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PDE Evans, 2nd edition, pages 105-106

Lemma 2 (Local invertibility). Assume we have the noncharacteristic condition $F_{p_n}(p^0,z^00,x^0) \not=0$. Then there exist an open interval $I \subseteq \mathbb{R}$ containing $0$, a neighborhood $W$ of $x^0$ in $\Gamma \subset \mathbb{R}^{n-1}$, and a neighborhood $V$ of $x^0$ in $\mathbb{R}^n$, such that for each $x \in V$ there exist unique $s \in I, y \in W$ such that $$x=\mathbf{x}(y,s).$$ The mappings $x \mapsto s, y$ are $C^2$.

Proof. We have $\mathbf{x}(x^0,0)=x^0$. Consequently, the Inverse Function Theorem gives the result, provided $\det D\mathbf{x}(x^0,0) \not=0$. Now \begin{align}\mathbf{x}(y,0)=(y,0) \qquad(y \in \Gamma)\end{align} and so if $i-1,\ldots,n-1,$ $$x_{y_i}^j(x^0,0) = \begin{cases}\delta_{ij} \qquad (j = 1,\ldots,n-1) \\ \, \, \,0 \qquad(j=n) \end{cases}$$

Furthermore equation $(31)(c)$, which on page 104 says $\mathbf{\dot{x}}(s)=D_p F (\mathbf{p}(s),z(s),\mathbf{x}(s))$, implies $$x_s^j(x^0,0)=F_{p_j}(p^0,z^0,x^0).$$ Thus, $$D\textbf{x}(x^0,0)=\begin{pmatrix} 1 & & 0 &F_{p_1}(p^0,z^0,x^0) \\ & \ddots & & \vdots \\ 0 & & 1 & \vdots \\ 0 & \cdots & 0 & F_{p_n}(p^0,z^0,x^0) \end{pmatrix}_{n \times n}$$

How is the matrix derived as an expression for $D\textbf{x}(x^0,0)$?

I only understood the last column of the matrix, which results from writing out all the terms of $F_{p_j}(p^0,z^0,x^0)$ for all $j = 1,\ldots,n$, and placing them collectively in a vertical array.

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  • Do you know the definition of $D\mathbf u$ for $\mathbf u \colon U \to \mathbb R^m$, $U \subset \mathbb R^n$? The definition is in the appendices of Evans, or https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant. – epimorphic Jun 20 '14 at 01:27
  • Yes. Is this how the matrix of $D\mathbf{x}(x^0,0)$ was derived? – Cookie Jun 20 '14 at 02:05
  • I only have the first edition handy, but shouldn't $x_{u_i}^j$ be $x_{y_i}^j$ instead, since we're differentiating by a component of $y$? – epimorphic Jun 21 '14 at 03:26
  • Yes you are right, I made a typo. :( – Cookie Jun 21 '14 at 04:11

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It looks like Evans is just applying the definition of the Jacobian. Each column of $D\mathbf x(y,s)$ is the partial derivative of $\mathbf x$ with respect to a component of one of the input variables $y$ and $s$. Explicitly, $$ D\mathbf x(x^0,0) = \begin{pmatrix} x_{y_1}^1(x^0,0) & \dotsm & x_{y_{n-1}}^1(x^0,0) & x_s^1(x^0,0) \\ \vdots & \ddots & \vdots & \vdots \\ x_{y_1}^{n-1}(x^0,0) & \dotsm & x_{y_{n-1}}^{n-1}(x^0,0) & x_s^{n-1}(x^0,0) \\ x_{y_1}^n(x^0,0) & \dotsm & x_{y_{n-1}}^n(x^0,0) & x_s^n(x^0,0) \end{pmatrix}. $$ Each entry can then be replaced with the equivalent expression given in the snippet you quoted (for example $x_{y_i}^j(x^0,0) = \delta_{ij}$), giving us the desired expression for $Dx(x^0,0)$.

Cookie
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    That matrix helps a lot. When you made your initial comment, the Jacobian concept seemed to fit really well into this IMO, but the final matrix written in Evans' book just threw me off. Also, I wish Evans explicitly stated in the textbook that he was applying the Jacobian as you pointed out. (Evans doesn't like to write a lot of details. :P But reading super-slick proofs (e.g. in baby Rudin) stinks if you're me who's self-studying in preparation for taking graduate PDE in the fall semester of this year or next year.) – Cookie Jun 21 '14 at 04:24
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    @glace Well, filling in details is a necessary skill for doing mathematics, and having to do it does force one to be more active while reading so it's not inherently a bad thing. I know how frustrating it sometimes can be though. Your initiative is admirable, by the way. :) – epimorphic Jun 22 '14 at 20:45