Are the following PDEs linear or nonlinear?

Click For Summary
The discussion focuses on determining the linearity of specific partial differential equations (PDEs). The wave equation is identified as linear, but confusion arises with the terms in two other PDEs: one involving a term $u/x$ and another with $u^3$. For the first PDE, an operator is proposed as $L = \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1/x$, though its validity is questioned. The second PDE's operator is more complex, as it needs to account for the non-linear $u^3$ term, prompting suggestions to define a cubic operator. The conversation emphasizes the importance of experimenting with operator definitions to clarify linearity.
Thomas Moore
Messages
12
Reaction score
2
Hi. I'm a bit confused on determining whether a certain PDE is linear or non-linear.

For example, for the wave equation, we have: u_{xx} + u_{yy} = 0, where a subscript denotes a partial derivative.
So, my textbook says to write:
$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$

And then it is easy to deduce that $L(u+v) = L(u) + L(v)$ and $L(c u) = cL(u)$.

But, I have no idea how to do this for the following PDEs:
1. $u_{t} - u_{xx} + u/x = 0$, the $u/x$ is throwing me off.
2. $u_{tt} - u_{xx} + u^3 = 0$, the $u^3$ term is throwing me off.
I don't know how to write this as $Lu = 0$, to determine linearity. Any help would be appreciated, thanks!
 
Physics news on Phys.org
Hi. That link was quite helpful. Well, this is where I got stuck though.

For the first example, I wrote:
1. $L = \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1/x$, which seems to work if you do $L u$. But, I'm not sure if adding 1/x is right like that.

2. But for this one, I have no idea! I want to write: $L = \frac{\partial^2}{\partial t^2} - \frac{\partial}{\partial x^2} $ plus something, but I don't know how to write this so Lu will give the u^3 term at the end!
 
When choosing your operator, you can let ( )^3 be in your operator expression. E.g. if G is the cubic operator, then G(2)=8. The cubic operator (the operator which cubes its input) is not linear, but that doesn't mean we can't define it.
 
If you double up the dollar signs, the TeX will render properly.
BrianT is correct ... the naive way would have been just to write it out as if you just divided through by u.
You should not be afraid to try out stuff.
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K