Partial Differential Equations with MATLAB

14 An Introduction to Partial Differential Equations with MATLAB©R

We will prove, in Exercise 8, thatLis linear if and only if

L[c 1 u 1 +c 2 u 2 ]=c 1 L[u 1 ]+c 2 L[u 2 ], (1.10)

for all constantsc 1 andc 2 , and all functions

u 1 andu 2 in the domain ofL.

We note that any discussion of linearity of PDEs is based upon the theorems
from calculus that tell us that first partial derivatives are linear (that is, that
∂
∂x(cu)=cuxand

∂
∂x(u^1 +u^2 )=u^1 x+u^2 x), from which it also follows that
all higher-order partial derivatives are linear (see Exercise 9).

In Examples 3–5, determine if the given PDE is linear or nonlinear.

Example 3ux+5u=x^2 y. The operator isL[u]=ux+5uand we have

L[c 1 u 1 +c 2 u 2 ]=(c 1 u 1 +c 2 u 2 )x+5(c 1 u 1 +c 2 u 2 )

=c 1 u (^1) x+c 2 u (^2) x+5c 1 u 1 +5c 2 u 2
=c 1 (u (^1) x+5u 1 )+c 2 (u (^2) x+5u 2 )
=c 1 L[u 1 ]+c 2 L[u 2 ]
andLis linear, so the PDE is linear, as well.
Example 4The eikonal equation,u^2 x+u^2 y=1. WehaveL[u]=u^2 x+u^2 y.
Consider, then,L[cu]:
L[cu]=(cu)^2 x+(cuy)^2
=c^2 u^2 x+c^2 u^2 y.
The question is, do we haveL[cu]=cL[u]forallconstantscandallfunctions
u(in the domain ofL)? That is, is
c^2 u^2 x+c^2 u^2 y=cu^2 x+cu^2 y?
Certainly, the answer isno. To be more precise, the equationmaybe true for
certain constantscand/or functionsu, but we need only findonecounterex-
ample, that is,onecase involving a particularcand a particularufor which
the equality doesn’t hold (e.g., tryc=2andu=x). Therefore, the PDE is
nonlinear.
Example 5y^2 uxx+uyy= 1. Here,L[u]=y^2 uxx+uyyand
L[c 1 u 1 +c 2 u 2 ]=y^2 (c 1 u 1 +c 2 u 2 )xx+(c 1 u 1 +c 2 u 2 )yy
=c 1 y^2 u (^1) xx+c 2 y^2 u (^2) xx+c 1 u (^1) yy+c 2 u (^2) yy
=c 1 (y^2 u 1 xx+u 1 yy)+c 2 (y^2 u 2 xx+u 2 yy)
=c 1 L[u 1 ]+c 2 L[u 2 ],
so this PDE is linear.