# Linear Operators

• Dec 26th 2010, 11:51 AM
dwsmith
Linear Operators
To be linear, $\displaystyle L(\alpha u+\alpha v)=\alpha (L(u)+ L(v)) \ \alpha\in\mathbb{R}$.

How is this equation (see below) linear?

$\displaystyle yu_{xxy}-e^xu_x+3=0$

$\displaystyle yu_{xxy}-e^xu_x+3=0 \ \mbox{and} \ yv_{xxy}-e^xv_x+3=0$

$\displaystyle L(u+v)=(yu_{xxy}-e^xu_x+yv_{xxy}-e^xv_x)+3\Rightarrow y(u_{xxy}+v_{xxy})-e^x(u_x+v_x)+3$

$\displaystyle \neq L(u)+L(v)=y(u_{xxy}+v_{xxy})-e^x(u_x+v_x)+6$
• Dec 27th 2010, 03:20 AM
Ackbeet
You have to distinguish between linear operators and linear equations. Typically, for the equation you've given, you'd throw the 3 over to the RHS and consider the operator to be everything else (that is, all the stuff with the dependent variable $\displaystyle u$ in it). So the operator is

$\displaystyle y\partial_{x}\partial_{x}\partial_{y}-e^{x}\partial_{x}.$

This is linear. So a differential equation is considered linear if the associated operator is linear. The DE can still be linear even if it's non-homogeneous, as in your case. Does that make sense?
• Dec 27th 2010, 12:22 PM
dwsmith
I understand you can have linear non-homogeneous equation.

Whenever there is a term that isn't associated with the partial derivatives, it is simply disregard?
• Dec 27th 2010, 04:52 PM
Ackbeet
I think you would ignore any term that doesn't have the dependent variable u in it.