# Check PDE for linearity

• frenchkiki
In summary, the problem involves determining the linearity of a linear operator, L, which is defined as d/dt + d^2/dx^2 + 1. After attempting the solution, it was found that the operator is not linear since it does not satisfy the linearity property. However, the given solution states that it is linear. After further analysis, it was determined that the issue lies in the inclusion of the constant term, +1, in the operator definition. After removing the constant term, the operator is found to be linear and inhomogeneous.

## 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

Are you sure the problem wants you to include the '+1' in the definition of the operator?

Mute said:
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.

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