$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 3.1.1: Normal Form in Two Variables

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

Consider the differential equation

\label{eqnplane}\tag{3.1.1.1}
a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}+\mbox{terms of lower order}=0

in $$\Omega\subset\mathbb{R}^2$$. The associated characteristic differential equation is

\label{planechar}\tag{3.1.1.2}
a\chi_x^2+2b\chi_x\chi_y+c\chi_y^2=0.

We show that an appropriate coordinate transform will simplify equation (\ref{eqnplane}) sometimes in such a way that we can solve the transformed equation explicitly.

Let $$z=\phi(x,y)$$ be a solution of (\ref{planechar}). Consider the level sets $$\{(x,y):\ \phi(x,y)=const.\}$$ and assume $$\phi_y\not=0$$ at a point $$(x_0,y_0)$$ of the level set. Then there is a function $$y(x)$$ defined in a neighborhood of $$x_0$$ such that $$\phi(x,y(x))=const.$$ It follows

$$y'(x)=-\dfrac{\phi_x}{\phi_y},$$

which implies, see the characteristic equation (\ref{planechar}),

ay'^2-2by'+c=0.

Then, provided $$a\not=0$$, we can calculate $$\mu:=y'$$ from the (known) coefficients $$a$$, $$b$$ and $$c$$:

\label{mu}\tag{3.1.1.4}
\mu_{1,2}=\dfrac{1}{a}\left(b\pm\sqrt{b^2-ac}\right).

These solutions are real if and only of $$ac-b^2\le0$$.

Equation (\ref{eqnplane}) is hyperbolic if $$ac-b^2<0$$, parabolic if $$ac-b^2=0$$ and ellipticif $$ac-b^2>0$$. This follows from an easy discussion of the eigenvalues of the matrix

$$\left(\begin{array}{cc} a&b\\ b&c \end{array}\right),$$

see an exercise.

## Normal form of a hyperbolic equation

Let $$\phi$$ and $$\psi$$ are solutions of the characteristic equation (\ref{planechar}) such that

\begin{eqnarray*}
y_1'\equiv\mu_1&=&-\dfrac{\phi_x}{\phi_y}\\
y_2'\equiv\mu_2&=&-\dfrac{\psi_x}{\psi_y},
\end{eqnarray*}

where $$\mu_1$$ and $$\mu_2$$ are given by (\ref{mu}). Thus $$\phi$$ and $$\psi$$ are solutions of the linear homogeneous equations of first order

\begin{eqnarray}
\label{phi}\tag{3.1.1.5}
\phi_x+\mu_1(x,y)\phi_y&=&0\\
\label{psi}\tag{3.1.1.6}
\psi_x+\mu_2(x,y)\psi_y&=&0.
\end{eqnarray}

Assume $$\phi(x,y)$$, $$\psi(x,y)$$ are solutions such that $$\nabla\phi\not=0$$ and $$\nabla\psi\not=0$$, see an exercise for the existence of such solutions.

Consider two families of level sets defined by $$\phi(x,y)=\alpha$$ and $$\psi(x,y)=\beta$$, see Figure 3.1.1.1.

Figure 3.1.1.1: Level sets

These level sets are characteristic curves of the partial differential equations (\ref{phi}) and (\ref{psi}), respectively, see an exercise of the previous chapter.

Lemma. (i) Curves from different families can not touch each other

(ii) $$\phi_x\psi_y-\phi_y\psi_x\not=0$$.

Proof. (i):
$$y_2'-y_1'\equiv\mu_2-\mu_1=-\dfrac{2}{a}\sqrt{b^2-ac}\not=0.$$
(ii):
$$\mu_2-\mu_1=\dfrac{\phi_x}{\phi_y}-\dfrac{\psi_x}{\psi_y}.$$

$$\Box$$

Proposition 3.1. The mapping $$\xi=\phi(x,y)$$, $$\eta=\psi(x,y)$$ transforms equation (\ref{eqnplane}) into

\label{normhyp}\tag{3.1.1.7}
v_{\xi\eta}=\mbox{lower order terms},

where $$v(\xi,\eta)=u(x(\xi,\eta),y(\xi,\eta))$$.}

Proof. The proof follows from a straightforward calculation.
\begin{eqnarray*}
u_x&=&v_\xi\phi_x+v_\eta\psi_x\\
u_y&=&v_\xi\phi_y+v_\eta\psi_y\\
u_{xx}&=&v_{\xi\xi}\phi_x^2+2v_{\xi\eta}\phi_x\psi_x+v_{\eta\eta}\psi_x^2+\mbox{lower order terms}\\
u_{xy}&=&v_{\xi\xi}\phi_x\phi_y+v_{\xi\eta}(\phi_x\psi_y+\phi_y\psi_x)+v_{\eta\eta}\psi_x\psi_y+\mbox{lower order terms}\\
u_{yy}&=&v_{\xi\xi}\phi_y^2+2v_{\xi\eta}\phi_y\psi_y+v_{\eta\eta}\psi_y^2+\mbox{lower order terms}.
\end{eqnarray*}
Thus
$$au_{xx}+2bu_{xy}+cu_{yy}=\alpha v_{\xi\xi}+2\beta v_{\xi\eta}+\gamma v_{\eta\eta}+l.o.t.,$$
where
\begin{eqnarray*}
\alpha:&=&a\phi_x^2+2b\phi_x\phi_y+c\phi_y^2\\
\beta:&=&a\phi_x\psi_x+b(\phi_x\psi_y+\phi_y\psi_x)+c\phi_y\psi_y\\
\gamma:&=&a\psi_x^2+2b\psi_x\psi_y+c\psi_y^2.
\end{eqnarray*}
The coefficients $$\alpha$$ and $$\gamma$$ are zero since $$\phi$$ and $$\psi$$ are solutions of the characteristic equation. Since
$$\alpha\gamma-\beta^2=(ac-b^2)(\phi_x\psi_y-\phi_y\psi_x)^2,$$
it follows from the above lemma that the coefficient $$\beta$$ is different from zero.

$$\Box$$

Example 3.1.1.1:

Consider the differential equation
$$u_{xx}-u_{yy}=0.$$
The associated characteristic differential equation is
$$\chi_x^2-\chi_y^2=0.$$
Since $$\mu_1=-1$$ and $$\mu_2=1$$, the functions $$\phi$$ and $$\psi$$ satisfy differential equations
\begin{eqnarray*}
\phi_x+\phi_y&=&0\\
\psi_x-\psi_y&=&0.
\end{eqnarray*}
Solutions with $$\nabla\phi\not=0$$ and $$\nabla\psi\not=0$$ are
$$\phi=x-y,\ \ \psi=x+y.$$
Then the mapping
$$\xi=x-y,\ \ \eta=x+y$$
leads to the simple equation
$$v_{\xi\eta} (\xi,\eta)=0.$$
Assume $$v\in C^2$$ is a solution, then $$v_\xi=f_1(\xi)$$ for an arbitrary $$C^1$$ function $$f_1(\xi)$$. It follows
$$v(\xi,\eta)=\int_0^\xi\ f_1(\alpha)\ d\alpha+g(\eta),$$
where $$g$$ is an arbitrary $$C^2$$ function. Thus each $$C^2$$-solution of the differential equation can be written as

$$(\star)$$ $$v(\xi,\eta)=f(\xi)+g(\eta)$$,

where $$f,\ g\in C^2$$. On the other hand, for arbitrary $$C^2$$-functions $$f$$, $$g$$ the function $$(\star)$$ is a solution of the differential equation $$v_{\xi\eta}=0$$. Consequently every $$C^2$$-solution of the original equation $$u_{xx}-u_{yy}=0$$ is given by
$$u(x,y)=f(x-y)+g(x+y),$$
where $$f,\ g\in C^2$$.

## Contributors

• Integrated by Justin Marshall.