1
$\begingroup$

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: $$xy\frac{\partial^2u}{\partial x^2}-e^{x-y}\frac{\partial^2 u}{\partial x\partial y} + \frac{\partial u} {\partial x} -2xu=cos(xy)$$

  • 3
    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 $u$. – 2017-02-15
  • 0
    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? – 2017-02-15
  • 2
    Because what matters is the dependence on $u$. Something that doesn't depend on $u$ is automatically linear in $u$. – 2017-02-15
  • 0
    I get it, big thanks! – 2017-02-15
  • 0
    Here are some videos that discuss the linear property of differential equations: https://youtu.be/VfJN9Px3KyM?list=PLlXfTHzgMRUK56vbQgzCVM9vxjKxc8DCr, https://youtu.be/sVz1hS_uXgk – 2017-02-15

0 Answers 0