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Check PDE for linearity

  • Thread starter frenchkiki
  • Start date
  • #1
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Homework Statement



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

Homework Equations



L is a linear operator if:

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

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
 

Answers and Replies

  • #2
Mute
Homework Helper
1,388
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Are you sure the problem wants you to include the '+1' in the definition of the operator?
 
  • #3
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Are you sure the problem wants you to include the '+1' in the definition of the operator?
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.
 
  • #4
Bacle2
Science Advisor
1,089
10
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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