12.3 Cauchy-Kovalevskaya Theorem

Definition 12.3.1 (Setting and Non-Characteristics).label Suppose

such that

then when $\exists! C^{\om}$ solution $u$ in a neighborhood $\OM$? By constant rank theorem, we can change coordinates so that locally, $S$ is given by $t=0$, $(\UBC{x_1,...,x_{n-1}}{x},t)$ coordinates. The Cauchy problem then becomes

We remark that Cauchy data determine all derivatives $\parens{\pd_x^\al\pd_t^j}u(x,0)$ as long as $j\leq k-1$. So the only derivative of $u$ which are unknown on $S$ are $\parens{\pd_t^ku}(x,0)$.

$S$ (given by $t=0$) is non-characteristic for

if one can solve for $\pd_{t}^{k}u$.

Example 12.3.2 (What can go wrong?).label

1. (1)

   Lack of Uniqueness

   \begin{align*}\begin{cases}\pd_{t}\pd_{x} u=0&u(x,0)=\p_{0}(x)\\ (x,t)\in\R^{2}&\pd_{t}u(x,0)=\p_{1}(x)\end{cases}\end{align*}

   
   

   $t=0$ in characteristic. Observe

   \begin{align*}(\pd_{t}\pd_{x}u)(x,0)=0\implies \pd_{x}\p_{1}(x)=0\implies \p_{1}\text{ const}\end{align*}

   
   

   So the Cauchy problem has no solutions unless $\p_{1}$ is constant. Suppose that $\p_{1}$ is constant then

   \begin{align*}u(x,t)=\p_{0}(x)+f(t)\end{align*}

   
   

   where $f$ is any function such that $f'(0)=\p_{1}$ then we have no uniqueness of solution.

2. (2)

   Dependence of Cauchy Data $\pd_{x}^{2}u-\pd_{t}u=0$, $t=0$ in characteristic. Suppose that $u$ is a solution with $u(x,0)=\p_{0}(x)$. Then $(\pd_{t}u)(x,0)=\pd_{xx}u(x,0)=\p_{0}''(x)$. Then $\p_{1}\define (\pd_{t}u)(x,0)$ is determined by $\p_{0}$ i.e. choosing $\p_{0}$ forces the choices of $\p_{1}$. Looking at characteristic tell us this also since

   \begin{align*}\chi=\xi_{x}^{2}=0\iff\xi_{x}=0\end{align*}

   
   

   so $(\xi_{x},\xi_{t})$ characteristic normal direction is given by $(0,const)$ in particular $t=const$ are characteristic curves.

Proposition 12.3.3.label If $G,\p_{0},...,\p_{k-1}$ are $C^{\om}$, then the PDE has at most one $C^{\om}$ solution.

Lemma 12.3.4.label The Cauchy Problem in Proposition 12.3.3 is equivalent to a quasi-linear system of $1^{st}$-order PDEs

where $Y=\begin{pmatrix}Y_{1}\\ \vdots\\ Y_{N}\end{pmatrix} =(y_{\al j})_{0\leq \abs{\al}+j\leq k,0\leq j\leq k-1},A_{i}=(a^{i}_{\el m})_{1\leq i\leq n-1;m,l=1,...,N},B=\begin{pmatrix}b_{1}\\ \vdots\\ b_{N}\end{pmatrix}$ and $\Phi$ is determined by $\p_{j}$.

Example 12.3.5 (Proof of Lemma 12.3.4 in Example).label Consider

Here $Y=(y_{00},y_{10},y_{01},y_{20},y_{11},y_{02})$ with notation $y_{xt}$. First we differentiate to get equation as in Lemma 12.3.4

Along with initial conditions

So we see that solution to Proposition 12.3.3 give solution to Lemma 12.3.4.

Now we show that solution to Lemma 12.3.4 give solution to Proposition 12.3.3.

By first and third equation $c_{20}(x)=0$ and similarly one can show $c_{11}(x)=0$. It remains to show $y_{10}=\pd_{x}y_{00},y_{01}=\pd_{t}y_{00}$

Compare with $y_{02}(x,0)$ give $c_{02}(x)=0$. Finally

OTOH, imposing again initial conditions

Similarly for $y_{01}$.

Lemma 12.3.6.label The Cauchy problem in Lemma 12.3.4 is equivalent to a Cauchy problem of the same form with $A_{j}$ and $B$ independent of $t$ and $\Phi=0$.

Lemma 12.3.7 (Facts about $C^{\om}$ functions).label

1. (1)

   $f$ is analytic near $x_{0}$ then there exists a cube $\OM=\curl{x:\abs{x_j-x_j^0}<r,1\leq j\leq n}$ such that $\sums{}{\al}\frac{1}{\al!}\parens{\pd^\al f}(x_{0})(x-x_{0})^{\al}$ converges to $f(x)$ for all $x\in\OM$ and convergence is uniform and absolute on compact subsets of $\OM$.

2. (2)

   Let $f(x)=\sums{}{}a_{\al}(x-x_{0})^{\al}$ be convergent enar $x_{0}$ and $x$ be an analyhtic function of $\xi$

   \begin{align*}x=\sums{}{\be}b_{\be}(\xi-\xi_{0})^{\be}\end{align*}

   
   

   with $x(\xi_{0})=x_{0}$. Then

   \begin{align*}F(\xi)=f(x(\xi))\end{align*}

   
   

   is analytic at $\xi_{0}$ and power series representation of $F$ is obtained by substitution

   \begin{align*}F(\xi)=\sums{}{\gam}c_{\gam}(\xi-\xi_{0})^{\gam}\end{align*}

   
   

   where

   \begin{align*}c_{\gam}=P_{\gam}(a_{\al},b_{\be})\end{align*}

   
   

   are universal polynomials with non-negative coefficients. (This is an algebraic fact about rings of power series)

3. (3)

   Given $M,r>0$ the function

   \begin{align*}f(x)=\frac{Mr}{r-\parens{\sums{n}{i=1}x_i}}\end{align*}

   
   

   is analytic in the rectangle $\curl{x:\MAX{i}\abs{x_j}<r/n}$ and

   \begin{align*}f(x)=M \sums{\infty}{k=0}\frac{\parens{\sums{n}{i=1}x_i}^{k}}{r^{k}}=M \sums{}{\abs{\al}\geq0}\frac{\abs{\al}!}{\al!r^{\abs{\al}}}x^{\al}\end{align*}

   
   
4. (4)

   We say that $\sums{}{\al}a_{\al}(x-x_{0})^{\al},a_{\al}\geq 0$ majorizes $\sums{}{\al }b_{\al}(x-x_{0}^{\al})$ if $a_{\al}\geq\abs{b_\al}$. In which case $\sums{}{\al }b_{\al}(x-x_{0}^{\al})$ converges absolutely whenever $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$ does.

5. (5)

   Suppose that $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$ in a rectangle $\curl{x:\MAX{i}\abs{x_i}<R }$ then $\exists$ geoemetric series as in (3) majorizing $\sums{}{\al }a_{\al}(x-x_{0}^{\al})$. Indeed let $x=(r,...,r),r<R$, $x\in$ rectangle. So $\exists M>0$ such that $\abs{a_\al r^{\abs{\al}}}\leq M$, $\fall \al$. So $\abs{a_\al}\leq\frac{M}{r^{\abs{\al}}}\leq \frac{M\abs{\al}!}{r\al!}$

Theorem 12.3.8 (Cauchy-Kovalevskaya).label If $G,\p_{0},...,\p_{k-1}$ are $C^{\om}$ then the PDE has a unique $C^{\om}$ solution.

Remark 12.3.9.label It may happen that the PDE have more than one solution in the sense that there exists non-analytic (say $C^{\infty}$) solution in addition to the $C^{\om}$ solution.

If there are points where $\pder{F}{\pd{\pd_t^k u}}=0$ there is a version of Cauchy-Kovalevskaya called the Ramified C-K theorem proved by J-Leray (1970’s) Garding wrote something on this.

Theorem 12.3.10 (Holmyer).label If the PDE is linear then the solution is unique.

Remark 12.3.11 (J. Hadamard).label Does the unique $C^{\om}$ solution of a CK equation depend “continuously” on the Cauchy data in any reasonable sense? No. Consider

$x_{2}=0$ we have $S$ is non characteristic

There exists unique $C^{\om}$ solution

As $k\to\infty$, Cauchy data and all derivatives $\to 0$. For $x_{2}>0$, $u(x_{1},x_{2})\to\infty$. OTOH $\pd_{1}^{2}u+\pd_{2}^{2}u=0$, $u(x_{1},0)=0,\pd_{2}u(x_{1},0)$ has unique $C^{\om}$ solution $u(x_{1},x_{2})=0$ for all $(x_{1},x_{2})\in\R^{2}$.

Example 12.3.12.label Let $S\suf\cali{B}(\cali{H})$ and consider $S'\define \curl{x\in\cali{B}(\cali{H}):xy=yx,\fall y\in S}$ which is WOT-closed, so if $S=S^{*}$ then $S'$ is a von Neumann algebra. Observe

1. (1)

   $S\suf S''$

2. (2)

   If $S_{1}\suf S_{2}$ then $S_{2}'\suf S_{1}'$

3. (3)

   $S'=S'''$

That is to say the bigger the underlying set $S$ is, the harder it is to “commute” that is belongs to $S'$.

Let $(X,\mu)$ be a probability space, $\cali{H}\define L^{2}(X,\mu)$, and $A=L^{\infty}(X,\mu)$. Consider the map $M:A\to \cali{B}(\cali{H})$ defined by $f\mto M_{f}=(\xi\mto f\cd \xi)$ then

1. (1)

   $\norm{f}{\infty}=\norm{M_f}{}$, in particular $A\suf\cali{B}(\cali{H})$

2. (2)

   $A=A'$ so $A$ is a von Neumann algebra and is a maximal abelian sub-algebra

3. (3)

   $M_{f}^{*}=M_{\cl{f}}$

Theorem 12.3.13 (Von Neumann Bicommutant).label Let $A\suf\cali{B}(\cali{H})$ be a $*$-algebra then TFAE:

1. (1)

   $A=A''$

2. (2)

   $A$ is weakly closed

3. (3)

   $A$ is strongly closed

Corollary 12.3.14.label For $S\suf\cali{B}(\cali{H})$, $S''$ is the von Neumann algebra generated by $S$ denoted $W^{*}(S)$.

Definition 12.3.15 (Group Von Neumann Algebra).label Let $G$ be a group and

then $\lam_{g}(\de_{t})=\de_{gt}$. The group von Neumann algebra of $G$ is $L(G)=\cl{Span(\lam(G))}$ under SOT.

Remark 12.3.16.label For a group $G$, let $\lam$ be left regular represention on $\el^{2}(G)$, $L(G)$ the SOT closure of the the linear space of $\lam$ in $B(\el^{2}(G))$ then $L(G)$ commutes with $R(G)$ and $L(G)=R(G)'$.

Definition 12.3.17 (States and Traces).label Let $\p$ be a linear functional $\p$ on a von Neumann algebra $M$ then $\p$ is

1. (1)

   positive if $\p(x^{*}x)\geq 0$ for all $x\in M$

2. (2)

   faithful if $\p(x^{*}x)=0\implies x=0$

3. (3)

   state if $\p(1)=1$

A positive linear functional $\p$ is a trace if $\p(xy)=\p(yx)$ for all $x,y\in M$. If $\p$ is also a trace then we call it a tracial state

For a vNA $M$ acting on $\cali{H}$, $\xi\in\cali{H}$ is cyclic for $M$ if $M\xi$ is dense in $\cali{H}$. $\xi\in\cali{H}$ is separating for $M$ if $x\xi\neq 0$ whenever $x\neq 0$.

We can observe that in $L(G)$, $\de_{e}$ is cyclic and separating^[3].

Definition 12.3.18 (Tracial vNA).label Let $\tau(x)\define \inner{\de_e}{x\de_e}$ for $x\in L(G)$ then $\tau(1)=\inner{\de_e}{\de_e}=1$ and $\tau(\lam(s_{1})\lam(s_{2}))=\inner{\de_e}{\de_{s_1s_2}}\neq0\iff s_{1}s_{2}=e$. $\tau$ is continuous w.r.t. the WOT.

A tracial von Neumann algebra is a vNA $M$ equipped with a faithful tracial state that is WOT continuous restricted to the unit ball.

Definition 12.3.19 (Factors).label We have finite dimensional tracial vNAs, $M_{n}(\com)$ equipped with $\frac{1}{n}Tr=:\tau$. A $II_{1}$ factor is an infinite dimensional tracial vNA whose centre is reduced to the scalar operator.

A factor is of type $I$ if it is isomoprhic to $\cali{B}(\cali{H})$

Let $(X,\mu)$ be a probability space, $E$ countable Borel equivalence relation on $X$ then $E$ has countable classes and $E\suf X^{2}$ is Borel. We define the measure $\nu$ on $E$ as

then $L^{2}(E,\nu$) is a Hilbert space and $\nu$ is $\s$-finite measure as a consequence of Luzin. We define $\cali{M}_{b}(E)$ to be bounded finitely supported Borel functions $f:E\to\com$ where $\exists C>0,\fall x,y$ we have $\abs{z\in X:F(z,y)\neq 0},\abs{z\in X:F(x,z)\neq 0}\leq C$. Denote $B_{b}(X)$ the bounded Borel functions on $X$ then $f\mto f(x)\indi{\la}$ is an injection $B_{b}(X)\inj \cali{M}_{b}(E)$. $E$ is probabiliity measure preserving (pmp) if for any Borel automorphism $f:X\to X$ such that $xEf(x),f_{*}\mu=\mu$.