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Proof of the linearity principle for a 2nd order PDE?

  1. Dec 31, 2014 #1
    1. The problem statement, all variables and given/known data
    My textbook (Advanced Engineering Mathematics, seventh edition, Kreyszig) indicates that if u1 and u2 are solutions to a second-order homogeneous partial differential equation, and c1 and c2 are constants, then u where

    u = c1u1 + c2u2

    is also a solution, this is the linearity principle (which it also calls the "Fundamental Theorem", not sure why though). I am asked to prove this theorem for second-order homogeneous partial differential equations of one and two variables. I am given as a hint that the proof is very similar to the same theorem for a second order homogeneous ordinary differential equation.

    2. Relevant equations
    First, the proof for the linearity principle for a 2nd-order homogeneous ODE of the form y'' + p(x)y' + q(x)y = 0. Substitute y = (c1y1 + c2y2). This results ultimately in:

    y'' + p(x)y' + q(x)y = c1(y''1 + py'1 + qy1) + c2(y''2 + py'2 + qy2) = 0

    I looked up also that the general form of a second-order homogeneous PDE is

    Auxx +Buxy + Cuyy + Dux + Euy + Fu + G = 0 where A, B, C, D, E, F, and G are functions of x and y.

    3. The attempt at a solution

    I tried the same approach with substitution. I let α and β be constants and u = αu1 + βu2

    Then

    Auxx +Buxy + Cuyy + Dux + Euy + Fu + G = 0

    Becomes

    A(αu1 + βu2)xx +B(αu1 + βu2)xy + C(αu1 + βu2)yy + D(αu1 + βu2)x + E(αu1 + βu2)y + F(αu1 + βu2) + G = 0

    Finally resulting in

    α(Au1xx +Bu1xy + Cu1yy + Du1x + Eu1y + Fu1 + G) + β(Au2xx +Bu2xy + Cu2yy + Du2x + Eu2y + Fu2 + G) = 0

    It is from here that I do not know how to continue. How do I prove from this, as in the case for homogeneous ODE's, that a linear combination of solutions is also a solution? Thank you in advance, and I seriously hope I didn't somehow wreck the formatting :)
     
  2. jcsd
  3. Dec 31, 2014 #2

    Stephen Tashi

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    You misunderstand the proof. You can't say "substitute (c1y1 + c2y2)", if you mean substitute this expression in place of y in the equation and keep the "=0". If you keep the "=0" you are assuming from the outset that the function (c1y1 + c2y2) is a solution to the equation. What you must do is prove that the expression is a solution.

    You don't get the "=0" just by substitution. Why does the expression on the left hand side "=0" ?
     
  4. Dec 31, 2014 #3

    SteamKing

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    Don't you now need to use the fact that u1 and u2, by definition, are solutions to the original PDE?

    In other words, if u1(x,y) = z(x,y), then

    Azxx +Bzxy + Czyy + Dzx + Ezy + Fz + G = 0

    and the same if u2(x,y) = z(x,y).
     
  5. Dec 31, 2014 #4


    I thought that was weird too, but it was just what came out of the book.
    Because it's homogeneous, right?

    Okay, I'll try that. Thanks:)
     
  6. Dec 31, 2014 #5

    LCKurtz

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    Of course, this result is false unless the PDE is linear.
     
  7. Dec 31, 2014 #6

    Stephen Tashi

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    "It's"?

    No, nothing about being homogeneous explains why the left hand side of the equation is zero. I think you didn't read the words that go along with the equations.
     
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