Analytic Majorants and the Cauchy–Kovalevskaya Theorem: Examples

For $\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }<1$, the geometric expansion and binomial formula give $[x^a{y}^b](1-x-y{)}^{-1}=\left(\genfrac{}{}{0px}{}{a+b}{a}\right)$. Hence $[x^a{y}^b]f=\left(\genfrac{}{}{0px}{}{a+b}{a}\right)+{\mathbf{1}}_{(a,b)=(2,0)}$. The proposed majorant has coefficient $2^{a+b+1}\left(\genfrac{}{}{0px}{}{a+b}{a}\right)$.

Outside $(a,b)=(2,0)$ the comparison follows from $2^{a+b+1}\ge 2>1$. At $(2,0)$ the coefficients are 2 and 8, respectively. In particular the constant coefficients are 1 and 2. Every coefficient is nonnegative, so these inequalities give the claimed majorisation by F1.

The map $(t,x)\mapsto x-at$ is linear and sends zero to zero, so F1 makes $u=g(x-at)$ analytic on a sufficiently small neighborhood. F2 gives $u_t=-a\cdot Dg(x-at)$ and $D_{x}u=Dg(x-at)$; thus $u_t+a\cdot D_{x}u=0$ and $u(0,x)=g(x)$.

The solved right side $F(t,x,u,p)=-a\cdot p$ is polynomial in its jet variables and therefore analytic at the required initial jet. F3 identifies the displayed solution with the unique analytic germ. For the explicit data $d=1,a=2,g(x)=x^2$, this gives $u=(x-2t{)}^2$, $u_t=-4(x-2t)$ and $2u_x=4(x-2t)$, displaying the cancellation directly.

For an analytic scalar solution put $v=u_t,w=u_x$. Then $u_t=v$, $v_t=u_{tt}=u_{xx}=w_x$ and $w_t=u_{xt}=u_{tx}=v_x$, with traces $(g,h,g^{\mathrm{\prime }})$. These are precisely the m=2 equations of F1.

Conversely F1 supplies an analytic system solution. Its error $e=w-u_x$ satisfies $e_t=v_x-{\mathrm{\partial }}_{x}v=0$ and $e(0,x)=g^{\mathrm{\prime }}(x)-g^{\mathrm{\prime }}(x)=0$, so e is identically zero. Thus $u_{tt}=v_t=w_x=u_{xx}$ and $u_t(0,x)=h$. For $g=x^2,h=1$, the explicit solution is $u=x^2+t^2+t$, $v=2t+1,w=2x$: both $v_t$ and $w_x$ equal 2, and both $w_t$ and $v_x$ equal 0.

For x nonzero define polynomials recursively by $P_0(z)=1$ and $P_{m+1}(z)=-z^2{P}_m^{\mathrm{\prime }}(z)+2z^3{P}_m(z)$. F1–F3 show by successive differentiation that $g^{(m)}(x)=P_m(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ off zero. Indeed $d(1\mathrm{/}x)\mathrm{/}dx=-1\mathrm{/}{x}^2$ and $d(-1\mathrm{/}{x}^2)\mathrm{/}dx=2\mathrm{/}{x}^3$, giving exactly that recurrence.

With $y=1\mathrm{/}{x}^2\to \mathrm{\infty }$, the absolute value of any polynomial in 1/x, and of that polynomial divided by x, is bounded by a constant times an integer power of y for y at least one. F4 makes both products with $\exp (-y)$ tend to zero. Starting with the continuity of g at zero, induction now gives $g^{(m)}(0)=0$: if the mth derivative equals the expression of step 1.1 off zero and is zero at zero, its difference quotient tends to zero, so the next derivative at zero exists and is zero. Its continuity follows from the same bound. Thus g is smooth and all its Taylor coefficients at zero vanish.

F5 gives g(x)>0 for x nonzero, arbitrarily close to zero. F6 therefore prevents g from being analytic at zero: its zero Taylor series could not equal those positive values. The function u(t,x)=g(x) is smooth, has u_t=0 and the required trace. If an analytic solution existed, substituting t=0 in its convergent two-variable series would give a convergent series for g with its derivative coefficients, contradicting the preceding conclusion. Hence no analytic germ has those data.

The principal symbol is $p({\xi }_x,{\xi }_t)={\xi }_x$, and the nonzero conormal to t=0 is dt=(0,1). Thus p(dt)=0, so the surface is characteristic by F1. For every real c, $u_c(x,t)=ct$ is analytic, satisfies $(u_c{)}_x=0$, and has zero initial trace. Distinct c give distinct germs and distinct normal derivatives.

If a differentiable u satisfied $u_x=0$ near zero and $u(x,0)=x$, differentiating the trace along x would give $u_x(x,0)=1$, contradicting the equation value zero. Therefore these analytic trace data are incompatible.

For $\mathrm{\mid }x\mathrm{\mid }<1$, the geometric identity gives $1\mathrm{/}(1+x^2)={\sum }_{k\ge 0}(-1{)}^k{x}^{2k}$, hence the data are analytic with $g^{(2k)}(0)=(-1{)}^k(2k)!$. If u were analytic, repeated differentiation of its equation and F1 would give ${\mathrm{\partial }}_t^{k}u={\mathrm{\partial }}_x^{2k}u$, beginning at k=0 and using ${\mathrm{\partial }}_t{\mathrm{\partial }}_x^{2k}u={\mathrm{\partial }}_x^{2k}{u}_t={\mathrm{\partial }}_x^{2k+2}u$ at each step. Evaluation on the initial trace gives the asserted derivatives.

The time coefficient is $a_k=(-1{)}^k(2k)!\mathrm{/}k!$. For every fixed t nonzero, $\mathrm{\mid }{a}_{k+1}{t}^{k+1}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{a}_k{t}^k\mathrm{\mid }=2(2k+1)\mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }$. Thus these terms eventually increase by a factor at least two and do not tend to zero. The time series diverges at every t nonzero and cannot represent an analytic germ. The PDE has total order two, with no u_tt term; its principal symbol is ${\xi }_x^2$ and vanishes at dt. It therefore lies outside the noncharacteristic CK hypothesis.

Use $\sinh s=(e^s-e^{-s})\mathrm{/}2$. F1 and F2 give $(u_k{)}_{tt}=k{e}^{-\sqrt{k}}\sinh (kt)\cos (kx)$ and $(u_k{)}_{xx}=-k{e}^{-\sqrt{k}}\sinh (kt)\cos (kx)$, so their sum is zero. At t=0, $\sinh 0=0$ and its derivative is one, giving $u_k(x,0)=0$ and $(u_k{)}_t(x,0)=e^{-\sqrt{k}}\cos (kx)$. The exponential and trigonometric series make each u_k analytic.

For every nonnegative integer j, each x-derivative of order j of the normal data has modulus at most $k^j{e}^{-\sqrt{k}}$ by F1 and F3. With $z=\sqrt{k}$ this is $z^{2j}{e}^{-z}\to 0$ by F4. The bound is uniform on the entire real line, hence on every compact interval. All value-data derivatives are already zero.

Fix t>0. Eventually $\sinh (kt)\ge e^{kt}\mathrm{/}4$ and $\sqrt{k}\le kt\mathrm{/}2$. Hence $u_k(0,t)\ge e^{kt\mathrm{/}2}\mathrm{/}(4k)\to \mathrm{\infty }$ by F4. The zero data give the zero solution, whereas these data converge to zero in every displayed seminorm and their solution values diverge. This disproves the asserted continuous dependence even though the Laplace principal coefficient on u_tt is one and each function is analytic.