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I just want to verify and check my understanding to see if I can do this right.

a)$ u_x + u_y = 0$

This is linear since: $(u+v)_x + (u+v)_y = u_x +u_y + v_x+v_y $

and $c(u_x+u_y) = cu_x + cu_y$

b) $u_x + yu_y = 0 $

This is linear since:

$(u+v)_x + y(u+v)_y = u_x+v_x + yu_y + yv_y $

and $c(u_x+yu_y) = cu_x+ ycu_y$

c) $u_x + uu_y = 0$

Linear again, I don't see the fail of linearity on this example as well.

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By $ux$ do you mean $u_x$, i.e. $du/dx$? If so, have you expanding out the linearity conditions for (c) like you did for (a) and (b)? –  Rahul Aug 25 '12 at 6:18
    
I formated the text using $\LaTeX$. Please take a look if this is how you intended the formulae to look. If you press edit, you can see the source-code and learn how to do the formatting yourself. –  Fabian Aug 25 '12 at 7:11
    
Yes, Rahul you are correct –  mary Aug 25 '12 at 19:15
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1 Answer 1

up vote 1 down vote accepted

The equation in c) is non-linear. For example, take any $\alpha \in \mathbb{R}$ and define the operator $$Lu = u_x + u u_y.$$ Now compute $$ L(\alpha u) = \alpha u_x + \alpha^2 u u_y, $$ which shows that $L(\alpha u)$ is in general different than $\alpha Lu$.

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