
# 3.1.1: Normal Form in Two Variables


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}), \label{quadratic}\tag{3.1.1.3} 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 and Attributions

• Integrated by Justin Marshall.