text: Comparing a linear PDE to a nonlinear PDE - Mathematics Stack Exchange
  Comparing a linear PDE to a nonlinear PDE
  I've got two PDE:s, one of them is linear, and the other is not. I have tried finding the answer to why this is, but I can't find a satisfying answer anywhere, would really appreciate some help.
  Nonlinear PDE: $$\frac{\partial^2u}{\partial x^2}-sin(\frac{\partial u}{\partial x}\frac{\partial u}{\partial y})=e^{xu}$$
  Linear PDE: $$xy\frac{\partial^2u}{\partial x^2}-e^{x-y}\frac{\partial^2 u}{\partial x\partial y} + \frac{\partial u} {\partial x} -2xu=cos(xy)$$
  asked Feb 15 '17 at 12:15
  $\begingroup$ A linear DE looks like $(Lu)(x,u)=f(x)$ where $L$ is a linear operator (and $x$ is a vector of independent variables). This means it satisfies $L(c_1 u_1 + c_2 u_2)=c_1 L(u_1) + c_2 L(u_2)$ whenever $c_1,c_2$ are scalars. A nonlinear DE is anything without this property. In your second one you have a bunch of complicated stuff going on but it is all in the form of coefficients of linear functions of $u$ which are themselves only functions of $x$ and $y$. By contrast in your first case the second and third terms are not linear in $u$. $\endgroup$ – Ian Feb 15 '17 at 12:17
  $\begingroup$ How come $e^{xu}$ is not linear, but $e^{x-y}$ is? Also, how come $cos(xy)$ is linear, but $sin(\frac{\partial u}{\partial x}\frac{\partial u}{\partial y})$ is not? $\endgroup$ – armara Feb 15 '17 at 12:25
  $\begingroup$ Because what matters is the dependence on $u$. Something that doesn't depend on $u$ is automatically linear in $u$. $\endgroup$ – Ian Feb 15 '17 at 12:28
  $\begingroup$ I get it, big thanks! $\endgroup$ – armara Feb 15 '17 at 12:28
  $\begingroup$ Here are some videos that discuss the linear property of differential equations: youtu.be/VfJN9Px3KyM?list=PLlXfTHzgMRUK56vbQgzCVM9vxjKxc8DCr, youtu.be/sVz1hS_uXgk $\endgroup$ – Lemma Feb 15 '17 at 13:22
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