3.1.1: Normal Form in Two Variables

Consider the differential equation

\begin{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
\end{equation}

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

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

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}),

\begin{equation}
\label{quadratic}\tag{3.1.1.3}
ay'^2-2by'+c=0.
\end{equation}

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

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

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

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.