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Linear operator or nonlinear operator?

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Verify whether or not the operator

    [tex]L(u) = u_x + u_y + 1[/tex]
    is linear.


    2. Relevant equations
    An operator L is linear if for any functions u, v and any constants c, the property

    [tex]L(c_1 u + c_2 v) = c_1 L(u) + c_2 L(v)[/tex]
    holds true.


    3. The attempt at a solution

    I feel as though this should be a linear operator, but the "+1" throws me off as I don't know what linear operator takes any function u, v into 1.

    [tex]L = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + ???[/tex]
     
    Last edited: Jan 19, 2013
  2. jcsd
  3. Jan 19, 2013 #2

    Zondrina

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    Alright so, really there's two things you want to verify, but I suppose you could combine them into one condition like that if you want.

    What is [itex]L(c_1 u + c_2 v) = ?[/itex]
     
  4. Jan 19, 2013 #3
    [tex] L = c_1 \frac{ \partial u}{\partial x} + c_1 \frac{\partial u}{\partial y} + c_2 \frac{ \partial v}{\partial x} + c_2 \frac{\partial v}{\partial y} + ??? [/tex]

    I'm still confused by that constant term.
     
    Last edited: Jan 19, 2013
  5. Jan 19, 2013 #4

    Zondrina

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    Just literally write out what the transform would give you, so :

    [itex]L(c_1 u + c_2 v) = c_1u_x + c_1u_y + c_1 + c_2v_x + c_2v_y + c_2[/itex]

    Can you continue from there? Get it into the form [itex]c_1 L(u) + c_2 L(v)[/itex]
     
  6. Jan 19, 2013 #5
    Sorry for the typos everywhere earlier (edited now).

    [tex]L(c_1 u + c_2 v) = c_1 u_x + c_1 u_y + c_1 + c_2 v_x + c_2 v_y + c_2[/tex]
    [tex]= c_1 (u_x + u_y + 1) + c_2 (v_x + v_y + 1)[/tex]
    [tex]= c_1 L(u) + c_2 L(v)[/tex]

    So is that what you were guiding me to? If I did that correctly, it makes a lot more sense now, thank you. If not...
     
  7. Jan 19, 2013 #6

    Dick

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    No, no, no. [tex]L(c_1 u + c_2 v) = (c_1 u + c_2 v)_x + (c_1 u + c_2 v)_y + 1[/tex]
     
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