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Are the following PDEs linear or nonlinear?

  1. Sep 11, 2015 #1
    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!
     
  2. jcsd
  3. Sep 11, 2015 #2

    Simon Bridge

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  4. Sep 11, 2015 #3
    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!
     
  5. Sep 12, 2015 #4
    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.
     
  6. Sep 13, 2015 #5

    Simon Bridge

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    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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