Skip to main content

# 2.2.1: A Linearization Method

[ "article:topic", "showtoc:no" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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

\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,

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

• Integrated by Justin Marshall.