# 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 :)

## Answers and Replies

Stephen Tashi
Science Advisor

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
Science Advisor
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
Science Advisor
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
Science Advisor

Because it's homogeneous, right?

"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.