I'm reading Zachmanoglou's book, and I'm having trouble interpreting how they conclude that $F(t,x,u,p)$ is analytic in their example:
Example. Let
$$ \begin{align} u_t &= uu_x \\ u(0,x) &= 1 + x^2 \tag{$\star$}. \end{align} $$
Use the Cauchy-Kovalevsky Theorem to find all terms $\leq 3$ in the Taylor series about the origin.
So, for a single variable function $f$, we say it is analytic about $x_0$ if
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n. \tag{1}$$
Clearly, $f(x) = 1 + x^2$, is analytic about the origin.
They then state that, $F(t,x,u,p) = up$, is analytic about the the neighborhood $(0,0,1,0)$, and so we can apply C-K. But what is the Taylor series analog of $(1)$ to actually show this? I don't yet see how this is trivial?
Edit 1: I will include the C-K Theorem here for reference.
Theorem. Suppose that $\phi$ is analytic in a neighborhood of the origin of $\mathbb{R}^n$ and suppose that the function $F$ is analytic in a neighborhood of the point $$(0, 0, \dots, \phi(0, \dots, 0), \phi_{x_1}(0, \dots, 0), \dots, \phi_{x_n}(0, \dots, 0))$$ of $\mathbb{R}^{2n + 2}$. Then the Cauchy problem $(\star)$ (it's really $\mathbb{R}^4$ for the example above) has a solution $u(t,x_1, \dots, x_n)$ which is defined and analytic in a neighborhood of the origin of $\mathbb{R}^{n+1}$ and this solution is unique in the class of analytic functions.