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?

  • $\begingroup$ 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? $\endgroup$ – Mattos Apr 15 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.