2.2.1: A Linearization Method
( \newcommand{\kernel}{\mathrm{null}\,}\)
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 ψ(x,y,u) such that the solution u=u(x,y) of Equation (2.2.1) is defined implicitly by ψ(x,y,u)=const. Assume there is such a function ψ and let u be a solution of (2.2.1), then
ψx+ψuux=0, ψy+ψuuy=0.
Assume ψu≠0, then
ux=−ψxψu, uy=−ψyψu.
From (2.2.1) we obtain
a1(x,y,z)ψx+a2(x,y,z)ψy+a3(x,y,z)ψz=0,
where z:=u.
We consider the associated system of characteristic equations
x′(t)=a1(x,y,z)y′(t)=a2(x,y,z)z′(t)=a3(x,y,z).
One arrives at this system by the same arguments as in the two-dimensional case above.
Proposition 2.2. (i) Assume w∈C1, w=w(x,y,z), is an integral, i. e., it is constant along each fixed solution of (2.2.1.1), then ψ=w(x,y,z) is a solution of (2.2.1.1).
(ii) The function z=u(x,y), implicitly defined through ψ(x,u,z)=const., is a solution of (2.2.1), provided that ψz≠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)=a1(x,y,u(x,y)), y′(t)=a2(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
Integrated by Justin Marshall.