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

## Homework Statement

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.

## Homework 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.

## 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 :)

Stephen Tashi

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:

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.

This results ultimately in:
y'' + p(x)y' + q(x)y = c1(y''1 + py'1 + qy1) + c2(y''2 + py'2 + qy2) = 0

You don't get the "=0" just by substitution. Why does the expression on the left hand side "=0" ?

SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

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.

## Homework 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.

## 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 :)

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).

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.

I thought that was weird too, but it was just what came out of the book.
You don't get the "=0" just by substitution. Why does the expression on the left hand side "=0" ?

Because it's homogeneous, right?

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).

Okay, I'll try that. Thanks:)

LCKurtz
Homework Helper
Gold Member

## Homework Statement

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).

Of course, this result is false unless the PDE is linear.

Stephen Tashi