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

[jack476]

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

Au_xx +Bu_xy + Cu_yy + Du_x + Eu_y + 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

Au_xx +Bu_xy + Cu_yy + Du_x + Eu_y + 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(Au_1xx +Bu_1xy + Cu_1yy + Du_1x + Eu_1y + Fu₁ + G) + β(Au_2xx +Bu_2xy + Cu_2yy + Du_2x + Eu_2y + 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]

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

Au_xx +Bu_xy + Cu_yy + Du_x + Eu_y + 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

Au_xx +Bu_xy + Cu_yy + Du_x + Eu_y + 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(Au_1xx +Bu_1xy + Cu_1yy + Du_1x + Eu_1y + Fu₁ + G) + β(Au_2xx +Bu_2xy + Cu_2yy + Du_2x + Eu_2y + 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

Az_xx +Bz_xy + Cz_yy + Dz_x + Ez_y + Fz + G = 0

and the same if u₂(x,y) = z(x,y).

 

[jack476]

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

Az_xx +Bz_xy + Cz_yy + Dz_x + Ez_y + Fz + G = 0

and the same if u₂(x,y) = z(x,y).

Okay, I'll try that. Thanks:)

 

[LCKurtz]

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]

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.