1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Sum of Second Order Linear PDEs

  1. Jan 31, 2013 #1
    Suppose we have two multivariate functions, [itex]u_{1}(x,t)[/itex] and [itex]u_{2}(x,t)[/itex]. These functions are solutions to second-order linear equations, which can be written as follows:


    Each of the coefficients are of the form [itex]A(x,y)[/itex]. Now, the linearity of these equations are undermined when any of the derivatives are altered by something other than their coefficients (a square, multiplied by another derivative, etc). Let's suppose that the previous linear model applies to [itex]u_{1}(x,t)[/itex] and [itex]u_{1}(x,t)[/itex] has the following format:


    The question is to determine whether [itex]u_{1}(x,t)+u_{2}(x,t)[/itex] is also a second degree linear PDE. If we were to compute this, we would find that the derivative of the sum would be the sum of the derivatives (i.e. [itex]\frac{\partial}{\partial x}=u_{1_{x}}+u_{2_{x}}[/itex]. However, in the long sum of the terms, the derivatives appear as a linear combination, so for example, our [itex]\frac{\partial}{\partial x}[/itex] term appears as [itex]Du_{1_{x}}+Ku_{2_{x}}[/itex]. Would the sum thus constitute as being non-linear?
    Last edited: Jan 31, 2013
  2. jcsd
  3. Feb 1, 2013 #2


    User Avatar
    Gold Member

    If you define ##v=u_{1}+u_{2}##, then by the linearity of differentiation, ##v## satisfies the PDE
    This is linear in ##v##. If I understand you correctly, this is what you're asking.
  4. Feb 3, 2013 #3
    I should have been more clear. I meant that the second equation would have variables [itex]w_{xx}[/itex] and so forth. This way, A+H could not reasonably be distributed as you have them.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook