Mat-F February 21, 2005 Separation of variables

Mat-FFebruary 21, 2005

Separation of variables

Åke Nordlund

Niels Obers, Sigfus Johnsen / Anders SvenssonKristoffer Hauskov Andersen

Peter Browne Rønne

Overview

Follow-up Computer problems Maple TA problems Maple TA changes New web pages

Chapter 19 Separation of variables Exercises today

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Follow-up:Maple TA

Maple TA improvements No longer accepts “0” or “whatever” as solutions

Remaining limitations Still accepts less than general PDE solutions

Use human judgment! I will !

Follow-up:Maple TA

Maple TA changes

For those who could not come Wednesday For continued exercises (computer problems) For additional training

Maple TA exercises will be (are) available in an open version from the day (Thursday) after the exercises.

Maple TA exercises will be (are) available in an open version from the day (Thursday) after the exercises.

New web pages

Maple Examples Examples / templates for turn-in assignments

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Rooms could possibly change next week But not this week!

19. Partial Differential Equations:Separation of variables …

Main principles Why?

Techniques How?

Main principles

Why? Because many systems in physics are separable In particular systems with symmetries Such as spherically symmetric problems

Example: Spherical harmonics

u(r,Θ,φ,t) = Ylmn(r,Θ,φ) eiωt

Example: Spherical harmonics

u(r,Θ,φ,t) = Ylmn(r,Θ,φ) eiωt

19. Partial Differential Equations:Separation of Variables …

Main principles Why?

Common in physics! Greatly simplifies things!

Techniques How?

19.1 Separation of variables:the general method

Suppose we seek a PDE solution u(x,y,z,t)

Ansatz:

u(x,y,z,t) = X(x) Y(y) Z(z) T(t)

Ansatz:


This ansatz may be correct or incorrect!

If correct we have


If correct we have

u/x = X’(x) Y(y) Z(z) T(t)

2u/x2 = X”(x) Y(y) Z(z) T(t)

u/x = X’(x) Y(y) Z(z) T(t)

2u/x2 = X”(x) Y(y) Z(z) T(t)


Suppose we seek a PDE solution u(x,y,z,t):

Ansatz:


Ansatz:



If correct we have


If correct we have

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t)

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t) u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t)

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t)



Ansatz:


Ansatz:



If correct we have


If correct we have

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t)

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t) u/z = X(x) Y(y) Z’(z) T(t)

2u/z2 = X(x) Y(y) Z”(z) T(t)

u/z = X(x) Y(y) Z’(z) T(t)

2u/z2 = X(x) Y(y) Z”(z) T(t)



Ansatz:


Ansatz:



If correct we have


If correct we have

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t)

u/y = X(x) Y’(y) Z(z) T(t)

2u/y2 = X(x) Y”(y) Z(z) T(t) u/t = X(x) Y(y) Z(z) T’(t)

2u/t2 = X(x) Y(y) Z(z) T”(t)

u/t = X(x) Y(y) Z(z) T’(t)

2u/t2 = X(x) Y(y) Z(z) T”(t)

Examples

u(x,y,z,t) = xyz2sin(bt)u(x,y,z,t) = xyz2sin(bt)

u(x,y,z,t) = xy + ztu(x,y,z,t) = xy + zt

u(x,y,z,t) = (x2+y2) z cos(ωt)u(x,y,z,t) = (x2+y2) z cos(ωt)

Examples




Separable?

No!

Yes!

Hm…?

Examples




Separable?

No!

Yes!

Hm…?

(x2+y2) p2(x2+y2) p2

Examples



u(p,z,t) = p2 z cos(ωt)u(p,z,t) = p2 z cos(ωt)

Separable?

No!

Yes!

Yes!

Example PDEs

The wave equation

where

c22u = 2u/t2 c22u = 2u/t2

2 2/x2 + 2/y2 + 2/z2 2 2/x2 + 2/y2 + 2/z2

Ansatz:


Ansatz:


Separation of variables in the wave equation

For the ansatz to work we must have (lets set c=1 for now to get rid of a triviality!)

Divide though with XYZT, for:

X”YZT + XY”ZT + XYZ”T = XYZT”X”YZT + XY”ZT + XYZ”T = XYZT”

X”/X + Y”/Y + Z”/Z = T”/TX”/X + Y”/Y + Z”/Z = T”/T

This can only work if all of these are constants!This can only work if all of these are constants!


Let’s take all the constants to be negative Negative bounded behavior at large x,y,z,t

X”/X + Y”/Y + Z”/Z = T”/TX”/X + Y”/Y + Z”/Z = T”/T

- 2 - 2 - 2 = - µ2 - 2 - 2 - 2 = - µ2

This is typical: If / when a PDE allows separation of variables, the partial derivatives are replaced with ordinary derivatives, and all that remains of the PDE is an algebraic equation and a set of ODEs – much easier to solve!

This is typical: If / when a PDE allows separation of variables, the partial derivatives are replaced with ordinary derivatives, and all that remains of the PDE is an algebraic equation and a set of ODEs – much easier to solve!

These are called separation constantsThese are called separation constants


Solutions of the ODEs:

Example: x-direction

X”/X = - 2


X”/X = - 2

General solution:

X(x) = A exp(i x) + B exp(-i x)

or

X(x) = A’ cos( x) + B’ sin( x)

General solution:


or



Combining all directions (and time)

Example:

u = exp(i x) exp(i y) exp(i x) exp(-i cµt)

or

u = exp(i ( x + y + x - ω t))

Example:

u = exp(i x) exp(i y) exp(i x) exp(-i cµt)

or

u = exp(i ( x + y + x - ω t))

This is a traveling wave, with wave vector {,,} and frequency ω. A general solution of the wave equation is a super-position of such waves.

This is a traveling wave, with wave vector {,,} and frequency ω. A general solution of the wave equation is a super-position of such waves.

Example PDEs

The diffusion equation

where

2u = u/t 2u = u/t

2 2/x2 + 2/y2 + 2/z2 2 2/x2 + 2/y2 + 2/z2

Ansatz:


Ansatz:


Separation of variables in the diffusion equation

For the ansatz to work we must have (lets set =1 for now to again get rid of that triviality)


X”YZT + XY”ZT + XYZ”T = XYZT’X”YZT + XY”ZT + XYZ”T = XYZT’

X”/X + Y”/Y + Z”/Z = T’/TX”/X + Y”/Y + Z”/Z = T’/T

Again, this can only work if all of these are constants!Again, this can only work if all of these are constants!


Again, we take all the constants to be negative Negative bounded behavior at large x,y,z,t

X”/X + Y”/Y + Z”/Z = T’/TX”/X + Y”/Y + Z”/Z = T’/T

- 2 - 2 - 2 = - µ - 2 - 2 - 2 = - µ

We see the typical behavior again: All that remains of the PDE is an algebraic equation and a set of ODEs – much easier to solve!

We see the typical behavior again: All that remains of the PDE is an algebraic equation and a set of ODEs – much easier to solve!




X”/X = - 2


X”/X = - 2

General solution:


or


General solution:


or


Time-direction:

T’/T = - µ

Time-direction:

T’/T = - µ


General solution:

T(t) = A exp(- µ t) + B exp(µ t)

General solution:

T(t) = A exp(- µ t) + B exp(µ t)


Combining all directions (and time)

Example:

u = exp(i x) exp(i y) exp(i x) exp(-µ t)

or

u = exp(i ( x + y + x )) exp(- µ t)

Example:

u = exp(i x) exp(i y) exp(i x) exp(-µ t)

or

u = exp(i ( x + y + x )) exp(- µ t)

This is a damped wave, with wave vector {,,} and damping constant µ. A general solution of the diffusion equation is a super-position of such waves.

This is a damped wave, with wave vector {,,} and damping constant µ. A general solution of the diffusion equation is a super-position of such waves.

Example PDEs

The Laplace equation

where

2u = 0 2u = 0

2 2/x2 + 2/y2 + 2/z2 2 2/x2 + 2/y2 + 2/z2

Ansatz:

u(x,y,z,t) = X(x) Y(y) Z(z)

Ansatz:

u(x,y,z,t) = X(x) Y(y) Z(z)

Separation of variables in the Laplace equation

For the ansatz to work we must have


X”YZT + XY”ZT + XYZ”T = 0X”YZT + XY”ZT + XYZ”T = 0

X”/X + Y”/Y + Z”/Z = 0X”/X + Y”/Y + Z”/Z = 0

Again, this can only work if all of these are constants!Again, this can only work if all of these are constants!


This time we cannot take all constants to be negative!

X”/X + Y”/Y + Z”/Z = 0X”/X + Y”/Y + Z”/Z = 0

- 2 - 2 - 2 = 0 - 2 - 2 - 2 = 0

At least one of the terms must be positive imaginary wave number; i.e., exponential rather than sinusoidal behavior!

At least one of the terms must be positive imaginary wave number; i.e., exponential rather than sinusoidal behavior!

Let’s assume for simplicity (as in the example in the book) that there is only x & y dependence!

Let’s assume for simplicity (as in the example in the book) that there is only x & y dependence!




X”/X = - 2 < 0 assumed


X”/X = - 2 < 0 assumed

General solution:


or


General solution:


or




Example: y-direction

Y”/Y = + 2 > 0 follows!

Example: y-direction

Y”/Y = + 2 > 0 follows!

General solution:

Y(y) = C exp( y) + D exp(- y)

or

Y(y) = C’ cosh( y) + B’ sinh( y)

General solution:

Y(y) = C exp( y) + D exp(- y)

or

Y(y) = C’ cosh( y) + B’ sinh( y)


Combining x & y directions

Example:u = [A exp(i x) + B exp(-i x)] [C exp( y) + D exp(- y)]

oru = [A’ cos( x) + B’ sin( x)] [C’ cosh( y) + D’ sinh(- y)]

Example:u = [A exp(i x) + B exp(-i x)] [C exp( y) + D exp(- y)]

oru = [A’ cos( x) + B’ sin( x)] [C’ cosh( y) + D’ sinh(- y)]

A general solution of the Laplace equation is a superposition of such functions (which everywhere have a ’saddle-like’ behavior – net curvature = 0)

A general solution of the Laplace equation is a superposition of such functions (which everywhere have a ’saddle-like’ behavior – net curvature = 0)

19.2 Superposition of separated solutions

If the PDE with separable solutions is linear … Wave equation Diffusion equation Laplace & Poisson eq. Schrødinger eq.

… then any linear combination (superposition) is also a solution Boundary conditions will make a limited selection!

Exercises today:

Exercise 19.1 Separation of variables

Exercise 19.2 Diffusion of heat

Exercise 19.3 Wave equation on a stretched membrane

Examples in the text of the book

Heat diffusion Stationary (u/t = 0)

equivalent to Laplace equation problem!

Various boundary conditions selects particular solutions NOTE: a diffusion (heat flow) equation may well have

a mixed sinusoidal & exponential behavior in space! consider the limit when the time dependence is weak!

2u = 0 2u = 0

2u = u/t 2u = u/t

2u = (1/)u/t → 0 2u = (1/)u/t → 0

- 2 - 2 - 2 = - µ → 0 - 2 - 2 - 2 = - µ → 0

Enough for today!

Good luck with the exercises 10:15-12:00Good luck with the exercises 10:15-12:00

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Mat-F February 21, 2005 Separation of variables