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Check PDE for linearity

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Check du/dt + d^2u/dx^2 + 1 = 0

    2. Relevant equations

    L is a linear operator if:

    cL(u)=L(cu) and L(u+v)=L(u)+L(v)

    3. The attempt at a solution

    L = d/dt + d^2/dx^2 + 1

    L(cu) = d(cu)/dt + d^2(cu)/dx^2 + 1 = c du/dt + c d^2(u)/dx^2 + 1 ≠ cL(u) = c du/dt + c d^2/dx^2 + c. So I found that it is not linear since it does not satisfy cL(u)=L(cu). However the solution tells me that it is. Can anyone spot my error?

    Thanks
     
  2. jcsd
  3. Aug 31, 2012 #2

    Mute

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    Homework Helper

    Are you sure the problem wants you to include the '+1' in the definition of the operator?
     
  4. Aug 31, 2012 #3
    So that would be du/dt - d^2u/dx^2 = -1?

    Then I have Lu = -1, L(cu)= d(cu)/dt - d^2(cu)/dx^2 = c du/dt + c d^2u/dx^2 = c L(u), L(u+v) = d(u+v)/dt - d^2(u+v)/dx^2 = du/dt + d^2u/dx^2 + dv/dt + d^2v/dx^2 = Lu + Lv, so it's linear and inhomogenous.

    Thank you.
     
  5. Sep 1, 2012 #4

    Bacle2

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    Science Advisor

    I think what Mute was suggesting is that the real problem you have with linearity is

    the -1 . For d(u+v),you will end up with the constant term -1 , but the same

    will be the case for each of du and dv .Then du+dv will have constant term equal to

    -1 -1 =-2≠ -1
     
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