How do I show that a PDE is linear?

Question

And more specifically, how can I show that the PDE $$u_{xx}+e^{-xy}u_y=cos(x)$$ is linear?

Do I first subtract the cos(x) to the left side?

I think I understand that in order for a PDE to be linear, it has to be able to be written in the form $L(u+v) = L(u) + L(v)$ where L is a linear operator and u,v are functions. And $L(cu) = cL(u)$, where c is a scalar.

The thing that also is confusing me is the $e^{-xy}$. Is it a constant?

Thank you in advance.

Answer 1

Take two solutions, $v$ and $w$, and two constants $\alpha$ and $\beta$, and substitute in the linear combination, $\alpha v+\beta w$, and see if it also a solution. This would give,

$$(\alpha v+\beta w)_{xx} + e^{-xy}(\alpha v+\beta w)_y=\cos(x)$$

Now try to use the linearity of the derivative operators $(\cdot)_{xx}$ and $(\cdot)_y$ to check if we still have a solution.

Answer 2

You're confusing a few things. The conditions $L(u+v)=L(u)+L(v)$ and $L(cu)=cL(u)$ are what it means for the operator $L$ to be linear.

A linear PDE is a PDE of the form $L(u)=g$ for some function $g$, and your equation is of this form with $L=\partial_x^2 + e^{-xy} \partial_y$ and $g(x,y)=\cos x$. (Sometimes this is called an inhomogeneous linear PDE if $g \neq 0$, to emphasize that you don't have superposition. Some may also say affine PDE but this is not very common.)

(And of course $e^{-xy}$ isn't constant – it depends on $x$ and $y$!)