# Simple linear(?) operator

1. Sep 25, 2010

### theneedtoknow

These 2 questions are somewhat related:

Is the operator linear: Lu = du/dx + du/dy + 1

I said it's non-linear since L(u+v) = du/dx + dv/dx + du/dy + dv/dy + 1 = Lu + Lv - 1 =/= Lu + Lv

and then:
State whether each of the following equations is nonlinear, linear inhomogeneous, or linear homogeneous.

du/dt - d^2u/dx^2 + 1 = 0
which can be rewritten as du/dt - d^2u/dx^2 = -1
so now we have pretty much the same situation as above, except since it was set to zero, i can just bring the 1 on the other side and the resulting equation is linear and non-homogeneous Lu = -1

So then, is my reasoning for the first question wrong, that the operator (which is almost identical except for the order of derivatives) is actually linear? And then, why is my reasoning wrong?

2. Sep 25, 2010

### fzero

Your reasoning is fine, there's just a difference between the conditions for an operator or equation to be linear. Given a linear operator L, there's always an associated linear equation L u = c. If we want to find the linear operator corresponding to a general linear equation, we need to separate any nonzero constant term.