Example 12.3.5 (Proof of Lemma 12.3.4 in Example).label Consider
\begin{align*}\begin{cases}u_{tt}=f(u_{xx},u_{xt})\\ u(x,0)=\p_{0}(x)&\parens{\pd_tu}(x,0)=\p_{1}(x)\end{cases}\end{align*}
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
\begin{align*}\pder{y_{00}}{t}&=y_{01}\\ \pder{y_{10}}{t}&=y_{11}\\ \pder{y_{01}}{t}&=y_{02}\\ \pder{y_{20}}{t}&=\pder{y_{11}}{x}\\ \pder{y_{11}}{t}&=\pder{y_{02}}{x}\\ \pder{y_{02}}{t}&=f_{1}(y_{20},y_{11})\pder{y_{20}}{t}+f_{2}(y_{20},y_{11})\pder{y_{11}}{t}=f_{1}(y_{20},y_{11})\pder{y_{11}}{x}+f_{2}(y_{20},y_{11})\pder{y_{02}}{x}\end{align*}
Along with initial conditions
\begin{align*}y_{00}(x,0)&=\p_{0}(x)\\ y_{10}(x,0)&=(\pd_{x}\p_{0})(x)\\ y_{01}(x,0)&=\p_{1}(x)\\ y_{20}(x,0)&=(\pd_{x}^{2}\p_{0})(x)\\ y_{11}(x,0)&=(\pd_{x}\p_{1})(x)\\ y_{02}(x,0)&=f((\pd_{2}^{2}\p_{0})(x),(\pd_{x}\p_{1})(x))\end{align*}
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.
\begin{align*}y_{20}(x,t)=\pd_{x}y_{10}+c_{20}(x)\\ y_{11}(x,t)=\pd_{x}y_{01}+c_{11}(x)\\ y_{20}(x,0)=\pd_{x}^{2}\p_{0}(x)=\pd_{x}(\pd_{x}\p_{0}(x))=(\pd_{x}y_{10})(x,0)\end{align*}
By first and third equation $c_{20}(x)=0$ and similarly one can show $c_{11}(x)=0$. It remains to show $y_{10}=\pd_{x}y_{00},y_{01}=\pd_{t}y_{00}$
\begin{align*}\pd_{t}y_{02}&=f_{1}\pd_{x}y_{11}+f_{2}\pd_{x}y_{02}\\&=f_{1}\pd_{t}y_{20}+f_{2}\pd_{t}y_{11}=\pd_{t}f\\ y_{02}=f(y_{20},_{11})+c_{02}(x)\end{align*}
Compare with $y_{02}(x,0)$ give $c_{02}(x)=0$. Finally
\begin{align*}\pder{y_{10}}{t}=y_{11}=\pd_{x}y_{01}=\pd_{x}\pd_{t}y_{00}=\pd_{t}\pd_{x}y_{00}\\ y_{10}=\pd_{x}y_{00}+c_{10}(x)\end{align*}
OTOH, imposing again initial conditions
\begin{align*}y_{10}(x,0)=\pd_{x}\p_{0}(x)=\pd_{x}y_{00}(x,0)\implies c_{10}(x)=0\end{align*}
Similarly for $y_{01}$.
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