauchy–Kovalevskaya Theorem

Noncharacteristic surfaces, majorants, the Cauchy–Kovalevskaya theorem, and Lewy’s unsolvable equation.

35 min read · Updated Oct 23, 2026

Read with Evans, Partial Differential Equations, section 4.6 (noncharacteristic surfaces, real analytic functions, the Cauchy–Kovalevskaya theorem); Taylor I, §6.4 ("The Cauchy–Kowalewsky theorem") and §3.10 ("Local solvability of constant-coefficient PDE"); Taylor III, §16.4, for the nonlinear version. This chapter proves the Cauchy–Kovalevskaya theorem by majorants in full.

In this chapter · 7 sections
  1. 5.1Continuing a potential field downward
  2. 5.2Real analytic functions
  3. 5.3Noncharacteristic surfaces
  4. 5.4The Cauchy–Kovalevskaya theorem
  5. 5.5What the theorem does not say
  6. 5.6History
  7. 5.7Exercises
  8. ↗Free resources

The simplest idea for solving a differential equation is to write the solution as a power series and determine its coefficients one at a time. For ODEs this works whenever the coefficients are analytic, and it is how Bessel's and Legendre's functions were first found (6B.4 Green’s Functions, Separation of Variables and Transforms). For PDEs it works too, under one condition on the surface carrying the data: the Cauchy–Kovalevskaya theorem says that a PDE with analytic coefficients and analytic data on a noncharacteristic surface has a unique analytic solution near that surface. It is the oldest general existence theorem for PDE, and the only one that applies to equations of every type at once.

It is also the clearest example of a theorem that answers the wrong question. It says nothing about stability: Hadamard's ill-posed problem for the Laplace equation (6A.1 What a PDE Is) satisfies all its hypotheses. It says nothing about non-analytic data: Hans Lewy found in 1957 a simple linear equation with smooth coefficients that has no solutions at all for most smooth right-hand sides. And for the heat equation with initial data it does not apply, because the initial surface is characteristic, and Sofia Kovalevskaya herself showed that the formal power series can diverge. These failures are why modern PDE theory works in Sobolev spaces with energy estimates. This chapter proves the theorem carefully and then explains, with examples, exactly what it does not do.

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

  • work with real analytic functions of several variables, multi-index notation and majorant series;
  • decide whether a surface is characteristic for a linear equation, and show how noncharacteristic data determine all derivatives on the surface;
  • prove the Cauchy–Kovalevskaya theorem for first-order systems by the method of majorants;
  • explain Kovalevskaya's heat-equation example, Hadamard's instability, Holmgren's uniqueness theorem and Lewy's unsolvable equation;
  • prove that solutions of the heat equation are analytic in space at positive times.

Continuing a potential field downward

In the world In use Downward continuation in geophysics

Aircraft and satellites measure the Earth's gravity and magnetic fields at altitude. Above the ground, where there are no sources, the gravitational potential is harmonic, so the field at flight height determines it everywhere above the ground. Geophysicists often want the field closer to its sources, at ground level or below, where the anomalies caused by buried rock or ore bodies are sharper. Computing it from the data at altitude is downward continuation: solving Laplace's equation with the potential and its vertical derivative given on a horizontal plane, and moving towards the sources.

This is a Cauchy problem for an elliptic equation, exactly Hadamard's example (6A.1 What a PDE Is). The data are analytic (a harmonic function is analytic, 6A.2 Harmonic Functions), and the plane is noncharacteristic, so the Cauchy–Kovalevskaya theorem guarantees that a solution exists near the plane and is unique. But a Fourier mode of wavenumber kk in the horizontal direction grows like e2π∣k∣he^{2\pi|k|h} when continued down a distance hh, so measurement noise at short wavelengths is amplified exponentially. In practice downward continuation is always regularised, by filtering out short wavelengths or by adding prior information, and it is reliable only over distances small compared with the wavelengths of interest. Upward continuation, the reverse operation, damps short wavelengths like e−2π∣k∣he^{-2\pi|k|h} and is perfectly stable: the boundary-value problem is well posed, the Cauchy problem is not.

Real analytic functions

A function f:U→Rf : U \to \mathbb{R} on an open set U⊆RnU \subseteq \mathbb{R}^n is real analytic near x0∈Ux_0 \in U if there are r>0r > 0 and constants fαf_\alpha such that

f(x)=∑αfα(x−x0)αfor ∣x−x0∣<r,f(x) = \sum_\alpha f_\alpha(x - x_0)^\alpha \qquad\text{for } |x - x_0| < r,

the series converging absolutely. Here α=(α1,…,αn)\alpha = (\alpha_1, \ldots, \alpha_n) runs over multi-indices of non-negative integers, xα=x1α1⋯xnαnx^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}, ∣α∣=α1+⋯+αn|\alpha| = \alpha_1 + \cdots + \alpha_n, α!=α1!⋯αn!\alpha! = \alpha_1!\cdots\alpha_n! and ∂α=∂1α1⋯∂nαn\partial^\alpha = \partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}. A convergent power series can be differentiated term by term inside its region of convergence, so ff is smooth and fα=∂αf(x0)α!f_\alpha = \frac{\partial^\alpha f(x_0)}{\alpha!}: the series is the Taylor series. Polynomials, exponentials, sin⁡\sin, cos⁡\cos, and 11−x\frac{1}{1 - x} near 00 are analytic; so are sums, products, quotients (where the denominator is non-zero) and compositions of analytic functions, and harmonic functions (6A.2 Harmonic Functions). The function e−1/x2e^{-1/x^2} (with value 00 at 00) is smooth but not analytic at 00: all its derivatives vanish there, so its Taylor series is zero, yet the function is not.

A useful test: ff is analytic near x0x_0 if and only if its derivatives satisfy Cauchy-type estimates ∣∂αf(x)∣≤C∣α∣!ρ∣α∣|\partial^\alpha f(x)| \leq C\frac{|\alpha|!}{\rho^{|\alpha|}} uniformly near x0x_0, for some CC and ρ>0\rho > 0. The "if" direction is the remainder estimate of 6A.2 Harmonic Functions's proof that harmonic functions are analytic.

Majorants. For two power series f=∑fαxαf = \sum f_\alpha x^\alpha and g=∑gαxαg = \sum g_\alpha x^\alpha, say gg majorises ff, written g≫fg \gg f, if gα≥∣fα∣g_\alpha \geq |f_\alpha| for every α\alpha. If gg converges for ∣x∣<r|x| < r and g≫fg \gg f, then ff converges there too (each term of ff is dominated in absolute value by the corresponding term of gg evaluated at (∣x1∣,…,∣xn∣)(|x_1|, \ldots, |x_n|)). The basic majorant is a geometric series.

Lemma 5.1 Every analytic function has a geometric majorant

Let f=∑fαxαf = \sum f_\alpha x^\alpha converge for ∣x∣<r|x| < r, and let 0<sn<r0 < s\sqrt n < r. Then there is CC such that

f≪Css−(x1+⋯+xn),and the right side converges for ∣x∣<sn.f \ll \frac{Cs}{s - (x_1 + \cdots + x_n)}, \qquad\text{and the right side converges for } |x| < \frac{s}{\sqrt n}.

Proof. The point y=(s,…,s)y = (s, \ldots, s) has ∣y∣=sn<r|y| = s\sqrt n < r, so ∑fαyα=∑fαs∣α∣\sum f_\alpha y^\alpha = \sum f_\alpha s^{|\alpha|} converges absolutely; its terms are bounded, ∣fα∣s∣α∣≤C|f_\alpha|s^{|\alpha|} \leq C. Expanding the right side,

Css−(x1+⋯+xn)=C∑k=0∞(x1+⋯+xns)k=C∑α∣α∣!α!xαs∣α∣,\frac{Cs}{s - (x_1 + \cdots + x_n)} = C\sum_{k=0}^\infty\Big(\frac{x_1 + \cdots + x_n}{s}\Big)^k = C\sum_\alpha\frac{|\alpha|!}{\alpha!}\frac{x^\alpha}{s^{|\alpha|}},

by the multinomial theorem. Since ∣α∣!α!≥1\frac{|\alpha|!}{\alpha!} \geq 1, its coefficients are at least Cs−∣α∣≥∣fα∣Cs^{-|\alpha|} \geq |f_\alpha|. It converges when ∣x1∣+⋯+∣xn∣<s|x_1| + \cdots + |x_n| < s, which holds if ∣x∣<s/n|x| < s/\sqrt n (Cauchy–Schwarz).

Noncharacteristic surfaces

Consider a linear equation of order kk,

∑∣α∣≤kaα(x) ∂αu=f,\sum_{|\alpha|\leq k}a_\alpha(x)\,\partial^\alpha u = f,

with Cauchy data on a smooth hypersurface Γ\Gamma with unit normal ν\nu: the values of uu and of its normal derivatives up to order k−1k - 1, that is, u,∂νu,…,∂νk−1uu, \partial_\nu u, \ldots, \partial_\nu^{k-1}u on Γ\Gamma. Can these data and the equation determine all the derivatives of uu on Γ\Gamma, so that the Taylor series of uu at each point of Γ\Gamma is known?

Flatten Γ\Gamma to the plane {xn=0}\{x_n = 0\} by a change of coordinates, so ν=en\nu = e_n. The data then give every derivative ∂βu\partial^\beta u on Γ\Gamma with βn≤k−1\beta_n \leq k - 1 (differentiate the data tangentially). The only kk-th order derivative not yet known is ∂nku\partial_n^ku, and the equation contains it with the coefficient a(0,…,0,k)a_{(0,\ldots,0,k)}. If that coefficient is non-zero, the equation can be solved for it:

∂nku=1a(0,…,0,k)(f−∑α≠(0,…,0,k)aα∂αu)on Γ.\partial_n^ku = \frac{1}{a_{(0,\ldots,0,k)}}\Big(f - \sum_{\alpha\neq(0,\ldots,0,k)}a_\alpha\partial^\alpha u\Big) \quad\text{on } \Gamma.

Differentiating the equation and repeating determines all higher normal derivatives too (Exercise 5.5). In general coordinates the coefficient is the principal symbol evaluated on the normal.

Definition 5.2 Noncharacteristic surface

Γ\Gamma is noncharacteristic for the operator at the point x0∈Γx_0 \in \Gamma if

∑∣α∣=kaα(x0) να≠0,\sum_{|\alpha|=k}a_\alpha(x_0)\,\nu^\alpha \neq 0,

where ν\nu is the unit normal to Γ\Gamma at x0x_0 and να=ν1α1⋯νnαn\nu^\alpha = \nu_1^{\alpha_1}\cdots\nu_n^{\alpha_n}. Otherwise it is characteristic there.

The expression is the principal symbol σ(x0,ν)\sigma(x_0, \nu) of 6A.1 What a PDE Is (up to factors of ii). So:

  • For an elliptic operator, σ(x,ν)≠0\sigma(x, \nu) \neq 0 for every ν≠0\nu \neq 0: no surface is characteristic. This is why Cauchy–Kovalevskaya applies to the Laplace equation with data on any surface, and also why the theorem can't tell good problems from bad ones.
  • For the wave equation utt−Δuu_{tt} - \Delta u, with ν=(νx,νt)\nu = (\nu_x, \nu_t), the symbol is νt2−∣νx∣2\nu_t^2 - |\nu_x|^2; the characteristic surfaces are those whose normal lies on the light cone, such as the cones ∣x∣=t|x| = t. Spacelike surfaces, such as t=0t = 0, are noncharacteristic.
  • For the heat equation ut−Δuu_t - \Delta u, the principal part is −Δ-\Delta (order 22 in xx only), and the symbol is −∣νx∣2-|\nu_x|^2. The initial surface t=0t = 0 has normal ν=(0,1)\nu = (0, 1) and symbol 00: it is characteristic. Data at t=0t = 0 give uu, and the equation then gives ut=Δuu_t = \Delta u, but not by solving for a second tt-derivative; the time derivatives are determined by differentiating the data in xx, which loses derivatives, and the resulting formal series can diverge (Exercise 5.8).

The Cauchy–Kovalevskaya theorem

By introducing all lower-order derivatives as new unknowns, any noncharacteristic problem for a kk-th order system can be reduced to a first-order system of the form below (Evans, section 4.6.3, carries out the reduction). So it suffices to prove:

Theorem 5.3 The Cauchy–Kovalevskaya theorem

Let u=(u1,…,um)\mathbf u = (u^1, \ldots, u^m) and consider

∂nu=∑j=1n−1Bj(u,x′) ∂ju+c(u,x′)near 0,u=0  on {xn=0},\partial_n\mathbf u = \sum_{j=1}^{n-1}\mathbf B_j(\mathbf u, x')\,\partial_j\mathbf u + \mathbf c(\mathbf u, x') \quad\text{near } 0, \qquad \mathbf u = 0 \ \text{ on } \{x_n = 0\},

where x′=(x1,…,xn−1)x' = (x_1, \ldots, x_{n-1}), the Bj\mathbf B_j are m×mm\times m matrices and c\mathbf c an mm-vector, all real analytic functions of (z,x′)∈Rm×Rn−1(\mathbf z, x') \in \mathbb{R}^m\times\mathbb{R}^{n-1} near 00. Then there is a real analytic solution near 00, and it is unique among real analytic solutions.

The form is less special than it looks. The data were made zero by subtracting any analytic function with the given data; the coefficients do not depend on xnx_n, which can be arranged by adding xnx_n as an extra unknown um+1u^{m+1} with ∂num+1=1\partial_nu^{m+1} = 1 and zero data; and the equation is solved for ∂nu\partial_n\mathbf u, which is what noncharacteristic means.

Proof. Step 1: the coefficients are determined, by polynomials with non-negative coefficients. Write uk,α=∂αuk(0)u_{k,\alpha} = \partial^\alpha u^k(0), bj,γδkl=∂zγ∂x′δbjkl(0,0)b^{kl}_{j,\gamma\delta} = \partial_{\mathbf z}^\gamma\partial_{x'}^\delta b_j^{kl}(0, 0) and cγδkc^k_{\gamma\delta} likewise. If αn=0\alpha_n = 0, then uk,α=0u_{k,\alpha} = 0, because u=0\mathbf u = 0 on {xn=0}\{x_n = 0\} and ∂α\partial^\alpha only differentiates tangentially. If αn≥1\alpha_n \geq 1, write α=β+en\alpha = \beta + e_n and apply ∂β\partial^\beta to the kk-th equation:

∂αuk=∂β(∑j,lbjkl(u,x′) ∂jul+ck(u,x′)).\partial^\alpha u^k = \partial^\beta\Big(\sum_{j,l}b_j^{kl}(\mathbf u, x')\,\partial_ju^l + c^k(\mathbf u, x')\Big).

Expand the right side at 00 by the Leibniz rule and the chain rule. Every term is a product of derivatives of bjklb_j^{kl} and ckc^k at (0,0)(0, 0) with derivatives ul,γu_{l,\gamma}, and every coefficient in the expansion is a non-negative integer. The derivatives of u\mathbf u that appear have order at most ∣α∣|\alpha| and, crucially, nn-th index at most βn=αn−1\beta_n = \alpha_n - 1 (the xnx_n-derivatives in ∂β\partial^\beta all fall on factors of u\mathbf u, and ∂jul\partial_ju^l with j<nj < n adds none). So

uk,α=pαk({bj,γδkl}, {cγδk}, {ul,γ:γn≤αn−1})u_{k,\alpha} = p^k_\alpha\big(\{b^{kl}_{j,\gamma\delta}\},\ \{c^k_{\gamma\delta}\},\ \{u_{l,\gamma} : \gamma_n \leq \alpha_n - 1\}\big)

for a polynomial pαkp^k_\alpha with non-negative coefficients. By induction on αn\alpha_n, every uk,αu_{k,\alpha} is determined. In particular any analytic solution must have these Taylor coefficients, which proves uniqueness.

Step 2: majorants control the coefficients. Suppose Bj∗≫Bj\mathbf B_j^* \gg \mathbf B_j and c∗≫c\mathbf c^* \gg \mathbf c (entry by entry, as power series in (z,x′)(\mathbf z, x')), and that the problem with starred coefficients has an analytic solution u∗\mathbf u^* near 00. Then u∗≫u\mathbf u^* \gg \mathbf u, where u\mathbf u is the formal series of Step 1. Indeed, by induction on αn\alpha_n, using that pαkp^k_\alpha has non-negative coefficients,

∣uk,α∣=∣pαk(b,c,u)∣≤pαk(∣b∣,∣c∣,∣u∣)≤pαk(b∗,c∗,u∗)=uk,α∗.|u_{k,\alpha}| = |p^k_\alpha(b, c, u)| \leq p^k_\alpha(|b|, |c|, |u|) \leq p^k_\alpha(b^*, c^*, u^*) = u^*_{k,\alpha}.

Since the series of u∗\mathbf u^* converges near 00, so does that of u\mathbf u.

Step 3: a majorant problem that can be solved. By Lemma 5.1, applied to all entries in the m+n−1m + n - 1 variables (z,x′)(\mathbf z, x'), there are CC and r>0r > 0 with

Bj≪Crr−(x1+⋯+xn−1)−(z1+⋯+zm)E,c≪Crr−(x1+⋯+xn−1)−(z1+⋯+zm)(1,…,1),\mathbf B_j \ll \frac{Cr}{r - (x_1 + \cdots + x_{n-1}) - (z_1 + \cdots + z_m)}\mathbf E, \qquad \mathbf c \ll \frac{Cr}{r - (x_1 + \cdots + x_{n-1}) - (z_1 + \cdots + z_m)}(1, \ldots, 1),

where E\mathbf E is the m×mm\times m matrix of ones. Use the right sides as Bj∗\mathbf B_j^* and c∗\mathbf c^*. Look for a solution of the starred problem of the form u∗=v(1,…,1)\mathbf u^* = v(1, \ldots, 1) with vv scalar. Writing s=x1+⋯+xn−1s = x_1 + \cdots + x_{n-1}, the system becomes the single equation

∂nv=Crr−s−mv(m∑j=1n−1∂jv+1),v=0  on {xn=0}.\partial_nv = \frac{Cr}{r - s - mv}\Big(m\sum_{j=1}^{n-1}\partial_jv + 1\Big), \qquad v = 0 \ \text{ on } \{x_n = 0\}.

Its solution is

v(x)=1mn(r−s−(r−s)2−2mnCr xn),v(x) = \frac{1}{mn}\Big(r - s - \sqrt{(r - s)^2 - 2mnCr\,x_n}\Big),

which you can verify by differentiating (Exercise 5.6); it is analytic near 00, where the expression under the square root is close to r2>0r^2 > 0, and it vanishes at xn=0x_n = 0.

Step 4: conclusion. By Steps 2 and 3, the formal series u\mathbf u of Step 1 converges near 00, defining an analytic function. Both sides of the equation are then analytic functions near 00, and by construction their Taylor coefficients at 00 agree, so they agree on a neighbourhood of 00 (two analytic functions with the same Taylor series at a point agree near it). And u=0\mathbf u = 0 on {xn=0}\{x_n = 0\} because all coefficients with αn=0\alpha_n = 0 vanish.

The proof is a comparison argument, like the maximum principle, but at the level of Taylor coefficients: an explicitly solvable problem with larger coefficients has a larger (convergent) solution, and so the original converges. The non-negativity of the recursion's coefficients plays the role that positivity of the heat kernel played in 6A.5 Maximum Principles.

What the theorem does not say

It is local. The solution exists in a neighbourhood whose size depends on the radii of convergence of the data and coefficients, and it may not extend further, as for the ODE y′=y2y' = y^2 (2B.10 Ordinary Differential Equations).

It gives no stability. Hadamard's problem, uxx+uyy=0u_{xx} + u_{yy} = 0 with u(x,0)=0u(x, 0) = 0 and uy(x,0)=1nsin⁡nxu_y(x, 0) = \frac1n\sin nx, has analytic data on a noncharacteristic line, so Cauchy–Kovalevskaya applies; the solution 1n2sin⁡nxsinh⁡ny\frac{1}{n^2}\sin nx\sinh ny exists, is analytic and is unique, and is huge at y=1y = 1 although the data are tiny (6A.1 What a PDE Is). Existence and uniqueness without continuous dependence is useless for computation, as the downward-continuation box explained.

Uniqueness extends beyond analytic solutions. Step 1 only shows uniqueness among analytic solutions. Erik Holmgren proved in 1901 that for linear equations with analytic coefficients, a CkC^k solution with zero Cauchy data on a noncharacteristic surface vanishes near it: there are no non-analytic solutions with the same data either. The proof uses Cauchy–Kovalevskaya for the adjoint equation to build enough test functions (Taylor I, §6.4; John, Partial Differential Equations). For the wave equation, Holmgren's theorem gives uniqueness in the domain of dependence, which 6B.3 The Wave Equation proved with energy.

It says nothing about non-analytic data, and smooth data can have no solution. Hans Lewy found in 1957 that the linear operator

Lu=∂xu+i∂yu−2i(x+iy)∂tuLu = \partial_xu + i\partial_yu - 2i(x + iy)\partial_tu

on R3\mathbb{R}^3, with simple polynomial coefficients, has the property that for most smooth (indeed C∞C^\infty) functions ff, the equation Lu=fLu = f has no solution, even a distributional one, in any open set. It is not a pathology of bad coefficients: LL is a natural operator in complex analysis (the tangential Cauchy–Riemann operator on the boundary of a domain in C2\mathbb{C}^2). By contrast, Bernard Malgrange and Leon Ehrenpreis proved in 1954–55 that every linear operator with constant coefficients has a fundamental solution, and so is locally solvable (Taylor I, §3.10). Lewy's example showed that local solvability for variable coefficients is a genuine question, answered for large classes of operators by Lars Hörmander, Louis Nirenberg and François Trèves in the 1960s and 70s.

It does not apply to the heat equation's initial-value problem, whose initial surface is characteristic, and there the formal series can diverge (Exercise 5.8). Yet the heat equation's solutions are as smooth as can be: for t>0t > 0 they are analytic in xx, for any bounded initial data (Exercise 5.10). Analyticity is produced by the equation, not assumed.

The lesson the subject drew from these examples is that the analytic category is the wrong setting: too rigid to contain the solutions of interest, and blind to stability. Existence should be proved in spaces defined by norms, with estimates that give continuous dependence: energy spaces for hyperbolic equations (6B.14 Hyperbolic Equations and Systems), Sobolev and Hölder spaces for elliptic and parabolic ones (6A.6 Weak Solutions and Elliptic Regularity, 6B.7 Schauder Theory and the Continuity Method).

Where this goes Analyticity and the Ricci flow

Cauchy–Kovalevskaya plays a small but real role in geometry. It proves that analytic Riemannian metrics admit local isometric embeddings into Euclidean space of high dimension (Janet and Cartan, 1926–27), and that Einstein's vacuum equations have local analytic solutions with analytic data. For the Ricci flow, the analytic side appears the other way round: like the heat equation, the Ricci flow produces analyticity, and Shigetoshi Bando showed in 1987 that a solution of the Ricci flow on a compact manifold is real analytic in space at every positive time, whatever its smooth initial metric. The proof combines Shi's derivative estimates (6A.7 Parabolic Regularity, 11A.5 Shi's Derivative Estimates) with bounds on how the constants grow with the order of the derivative, which is the same mechanism as Exercise 5.10.

History

Augustin-Louis Cauchy proved existence for analytic ODE and for some first-order PDE by majorants in the 1840s ("calcul des limites"). Sofia Kovalevskaya proved the general theorem in her 1874 doctoral thesis at Göttingen, supervised by Karl Weierstrass, and gave the example showing that the heat equation's formal series can diverge; she was among the first women to earn a doctorate in mathematics, and became the first woman to hold a full professorship in mathematics in northern Europe (Stockholm, 1889). Erik Holmgren's uniqueness theorem dates from 1901. Jacques Hadamard's analysis of the ill-posed Cauchy problem appeared in 1902 and in his 1923 lectures. Hans Lewy's example was published in 1957, and the Malgrange–Ehrenpreis theorem in 1954–55.

Recall Where we stand

Real analytic functions are given locally by convergent power series, and every one has a geometric majorant Css−(x1+⋯+xn)\frac{Cs}{s - (x_1 + \cdots + x_n)}. A surface is noncharacteristic when the principal symbol is non-zero on its normal; then Cauchy data and the equation determine every derivative on it. For elliptic operators every surface is noncharacteristic; for the heat equation t=0t = 0 is characteristic. The Cauchy–Kovalevskaya theorem gives a unique local analytic solution of a noncharacteristic problem with analytic data: the Taylor coefficients are non-negative polynomials of lower ones, and comparison with an explicitly solvable majorant problem proves convergence. It gives no stability (Hadamard), no information for smooth data (Lewy's unsolvable equation), and does not apply to the heat equation's initial-value problem (Kovalevskaya's divergent series), whose solutions are nonetheless analytic in space for t>0t > 0. Holmgren extends uniqueness to CkC^k solutions of linear equations. 6B.6 Perron’s Method and Barriers returns to the Dirichlet problem, and solves it on general domains by Perron's method.

Exercises

Exercise 5.4 Majorants in one variable

(a) Show that 11−x\frac{1}{1 - x} majorises sin⁡x\sin x, cos⁡x\cos x and exe^x near 00, and find a majorant of the form Css−x\frac{Cs}{s - x} for 12−x\frac{1}{2 - x}. (b) Show that if f≪gf \ll g and h≪kh \ll k then fh≪gkfh \ll gk and f+h≪g+kf + h \ll g + k.

Solution

(a) The coefficients of sin⁡\sin, cos⁡\cos and exe^x are at most 11 in absolute value, and those of 11−x\frac{1}{1 - x} are all 11. 12−x=∑xk2k+1\frac{1}{2 - x} = \sum\frac{x^k}{2^{k+1}}, which equals Css−x\frac{Cs}{s - x} with s=2s = 2, C=12C = \frac12. (b) The coefficients of fhfh are ∑β+γ=αfβhγ\sum_{\beta+\gamma=\alpha}f_\beta h_\gamma, bounded in absolute value by ∑gβkγ\sum g_\beta k_\gamma, the coefficients of gkgk; sums are immediate.

Exercise 5.5 Determining derivatives on a noncharacteristic surface

For the Laplace equation uxx+uyy=0u_{xx} + u_{yy} = 0 with data u(x,0)=f(x)u(x, 0) = f(x) and uy(x,0)=g(x)u_y(x, 0) = g(x) on the line y=0y = 0, show that ∂y2ku(x,0)=(−1)kf(2k)(x)\partial_y^{2k}u(x, 0) = (-1)^kf^{(2k)}(x) and ∂y2k+1u(x,0)=(−1)kg(2k)(x)\partial_y^{2k+1}u(x, 0) = (-1)^kg^{(2k)}(x). Write down the formal power series in yy, and check it for f=0f = 0, g=1nsin⁡nxg = \frac1n\sin nx against Hadamard's solution.

Solution

uyy=−uxxu_{yy} = -u_{xx}, so ∂y2ku=(−∂x2)ku\partial_y^{2k}u = (-\partial_x^2)^ku and ∂y2k+1u=(−∂x2)k∂yu\partial_y^{2k+1}u = (-\partial_x^2)^k\partial_yu; evaluate at y=0y = 0. The series is u=∑k(−1)k(f(2k)(x)y2k(2k)!+g(2k)(x)y2k+1(2k+1)!)u = \sum_k(-1)^k\big(f^{(2k)}(x)\frac{y^{2k}}{(2k)!} + g^{(2k)}(x)\frac{y^{2k+1}}{(2k+1)!}\big). With g=1nsin⁡nxg = \frac1n\sin nx, g(2k)=(−1)kn2k−1sin⁡nxg^{(2k)} = (-1)^kn^{2k-1}\sin nx, giving 1nsin⁡nx∑n2ky2k+1(2k+1)!=1n2sin⁡nxsinh⁡ny\frac1n\sin nx\sum\frac{n^{2k}y^{2k+1}}{(2k + 1)!} = \frac{1}{n^2}\sin nx\sinh ny.

Exercise 5.6 The majorant problem

Verify that v=1mn(r−s−(r−s)2−2mnCrxn)v = \frac{1}{mn}\big(r - s - \sqrt{(r - s)^2 - 2mnCrx_n}\big), with s=x1+⋯+xn−1s = x_1 + \cdots + x_{n-1}, solves ∂nv=Crr−s−mv(m∑j<n∂jv+1)\partial_nv = \frac{Cr}{r - s - mv}\big(m\sum_{j<n}\partial_jv + 1\big) with v=0v = 0 at xn=0x_n = 0.

Hint

Write Q=(r−s)2−2mnCrxnQ = \sqrt{(r - s)^2 - 2mnCrx_n}. Then r−s−mv=(n−1)(r−s)+Qnr - s - mv = \frac{(n - 1)(r - s) + Q}{n}, ∂nv=CrQ\partial_nv = \frac{Cr}{Q}, and each ∂jv=1mn(−1+r−sQ)\partial_jv = \frac{1}{mn}\big(-1 + \frac{r - s}{Q}\big) for j<nj < n.

Solution

Using the hint, m∑j<n∂jv+1=n−1n(r−sQ−1)+1=(n−1)(r−s)+QnQm\sum_{j<n}\partial_jv + 1 = \frac{n - 1}{n}\big(\frac{r - s}{Q} - 1\big) + 1 = \frac{(n - 1)(r - s) + Q}{nQ}. So the right side is Cr n(n−1)(r−s)+Q⋅(n−1)(r−s)+QnQ=CrQ=∂nv\frac{Cr\,n}{(n - 1)(r - s) + Q}\cdot\frac{(n - 1)(r - s) + Q}{nQ} = \frac{Cr}{Q} = \partial_nv. At xn=0x_n = 0, Q=r−sQ = r - s and v=0v = 0.

Exercise 5.7 Characteristic or not

Decide whether each surface is characteristic for each operator: (a) the line x+y=0x + y = 0 for uxx−uyyu_{xx} - u_{yy}; (b) the line y=0y = 0 for uxx−uyyu_{xx} - u_{yy}; (c) the plane t=0t = 0 for utt−Δuu_{tt} - \Delta u; (d) the cone t=∣x∣t = |x| for utt−Δuu_{tt} - \Delta u; (e) the plane x1=0x_1 = 0 for ut−Δuu_t - \Delta u.

Solution

(a) Normal (1,1)2\frac{(1, 1)}{\sqrt2}, symbol νx2−νy2=0\nu_x^2 - \nu_y^2 = 0: characteristic. (b) Normal (0,1)(0, 1): symbol −1≠0-1 \neq 0: noncharacteristic. (c) Normal (0,1)(0, 1) in (x,t)(x, t): νt2−∣νx∣2=1\nu_t^2 - |\nu_x|^2 = 1: noncharacteristic. (d) Normal proportional to (−x/∣x∣,1)(-x/|x|, 1): 1−1=01 - 1 = 0: characteristic. (e) Normal e1e_1: the symbol of the principal part −Δ-\Delta is −∣νx∣2=−1≠0-|\nu_x|^2 = -1 \neq 0: noncharacteristic (data on a timelike plane are fine; it is the initial plane t=0t = 0 that is characteristic).

Exercise 5.8 Kovalevskaya's example

Consider ut=uxxu_t = u_{xx} with u(x,0)=g(x)=11+x2u(x, 0) = g(x) = \frac{1}{1 + x^2}. (a) Show that a formal power series solution must satisfy ∂tku(x,0)=g(2k)(x)\partial_t^ku(x, 0) = g^{(2k)}(x). (b) Using g(x)=∑j(−1)jx2jg(x) = \sum_j(-1)^jx^{2j} near 00, show g(2k)(0)=(−1)k(2k)!g^{(2k)}(0) = (-1)^k(2k)!. (c) Deduce that the Taylor series of uu in tt at (0,0)(0, 0) is ∑k(−1)k(2k)!k!tk\sum_k(-1)^k\frac{(2k)!}{k!}t^k, which diverges for every t≠0t \neq 0. So no analytic solution exists near (0,0)(0, 0), although gg is analytic. (The heat kernel solution exists and is smooth; it is just not analytic in tt at t=0t = 0.)

Solution

(a) ∂tku=∂x2ku\partial_t^ku = \partial_x^{2k}u, evaluated at t=0t = 0. (b) The coefficient of x2kx^{2k} is (−1)k(-1)^k, so g(2k)(0)=(−1)k(2k)!g^{(2k)}(0) = (-1)^k(2k)!. (c) The coefficients (2k)!k!\frac{(2k)!}{k!} grow faster than any geometric sequence ((2k+2)!/(k+1)!(2k)!/k!=2(2k+1)→∞\frac{(2k + 2)!/(k + 1)!}{(2k)!/k!} = 2(2k + 1) \to \infty), so by the ratio test the series diverges for all t≠0t \neq 0.

Exercise 5.9 Holmgren and the wave equation

Explain why Holmgren's theorem implies that a C2C^2 solution of utt=uxxu_{tt} = u_{xx} with u=ut=0u = u_t = 0 on the segment {t=0,∣x∣<1}\{t = 0, |x| < 1\} vanishes near that segment, and compare with the conclusion of 6B.3 The Wave Equation's finite speed of propagation, which gives vanishing on the whole triangle ∣x∣<1−t|x| < 1 - t.

Solution

The equation is linear with constant (hence analytic) coefficients, and t=0t = 0 is noncharacteristic, so Holmgren gives vanishing in a neighbourhood of the segment. Applied repeatedly on a family of noncharacteristic surfaces sweeping out the triangle (curves that bend up towards its apex while staying spacelike), it gives the whole domain of dependence, the same conclusion as the energy argument. The energy argument is more robust: it works for non-analytic coefficients.

Exercise 5.10 Rehearsal: heat flow makes functions analytic

Let g∈L2(R)g \in L^2(\mathbb{R}) and u(⋅,t)=Φ(⋅,t)∗gu(\cdot, t) = \Phi(\cdot, t) * g. (a) Using the Fourier transform and Plancherel (4A.6 The Fourier Transform), show ∥∂xku(⋅,t)∥L2≤sup⁡y≥0(yke−y2t)∥g∥L2=(k2et)k/2∥g∥L2\|\partial_x^ku(\cdot, t)\|_{L^2} \leq \sup_{y\geq0}\big(y^ke^{-y^2t}\big)\|g\|_{L^2} = \big(\frac{k}{2et}\big)^{k/2}\|g\|_{L^2}, where y=2π∣ξ∣y = 2\pi|\xi|. (b) Using kk≤ekk!k^k \leq e^kk!, deduce ∥∂xku(⋅,t)∥L2≤(2t)−k/2k! ∥g∥L2\|\partial_x^ku(\cdot, t)\|_{L^2} \leq (2t)^{-k/2}\sqrt{k!}\,\|g\|_{L^2}, and use the one-dimensional Sobolev bound sup⁡∣w∣≤∥w∥2+∥w′∥2\sup|w| \leq \|w\|_2 + \|w'\|_2 to show sup⁡∣∂xku(⋅,t)∣≤2(2t)−(k+1)/2(k+1)! ∥g∥L2\sup|\partial_x^ku(\cdot, t)| \leq 2(2t)^{-(k+1)/2}\sqrt{(k + 1)!}\,\|g\|_{L^2} for 0<t≤120 < t \leq \frac12. (c) Conclude that the Taylor series of u(⋅,t)u(\cdot, t) at any point converges on the whole line, so u(⋅,t)u(\cdot, t) is real analytic (even entire) for every t>0t > 0. Bando's theorem for the Ricci flow is the geometric version of this: derivative bounds whose constants grow no faster than k!k! give analyticity.

Solution

(a) ∂xku^=(2πiξ)ke−4π2ξ2tg^\widehat{\partial_x^ku} = (2\pi i\xi)^ke^{-4\pi^2\xi^2t}\hat g, and sup⁡yyke−y2t\sup_y y^ke^{-y^2t} is attained at y2=k2ty^2 = \frac{k}{2t}, with value (k2t)k/2e−k/2\big(\frac{k}{2t}\big)^{k/2}e^{-k/2}. (b) From kk≤ekk!k^k \leq e^kk!, kk/2≤ek/2k!k^{k/2} \leq e^{k/2}\sqrt{k!}, so (k2et)k/2≤(2t)−k/2k!\big(\frac{k}{2et}\big)^{k/2} \leq (2t)^{-k/2}\sqrt{k!}. The Sobolev bound holds because w(x)2=∫−∞x2ww′≤2∥w∥2∥w′∥2≤∥w∥22+∥w′∥22w(x)^2 = \int_{-\infty}^x2ww' \leq 2\|w\|_2\|w'\|_2 \leq \|w\|_2^2 + \|w'\|_2^2. Then sup⁡∣∂xku∣≤∥∂xku∥2+∥∂xk+1u∥2≤((2t)−k/2k!+(2t)−(k+1)/2(k+1)!)∥g∥2\sup|\partial_x^ku| \leq \|\partial_x^ku\|_2 + \|\partial_x^{k+1}u\|_2 \leq \big((2t)^{-k/2}\sqrt{k!} + (2t)^{-(k+1)/2}\sqrt{(k + 1)!}\big)\|g\|_2, and for 2t≤12t \leq 1 each term is at most (2t)−(k+1)/2(k+1)!∥g∥2(2t)^{-(k+1)/2}\sqrt{(k + 1)!}\|g\|_2. (For larger tt, start the flow at time t−12t - \frac12 instead.) (c) The Taylor coefficients satisfy ∣∂ku(x0)∣k!≤C(k+1)!k!(2t)−(k+1)/2\frac{|\partial^ku(x_0)|}{k!} \leq C\frac{\sqrt{(k + 1)!}}{k!}(2t)^{-(k+1)/2}, and (k+1)!k!Rk→0\frac{\sqrt{(k+1)!}}{k!}R^k \to 0 for every RR, so the series converges everywhere; the remainder in Taylor's theorem is bounded by the next term's size on any interval, so the series converges to uu.

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