# PDE Linearity Textbook Problems: Deducing Correct Operators For Determining Linearity

I am doing PDE textbook problems that require me to determine whether an operator is linear or not.

In a previous problem, I had $$\mathscr{L} u = u_x + u_y + 1$$. I determined that the operator is $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + 1$$, and so $$\mathscr{L} (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1 = \dots$$. I think this is the correct operator, although I am not sure.

I now have the PDE $$u_{tt} - u_{xx} + x^2 = 0$$. It seems to me that the operator is $$\mathscr{L} = \dfrac{\partial^2}{\partial{t}^2} - \dfrac{\partial^2}{\partial{x}^2} + x^2$$; this way, we have that $$\mathscr{L}u = \dfrac{\partial^2 u}{\partial{t}^2} - \dfrac{\partial^2u}{\partial{x}^2} + x^2$$. However, I have some third-party solutions (see part d), bottom of page 3) that claim that the operator is $$\mathscr{L} = \dfrac{\partial^2}{\partial{t}^2} - \dfrac{\partial^2}{\partial{x}^2}$$, since we can have $$\mathscr{L}u = -x^2.$$ This would then classify the PDE as linear inhomogeneous, instead of linear homogeneous, which is what my operator would indicate.

Although I haven't taken functional analysis yet, my understanding of operators is that they are mappings, and so I think it is valid to have have added constants, as in $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + 1$$, or added independent variables, as in $$\mathscr{L} = \dfrac{\partial^2}{\partial{t}^2} - \dfrac{\partial^2}{\partial{x}^2} + x^2$$, right?

Are the (third-party) solutions incorrect, or am I misunderstanding something?

• In both cases, the PDEs determined by $$\mathcal{L}u = 0$$ are definitely linear inhomogeneous. Does that help you determine what the operators should be? – Mattos Apr 15 at 13:23