2.2.1: A Linearization Method

We can transform the inhomogeneous Equation (2.2.1) into a homogeneous linear equation for an unknown function of three variables by the following trick.

We are looking for a function \(\psi(x,y,u)\) such that the solution \(u=u(x,y)\) of Equation (2.2.1) is defined implicitly by \(\psi(x,y,u)=const.\) Assume there is such a function \(\psi\) and let \(u\) be a solution of (2.2.1), then

$$
\psi_x+\psi_uu_x=0,\ \ \psi_y+\psi_uu_y=0.
$$
Assume \(\psi_u\not=0\), then
$$
u_x=-\frac{\psi_x}{\psi_u},\ \ u_y=-\frac{\psi_y}{\psi_u}.
$$
From (2.2.1) we obtain
\begin{equation}
\label{homthree}\tag{2.2.1.1}
a_1(x,y,z)\psi_x+a_2(x,y,z)\psi_y+a_3(x,y,z)\psi_z=0,
\end{equation}
where \(z:=u\).

We consider the associated system of characteristic equations
\begin{eqnarray*}
x'(t)&=&a_1(x,y,z)\\
y'(t)&=&a_2(x,y,z)\\
z'(t)&=&a_3(x,y,z).
\end{eqnarray*}

One arrives at this system by the same arguments as in the two-dimensional case above.

Proposition 2.2. (i) Assume \(w\in C^1\), \(w=w(x,y,z)\), is an integral, i. e., it is constant along each fixed solution of (\ref{homthree}), then \(\psi=w(x,y,z)\) is a solution of (\ref{homthree}).

(ii) The function \(z=u(x,y)\), implicitly defined through \(\psi(x,u,z)=const.\), is a solution of (2.2.1), provided that \(\psi_z\not=0\).

(iii) Let \(z=u(x,y)\) be a solution of (2.2.1) and let \((x(t),y(t))\) be a solution
of
$$
x'(t)=a_1(x,y,u(x,y)),\ \ y'(t)=a_2(x,y,u(x,y)),
$$
then \(z(t):=u(x(t),y(t))\) satisfies the third of the above characteristic equations.

Proof. Exercise.

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.