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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 ψu0, 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 wC1, 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 ψz0.

(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


This page titled 2.2.1: A Linearization Method is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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