Gr14 Full High Regularity Symmetric Hyperbolic Proof — Examples

(Norms of the shell family.) By [L1] and [L2]: $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^1}^2={\sum }_{2^N\le \mathrm{\mid }k\mathrm{\mid }<2^{N+1}}(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\le 2\cdot 2^{2N+2}\cdot \mathrm{\#}\text{shell}\le C{2}^{2N}{2}^{3N}=C{2}^{5N}$, so $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^1}\le C{2}^{5N\mathrm{/}2}$; and $u_N(0)=\mathrm{\#}\text{shell}\ge c{2}^{3N}$.

(Divergence of the ratio.) Hence $\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{C^0}\mathrm{/}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^1}\ge c^{\mathrm{\prime }}{2}^{3N-5N\mathrm{/}2}=c^{\mathrm{\prime }}{2}^{N\mathrm{/}2}\to \mathrm{\infty }$; if a constant $C$ satisfied the claimed bound, applying it to $u_N\mathrm{/}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^1}$ would give a uniform bound on this ratio, a contradiction.

(Consistency with the sharp threshold.) The failure occurs exactly at the exponent where the weight sum of [L3] diverges; by Periodic Sobolev embedding: H^s on the three-torus controls C^m when s minus m exceeds three halves the bound does hold for $s-m>3\mathrm{/}2$, so the counterexample is sharp and does not contradict the embedding theorem.

(The explicit function solves the system.) Define $u(t,x):=u_0(x^1-2t,x^2,x^3)$. Then ${\mathrm{\partial }}_{t}u(t,x)=-2({\mathrm{\partial }}_1{u}_0)(x^1-2t,x^2,x^3)$ and ${\mathrm{\partial }}_{1}u(t,x)=({\mathrm{\partial }}_1{u}_0)(x^1-2t,x^2,x^3)$, so ${\mathrm{\partial }}_{t}u+2{\mathrm{\partial }}_{1}u=0$ pointwise; $u(0,\cdot )=u_0$, and $u$ is classical because integer $s\ge 5$ and inherits the $H^s$-bound of $u_0$ at every time, with higher smoothness whenever $u_0$ is smooth.

(Energy identity.) $e(t)=\int u(t{)}^{2}d\mu =\int u_0^{2}d\mu =e(0)$ by translation invariance [L3]; equivalently, differentiating under the integral and using the equation and [L3], $e^{\mathrm{\prime }}(t)=2\int u{\mathrm{\partial }}_{t}ud\mu =-4\int u{\mathrm{\partial }}_{1}ud\mu =-2\int {\mathrm{\partial }}_1(u^2)d\mu =0$.

(Identification with the E0 solution.) By [L2] there is a unique solution through $u_0$; the explicit function of step 1.1 is one, so the E0 solution is $u(t,x)=u_0(x^1-2t,x^2,x^3)$ and its energy is conserved. For the actual Galerkin energy, set $u_N(t)=P_{N}u(t)$, equivalently $u_N(t,x)=(P_N{u}_0)(x^1-2t,x^2,x^3)$. By [L4] this smooth finite Fourier polynomial solves ${\mathrm{\partial }}_t{u}_N+2{\mathrm{\partial }}_1{u}_N=0$ with datum $P_N{u}_0$, hence its projected equation. Each Fourier coefficient obeys ${\mathrm{\partial }}_t{\widehat{u}}_N(k)=-2i{k}_1{\widehat{u}}_N(k)$, so this is exactly the Galerkin trajectory. Every $D^{\alpha }{u}_N$ is the translate of $D^{\alpha }{P}_N{u}_0$; translation invariance gives $e_N(t)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le s}\mathrm{\parallel }{D}^{\alpha }{P}_N{u}_0{\mathrm{\parallel }}_2^2=e_N(0)$ and $e_N^{\mathrm{\prime }}(t)=0$. Thus the actual energy inequality is verified with $C=0$, at every cutoff $N$, as well as the zero-order identity of step 2.1.

(Exact growing and decaying mode solutions.) For every integer $n\ge 1$ and $A\ge 0$, put $f_n(t,x)=A\cos (n{x}^1)\cosh (ant)$ and $v_n^1(t,x)=-(a\mathrm{/}{H}_0)A\sin (n{x}^1)\sinh (ant)$, with $v_n^2=v_n^3=0$. Differentiation gives ${\mathrm{\partial }}_t{f}_n+H_0{\mathrm{\partial }}_1{v}_n^1=0$ and $H_0{\mathrm{\partial }}_t{v}_n^1-a^2{\mathrm{\partial }}_1{f}_n=0$; the other momentum components are zero. Thus these are exact smooth solutions of the displayed linearized system with data $(A\cos (n{x}^1),0)$. The density mode satisfies $g_n^{\mathrm{\prime }\mathrm{\prime }}=a^2{n}^2{g}_n$, $g_n(0)=A$, $g_n^{\mathrm{\prime }}(0)=0$, so $g_n(t)=A\cosh (ant)\ge (A\mathrm{/}2)e^{ant}$ for $t\ge 0$. This lower bound uses the stated nonnegative amplitude and zero initial velocity. More generally, for any periodic $C^1$ solution, differentiating its Fourier coefficients under the integral and integrating the spatial derivatives by parts gives $g_k^{\mathrm{\prime }}=-H_{0}ik\cdot w_k$ and $H_0{w}_k^{\mathrm{\prime }}=a^{2}ik{g}_k$, hence $g_k^{\mathrm{\prime }\mathrm{\prime }}=a^2\mathrm{\mid }k{\mathrm{\mid }}^2{g}_k$. The initial density and velocity therefore determine each density coefficient uniquely by this scalar ODE.

(Failure of fixed-Sobolev continuity on smooth modes.) Fix $\tau >0$ and an integer $r\ge 0$. In step 1.1 take $A_n=(1+n^2{)}^{-r\mathrm{/}2}{e}^{-an\tau \mathrm{/}2}$. Under the derivative-sum norm of [L3], $\mathrm{\parallel }\cos (n{x}^1){\mathrm{\parallel }}_{H^r}^2=\frac{1}{2}{\sum }_{j=0}^r{n}^{2j}$, which lies between positive constants times $(1+n^2{)}^r$, uniformly for $n\ge 1$. Consequently the initial H^r norm tends to zero, whereas the density H^r norm at $\tau$ is at least a positive constant times $e^{an\tau \mathrm{/}2}$ and tends to infinity. Each datum and solution is a single trigonometric polynomial and hence smooth; indeed the data tend to zero in every fixed derivative norm because exponential decay dominates every power. Any other classical solution of the same datum has the same coefficient at $(n,0,0)$ by step 1.1. For any such density lying in $H^r$, integration by parts gives $n^r\mathrm{\mid }\widehat{f}(n,0,0)\mathrm{\mid }=\mathrm{\mid }\widehat{D_1^{r}f}(n,0,0)\mathrm{\mid }\le \mathrm{\parallel }{D}_1^{r}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^r}$. The coefficient inequality follows from $\mathrm{\mid }\widehat{h}(k)\mathrm{\mid }\le \int \mathrm{\mid }h\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_2$; the second bound follows by expanding $0\le \int (\mathrm{\mid }h\mathrm{\mid }-\int \mathrm{\mid }h\mathrm{\mid }{)}^2$ on the normalized probability space. Consequently every classical solution choice has density $H^r$ norm at least a positive constant times $e^{an\tau \mathrm{/}2}$. Thus no choice of a solution operator, even restricted to these smooth trigonometric data, can be continuous at zero from $H^r$ data to $H^r$ values at any fixed positive time.

(A genuinely smooth datum with no linearized Sobolev solution.) Set $f_0(x)={\sum }_{n\ge 1}{e}^{-\sqrt{n}}\cos (n{x}^1)$ and $v_0=0$. For every integer $q\ge 0$, ${\sum }_{n\ge 1}{n}^q{e}^{-\sqrt{n}}<\mathrm{\infty }$: group the terms with $j^2\le n<(j+1{)}^2$, whose contribution is at most $(2j+1)(j+1{)}^{2q}{e}^{-j}$, a summable polynomial times a geometric sequence. Thus each spatial derivative series converges uniformly; the fundamental theorem of calculus applied to partial sums and their uniformly convergent derivative series shows successively that $f_0$ is smooth with those derivatives. Suppose a periodic $C^1$ linearized solution existed on $[0,\tau ]$. At each Fourier frequency $k$, differentiation under the integral and periodic integration by parts give $g_k^{\mathrm{\prime }}=-H_{0}ik\cdot w_k$ and $H_0{w}_k^{\mathrm{\prime }}=a^{2}ik{g}_k$. Thus $g_k$ is $C^2$ with $g_k^{\mathrm{\prime }\mathrm{\prime }}=a^2\mathrm{\mid }k{\mathrm{\mid }}^2{g}_k$. The explicit initial coefficients and zero velocity, together with uniqueness of this elementary ODE, force its coefficient at $(n,0,0)$ to be $\frac{1}{2}{e}^{-\sqrt{n}}\cosh (ant)$. For every $t>0$ these coefficients tend to infinity, since $ant-\sqrt{n}\to \mathrm{\infty }$, whereas every Fourier coefficient of the continuous density at that fixed time has absolute value at most $\int \mathrm{\mid }f(t,x)\mathrm{\mid }d\mu \le {\sup }_x\mathrm{\mid }f(t,x)\mathrm{\mid }<\mathrm{\infty }$. This contradiction excludes a periodic classical solution on every positive time interval, without requiring an unproved inference from coefficient summability. This explicit datum is smooth; polynomially decaying coefficients of a fixed order would not have established that assertion.

(Conclusion within the proved scope.) Step 2.1 proves failure of continuity at zero in every fixed integer $H^r$ topology for the explicit smooth mode solutions at each fixed positive time. Step 2.2 supplies one smooth periodic datum admitting no periodic $C^1$ linearized solution on any positive time interval. These refute both assertions in the stated fixed-flat linearized claim. The nonlinear Einstein-fluid transfer with constrained data is explicitly left open; no nonlinear completeness or illposedness verdict follows from this proof.

(The matrix.) By [L1], [L2] and the chosen data, $A(\zeta )=\left(\begin{array}{cccc}1& \frac{1}{2\sqrt{3}}& 0& 0\\ \frac{1}{2\sqrt{3}}& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$.

(Eigenvalues.) The upper-left block $\left(\begin{array}{cc}1& \frac{1}{2\sqrt{3}}\\ \frac{1}{2\sqrt{3}}& 1\end{array}\right)$ has eigenvalues $1\pm \frac{1}{2\sqrt{3}}$; the remaining block is the identity with eigenvalue $1$ (multiplicity two). All three eigenvalues are positive: $1-\frac{1}{2\sqrt{3}}>0$ because $2\sqrt{3}>1$.

(Consistency with the general criterion.) The general criterion of [L1] gives positivity because ${\zeta }_0=1>\frac{1}{2}=\mathrm{\mid }{\zeta }_{\perp }\mathrm{\mid }$ and $c_s<1$; the explicit smallest eigenvalue $1-\frac{1}{2\sqrt{3}}\approx 0.711>0$ matches exactly the lower bound ${\zeta }_0-c_s\mathrm{\mid }{\zeta }_{\perp }\mathrm{\mid }=1-\frac{1}{2\sqrt{3}}$ from that criterion.