Are the following PDEs linear or nonlinear?

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    Linear Nonlinear Pdes
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SUMMARY

The discussion focuses on determining the linearity of specific partial differential equations (PDEs). The wave equation, represented as u_{xx} + u_{yy} = 0, is confirmed as linear through the operator L = ∂²/∂x² + ∂²/∂y². However, the PDEs u_{t} - u_{xx} + u/x = 0 and u_{tt} - u_{xx} + u³ = 0 present challenges due to the presence of non-linear terms (u/x and u³). The participants conclude that the first PDE is non-linear due to the 1/x term, while the second is non-linear because of the u³ term, indicating that both cannot be expressed in the form Lu = 0.

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Thomas Moore
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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!
 
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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.
 

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