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(1) $u_x + u_y = 0$
(2) $u_x + yu_y = 0$
(3) $u_x + uu_y = 0$
(4) $u_{xx} + u_{yy} = 0$
(5) $u_{tt} − u_{xx} + u^3 = 0$
(6) $u_t + uu_x + u_{xxx} = 0$
(7) $u_{tt} + u_{xxxx} = 0$
(8) $u_t − iu_{xx} = 0$

Can someone provide an explanation from definition of linearity why (3), (5), and (6) are NOT linear PDE's?

I.e., I tried $L(cu) = cL(u)$ and $L(u+v) = Lu + Lv$ but can't get this definition to work out nicely for examples 3, 5, and 6 above.

Thanks

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    $\begingroup$ (3) was already explained to you in your previous question, (6) is pretty much the same (up to the linear "uxxx" term). By the way: Still no LaTeX after (more than) 70 questions? $\endgroup$
    – t.b.
    Aug 25, 2012 at 19:24
  • $\begingroup$ Please verify that I interpreted your equations correctly. $\endgroup$ Aug 25, 2012 at 19:25
  • $\begingroup$ Brian, that is correct. Thanks for your help $\endgroup$
    – mary
    Aug 25, 2012 at 19:32

2 Answers 2

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Let's look at number 3 in full detail. Define

$$L(u)=u_x+u u_y$$ Then for arbitrary constant c, we have

$$L(c u)=(c u)_x+(c u )(c u)_y=c u_x+c u \left(c u_y\right)=c u_x+c^2 u u_y$$

Compare this with

$$c L(u)=c\left(u_x+u u_y\right)=c u_x+c u u_y$$

So

$$L(c u) - c L(u) = c u_x + c^2 u u_y -\left(c u_x+c u u_y\right)=\left(c^2-c\right)u u_y\neq 0$$ and therefore $L(u)$ is not linear. Similarly, examples 5 and 6 are also seen to have non-linear terms.

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A differential equation is linear if the dependent variable(s) and its derivatives appear linearly in the DEqn. In other words, if the DEqn is a linear combination of $u,u_x,u_y, u_{xy}, u_{xx},...$ then the DEqn will be linear. Here the coefficients of the combination can be nonlinear functions of the independent variables $x,y,...$ in your current notation.

For example,

$$uu_x+u_y=0$$

is not linear because it has a term with a product of dependent terms $u$ and $u_x$. On the other hand,

$$ u_x+x^2u_{yy}+\sin(xy)u_{yy}=0$$

is a linear PDE and as you can perhaps see the dependent terms appear linearly.

What is a linear combination? Given $v_1,v_2,...v_k$ a linear combination of the $v_j$'s over $S$ is

$$ s_1v_1+s_2v_2+ \dots s_kv_k $$

where $s_1,s_2,\dots s_k \in S$. All of this said, the pattern I point out here is merely given to insure that the PDE can be written as a linear operator. I think Robert Miller's answer goes to this point.

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