viii.2 method of separation of variables – stationary ...vps/me505/iem/08 02.pdf · chapter viii...

Chapter VIII PDE VII.2 Separation of Variables - Stationary Boundary Value Problems October 30, 2019

573

VIII.2 Method of Separation of Variables –

Stationary Boundary Value Problems

VIII. 2.1 The Concept of Separation of Variables

2.2 Laplace’s Equation

2.3. Solution of the non-basic case: more than one of the boundary conditions are non-homogeneous

2.4. Poisson’s Equation – Eigenfunction Expansion Method

2.5 Laplace Equation in Cylindrical Coordinates 2.6 Laplace Equation in Spherical Coordinates (to be done) 2.7 Review Questions and Exercises

Chapter VIII PDE VII.2 Separation of Variables - Stationary Boundary Value Problems October 30, 2019 574


575

VIII.2.1 Concept of Separation of Variables

The classical analytical approach to the solution of initial-boundary value problems of equations of mathematical physics is based on the method of separation of variables. This method consists in building the set of basic functions which is used in developing solutions in the form of an infinite series expansion over the basic functions. This method is applicable for solution of homogeneous equations such as the Laplace equation

0yu

xu

2

2

2

2

=∂∂

+∂∂

Zero function is always a solution of homogeneous equation – zero solution is called a trivial solution. We are looking for non-trivial solution ( )y,xu , which is not identically equal to zero function. The unknown function ( )y,xu is assumed to be represented as a product of two functions each of a single variable:

assumption of separation of variables ( ) ( ) ( )yYxXy,xu =

where both ( )X x and ( )Y y are non-zero functions. Substitution of the assumed form of the solution into the Laplace Equation yields

( )( )

( )( )

X x Y yX x Y y

′′ ′′= −

where the left hand side and right hand side depend on different independent variables, and therefore, do not depend on either of them, that means that they are the constants Indeed, differentiation of both sides of equation with respect to x yields

X 0x X

′′∂ = ∂

Integration yields XX

µ′′

= with µ as a constant of integration.

Therefore,

Separated equation X YX Y

µ′′ ′′

= − =

where µ is called a separation constant. It yields two ordinary differential equations:

X X 0µ′′ − = Y Y 0µ′′ + =

If for one of these equations, the corresponding boundary conditions are also homogeneous, then the Sturm-Liouville Theorem will provide the existence of the set of eigenvalues nµ and eigenfunctions, either

( )nX x or ( )nY y , which will be a basis for construction of the solution in the form of infinite series:

( ) ( ) ( )n n nn

u x, y c X x Y y= ∑

where coefficients nc should be determined in the process of solution of the corresponding boundary value problem.

Remark Assumption of separated variables is a restriction applied at the initial step of construction of solution. However, the final form of solution is not necessarily a separable function.


2.2 2-D Laplace’s Equation (LE) BASIC CASE: 1 non-homogeneous and 3 homogeneous boundary conds

0yu

xu

2

2

2

2

=∂∂

+∂∂ ( )u x, y : ( ) ( ) ( )x, y D 0,L 0,M∈ = ×

x 0= [ ] ( )3x 0u f y=

=

x L= [ ]x Lu 0=

=

y 0= [ ]y 0u 0=

=

y M= [ ]y Mu 0=

=

Separation of variables We assume that the function ( )y,xu can be represented as a product of two

functions each of a single variable: ( ) ( ) ( )yYxXy,xu =

where both ( )X x and ( )Y y are non-zero functions. Differentiate the function ( )y,xu consequently with respect to x and y:

YXxu ′=

∂∂ YX

xu2

2

′′=∂∂

YXyu ′=

∂∂ YX

yu2

2

′′=∂∂

and substitute the second order derivatives into LE: 0YXYX =′′+′′ Divide this equation by the product XY and separate the terms

YY

XX ′′

−=′′

It yields an equation with separated variables: the left hand side of this equation is a function of the independent variable x only, and the right hand side is a function of the independent variable y only. The equality for all values of x and y is possible only if both sides are equal to the same constant (call it µ ). Indeed, differentiate the equation with respect to x:

X 0x X

′′∂ = ∂

then by integration we obtain

XX

µ′′

=

Therefore,

X YX Y

µ′′ ′′

= − =

where µ is called a separation constant. It yields two ordinary differential equations:

X X 0µ′′ − = and Y Y 0µ′′ + =

x

[ ]x Lu 0=

=[ ] ( )3x 0u f y=

=

y

M

L0

[ ]y Mu 0=

=

[ ]y 0u 0=

=

2u 0∇ =


577

Boundary conditions: Boundary conditions are given for the function ( )y,xu . Determine, what conditions should be satisfied by functions X and Y (assuming that both are non-trivial):

x 0= [ ] ( )3x 0u f y=

= [ ] ( ) ( )3x 0X Y y f y=

=

x L= [ ]x Lu 0=

= [ ] ( )x LX Y y 0=

= ⇒ [ ]x LX 0=

=

y 0= [ ]y 0u 0=

= ( )[ ]y 0X x Y 0=

= ⇒ [ ]y 0Y 0=

=

y M= [ ]y Mu 0=

= ( )[ ]y MX x Y 0=

= ⇒ [ ]y MY 0=

=

Solution of SLP: Start with the equation for which both boundary conditions are homogeneous:

Y Y 0µ′′ + =

[ ]y 0Y 0=

= Sturm-LiouvilleTheorem (Table9)

⇒ 2nµ λ= n 1,2,...= eigenvalues

[ ]y MY 0=

= ( )nY y eigenfunctions

(in a case of N-N: 0 0λ = , 0Y 1= )

Solution for X : We determined that separation constant is 2nµ λ= . Then equation for X is

2

nX X 0λ′′ − =

Auxiliary equation 2 2nm 0λ− = yields the roots nm λ= ± n 1,2,...=

(for N-N: 0m 0= )

Therefore, the solution for X is (in the Table 9, choose the shifted form of solution with hyperbolic functions, because at the boundary x=L, we have a homogeneous boundary condition, and the interval is finite)

( ) ( ) ( )n 1 n 2 nX x c cosh x L c sinh x Lλ λ = − + − ,...3,2,1n =

Consider first homogeneous boundary condition at Lx = :

( ) ( ) ( )n 1 n 2 n 1X L 0 c cosh 0 c sinh 0 cλ λ = = + = ⇒ 0c1 =

( ) ( )n 2 nX x c sinh x Lλ = − ,...3,2,1n = (choose 2c 1= )

In a case of N-N boundary conditions, there will be an additional solution: ( ) ( )0 1 2X x c c x L= + −

( )n 1 2 1X L 0 c c 0 c= = + ⋅ = ⇒ 0c1 =

( ) ( )0X x x L= −

[ ]

[ ][ ]

[ ]

Here, notation u meansthat boundary conditions can be of any type:I Dirichlet u f

II Neumann u f

III Robin u +Hu f

=

′ =

′ =


Basic solutions According to the assumed form of solution, we may construct a set of basic solutions:

( )0u x, y = ( ) ( )0 0X x Y y = x L− (only for N-N conditions)

( )y,xu1 = ( ) ( )yYxX 11 = ( ) ( )1 1sinh x L Y yλ −

( )2u x, y = ( ) ( )2 2X x Y y = ( ) ( )2 2sinh x L Y yλ −

( )nu x, y = ( ) ( )n nX x Y y = ( ) ( )n nsinh x L Y yλ − All these solutions satisfy the LE and 3 homogeneous boundary conditions.

Any linear combination of the basic solutions is also a solution. Construct a solution of LE in the form of a linear combination with coefficients na :

( ) ( ) ( )n n n n nn 1 n 1

u( x, y ) a u x, y a X x Y y∞ ∞

= =

= =∑ ∑

( ) ( ) ( )n n n n nn 1 n 1

u( x, y ) a u x, y a X x Y y∞ ∞

= =

= =∑ ∑ (only for N-N conditions)

This solution also satisfies LE and 3 homogeneous boundary conditions. Determine coefficients na in such a way that the last boundary condition (non-homogeneous) is also satisfied (it will yield the solution of the problem). Boundary condition at 0x = :

[ ] ( ) ( ) ( )3 n n nx 0 x 0n 1

u f 3 a X x Y y∞

= ==

= = ∑

If we treat this sum as a Generalized Fourier series expansion of the function

( )3f y on the interval [ ]My ,0∈ with coefficients

[ ]( ) ( )

( )

M

n0

n Mx 02

n0

f y Y y dya X

Y y dy=

⋅ =∫

∫

then coefficients na are determined as:

[ ]

( ) ( )

( )

M

n0

n M2x 0

n0

f y Y y dy1a

XY y dy=

=∫

∫

and the solution now is completely determined. Solution of BVP

( )( ) ( )

( )

( )[ ] ( )

M

nn0

n n nMn 1 n 1 2 n x 0

n0

f y Y y dyX x

u( x, y ) a u x, y Y yX

Y y dy

∞ ∞

= = =

= =

∫∑ ∑

∫

In the case of the Neumann-Neumann boundary conditions the solution is:

( )( ) ( )

( )

( )[ ] ( )

M

nn0

n n nMn 0 n 0 2 n x 0

n0

f y Y y dyX x

u( x, y ) a u x, y Y yX

Y y dy

∞ ∞

= = =

= =

∫∑ ∑

∫


579

Example Dirichlet Problem for Laplace’s Equation

x 0= ( ) ( )3u 0, y f y=

x L= ( )u L, y 0= ( ) ( )X L Y y 0= ⇒ ( )X L 0=

y 0= ( )u x,0 0= ( ) ( )X x Y 0 0= ⇒ ( )Y 0 0=

y M= ( )u x,M 0= ( ) ( )X x Y M 0= ⇒ ( )Y M 0=

Y Y 0µ′′ + =

( )Y 0 0= Sturm-LiouvilleTheorem (T.13)

⇒ 2nµ λ= n

nMπλ = n 1,2,...= eigenvalues

( )Y M 0= ( )nY y eigenfunctions

( ) ( )n nX x sinh x Lλ = − ,...3,2,1n =

u( x, y ) ( ) ( )

( )

( )[ ] ( )

M

nn0

nMn 1 2 n x 0

n0

f y Y y dyX x

Y yX

Y y dy

∞

= =

=

∫∑

∫

( ) ( )

( )

M

n0

n 1 n

nf y sin y dy sinh x LM nsin yM sinh L M2

πλ π

λ

∞

=

− = −

∫∑

( )( )

( )M

n 1 n0

nsinh x L2 n nMf y sin y dy sin yM M sinh L M

ππ π

λ

∞

=

− = − − ∑ ∫

Let ( )3f y 1= , then ( )nMM

00

1 1n n cos n Msin ydy cos y M MM M n n nπ π π

π π π− −

= − = − + = −∫

( ) ( )n

n 1

nsinh L x1 12 nMu( x, y ) sin ynM n MsinhM

ππ

ππ

∞

=

− − − =

∑

( )3f y

0

0

0


Visualization of the solution can be performed in the following ways: Observations: 1. The solution is in the form of an infinite series. It represents the function from the

functional vector space spanned by the obtained eigenfunctions ( )y,xun (basis).

2. The eigenvalue problem form of the equation for this case is LY Yµ=

where operator L is defined by equation ( )LY Y ′′≡ −

3. Assumption of separation of variables was used to obtain the basic functions (eigenfunctions). But the obtained solution is not an approximation – it is an exact solution (this fact follows from the uniqueness of the solution of the Dirichlet problem for Laplace’s Equation). The same result may be obtained by the other methods without separation of variables (for example, using the finite integral transform).

4. Solution of the example problem may be treated:

as a stationary temperature field in the rectangular domain with a fixed temperature at the boundaries;

as the equilibrium shape of a membrane stretched on the fixed frame;

as the solution of the problem from differential geometry on optimization: find the surface with fixed boundaries, which has the minimal area.

5. The Maximum Principle for Laplace’s Equation:

Solution of the Dirichlet Problem for Laplace’s Equation attains its extreme values only at the boundaries of the domain:

( ){ } ( )

( ){ }1 2 3 4 1 2 3 4x,y S x ,y S

min f , f , f , f u x, y max f , f , f , f∈ ∈

≤ ≤

6. Solution of the Dirichlet Problem for Laplace’s Equation is unique.

( )u x, y ( )u x, y

stationary distribution of the density field(diffusion of white particles)

contour plot(isotherms)

surface with level curves(isotherms)

Temperature field(minimal surface with fixed boundaries)


581

4.3. Solution of the non-basic case: more than one of the boundary conditions are non-homogeneous Superposition Principle: If all boundary conditions for the Laplace Equation are non-

homogeneous then the problem can be subdivided into four supplemental basic problems for functions 1u , 2u , 3u , and 4u in such a way that for each of these functions the BVP includes only one non-homogeneous boundary condition.

After that the superposition principle for the linear equation can be used to obtain the solution:

( ) ( ) ( ) ( ) ( )y,xuy,xuy,xuy,xuy,xu 4321 +++=

The complete solution of the Laplace Equation with all non-homogeneous equations is given below. Moreover, the case of the non-homogeneous equation (called the Poisson Equation) is also solved with application of the superposition principle with the help of a supplemental problem for non-homogeneous equations with all homogeneous boundary conditions.

superposition ofBASIC CASES

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Example – Non-homogeneous boundary conditions (superposition principle) – Dirichlet Problem ( )xf2 M

( )yf3 0u2 =∇ ( )yf4 0 L

( )xf1 Supplemental problems: ↓ 0 ( )xf2 0 0 0 0u1

2 =∇ 0 0 0u22 =∇ 0 ( )yf3 0u3

2 =∇ 0 0 0u42 =∇ ( )yf4

( )xf1 0 0 0

Solution of supplemental basic problems:

( ) ( )1 nn 1

n nu x, y a sin x sinh y ML Lπ π∞

=

= − ∑

( )L

10

n

2 nf x sin x dxL L

ansinh ML

π

π

− =

∫

( )2 nn 1

n nu x, y b sin x sinh yL Lπ π∞

=

=

∑ ( )

L

20

n

2 nf x sin x dxL L

bnsinh ML

π

π

=

∫

( ) ( )3 nn 1

n nu x, y c sinh x L sin yM Mπ π∞

=

= − ∑

( )M

30

n

2 nf y sin y dyM M

cnsinh LM

π

π

− =

∫

( )4 nn 1

n nu x, y d sinh x sin yM Mπ π∞

=

=

∑ ( )

M

40

n

2 nf y sin y dyM M

dnsinh LM

π

π

=

∫

Solution of Dirichlet problem:



583

Non-homogeneous equation Poisson’s Equation – Dirichlet Problem ( )xf2 M

( )yf3 ( )2u F x, y 0∇ + = ( )yf4 0 L ( )xf1

Supplemental problems: ↓ ( )xf2 0 ( )yf3 0u5

2 =∇ ( )yf4 0 26u F 0∇ + = 0

( )xf1 0

Solution of supplemental problems:

Solution of Dirichlet problem (Laplace’s homogeneous equation):


Solution of Poisson’s equation with homogeneous boundary conditions

( )6 nmn 1 m 1

n mu x, y A sin x sin yL Mπ π∞ ∞

= =

=

∑ ∑

( )M L

nm 2 20 02

2 2

4 n mA F x, y sin x sin y dxdyL Mn mLM

L M

π π

π

= +

∫ ∫

Solution of Poisson’s Equation (superposition principle): ( ) ( ) ( )y,xuy,xuy,xu 65 +=


4.4. Solution of Poisson’s Equation – homogeneous boundary conditions – Eigenfunction Expansion Method ( )2u F x, y 0∇ + = ( )PE

x 0= [ ]x 0u 0=

=

x L= [ ]x Lu 0=

=

y 0= [ ]y 0u 0=

=

y M= [ ]y Mu 0=

=

Supplemental Eigenvalue Problems with boundary conditions corresponding to boundary conditions of ( )PE :

′′ − =X X 0µ ′′ = − 2n n nX Xλ ′′ − =Y Y 0η ′′ = − 2

m m mY Yν

[ ]x 0X 0=

= ⇒

Sturm-Liouville Theorem

= − 2nη λ =n 1,2,... [ ]y 0Y 0

== ⇒

Sturm-Liouville Theorem

= − 2mη ν =m 1,2,...

[ ]x LX 0=

= ( )nX x [ ]y MY 0=

= ( )mY y

( ) ( )( ) =n kX x ,X x 0 if ≠n k ( ) ( )( ) =m kY y ,Y y 0 if ≠m k

Assume the solution of the BVP ( )PE in the form of double series expansion over the eigenfunctions:

( )u x, y ( ) ( )∞ ∞

= =

= ∑ ∑ nm n mn 1 m 1

A X x Y y and substitute it into Poisson’s Equation ( )PE

( ) ( ) ( ) ( ) ( )∞ ∞ ∞ ∞

= = = =

∂ ∂ + + = ∂ ∂ ∑ ∑ ∑ ∑

2 2

nm n m nm n m2 2n 1 m 1 n 1 m 1

A X x Y y A X x Y y F x,y 0x y

( ) ( ) ( ) ( ) ( )nm n m nm n mn 1 m 1 n 1 m 1

A X x Y y A X x Y y F x, y∞ ∞ ∞ ∞

= = = =

′′′′ + = −∑ ∑ ∑ ∑

( ) ( ) ( ) ( ) ( )2 2nm n n m nm n m m

n 1 m 1 n 1 m 1A X x Y y A X x Y y F x, yλ ν

∞ ∞ ∞ ∞

= = = =

− − = −∑ ∑ ∑ ∑

( ) ( ) ( ) ( )2 2nm n m n m

n 1 m 1A X x Y y F x, yλ ν

∞ ∞

= =

+ =∑ ∑

( ) ( ) ( ) ( )2 2nm n m m n

n 1 m 1A Y y X x F x, yλ ν

∞ ∞

= =

+ = ∑ ∑ ⇒ ( ) ( )

( ) ( )

( )

∞

=

+ =∫

∑

L

n2 2 0

nm n m m 2m 1 n

F x,y X x dxA Y y

X xλ ν

( )( ) ( ) ( )

( ) ( )+ =

∫ ∫M L

n m2 2 0 0

nm n m 2 2n m

F x,y X x Y y dxdyA

X x Y yλ ν

( )u x, y ( ) ( )∞ ∞

= =

= ∑ ∑ nm n mn 1 m 1

A X x Y y ( ) ( ) ( )

( ) ( ) ( )

M L

n m0 0

nm 2 2 2 2n m n m

F x, y X x Y y dx dyA

X x Y y λ ν=

+

∫ ∫

x

[ ]x Lu 0=

=[ ]x 0u 0=

=

y

M

L0

[ ]y Mu 0=

=

[ ]y 0u 0=

=

( )2u F x, y∇ =

eigenfunctionexpansion method


585

L

[ ] ( )3x 0u f y=

=

M

y

x0

[ ] ( )4x Lu f y=

=

[ ] ( )2y Mu f x=

=

[ ] ( )1y 0u f x=

=

2u F 0∇ + =

POISSON'S EQUATION

Solution of Basic Caseof Poisson's Eqn(homogeneous b.c.'s)

Solution of Basic Casesof Laplace's Equationwith one non-homegeneous b.c.

( )Separation of Variables SV

( )Eigenvalue Expansion EE

( )Superposition Principle SP

( )2 2

2 2

u u F x, y 0x y

∂ ∂+ + =

∂ ∂( ) ( ) ( ) 2x, y 0,L 0,M∈ × ⊂

21u 0∇ =0

0

0

1f

22u 0∇ =0

2f

0

0

23u 0∇ =3f

0

0

0

24u 0∇ =0

0

4f

0

26u F∇ = −0

0

0

0

25u 0∇ =3f

2f

4f

1f

5 1 2 3 4u u u u u= + + + ( )Superposition Principle SP

( )6 nm n mn m

u x, y A X Y= ∑∑

( ) ( )11 n n n

nu x, y a X Y= ∑ ( ) ( )

L

1 n0

n 21n n

f X dx1a

Y 0 X=

∫nX

( )1nY

[ ]Su f=st nd rd

brackets mean that conditionat the boundary can be either of I , II or III kind

( ) ( )13 m m m

mu x, y c X Y= ∑ ( ) ( )

M

3 m0

m 21m m

f Y dy1c

X 0 Y=

∫( )1mX

mY

( ) ( )22 n n n

nu x, y b X Y= ∑ ( ) ( )

L

2 n0

n 22n n

f X dx1b

Y M X=

∫nX

( )2nY

( ) ( )24 m m m

mu x, y d X Y= ∑ ( ) ( )

M

4 m0

m 22m m

f Y dy1d

X L Y=

∫( )2mX

mY

LAPLACE'S EQUATION

( )

M L

n m0 0

mn 2 22 2n m n m

FX Y dxdyA

X Yλ ν=

+

∫ ∫

Solution of Poisson's Equation with non-homogeneous b.c.'s

(homogeneous eqn,non-homogeneous boundary conditions)

nX

mY

Z Z 0γ′′ − =

2k kγ ω= −

2k k kZ Zω′′ = −

( )kZ z

[ ]z 0Z 0=

=

[ ]z KZ 0=

=

SLP⇒

Y Y 0η′′ − =

2m mη ν= −

2m m mY Yν′′ = −

( )mY y

[ ]y 0Y 0=

=

[ ]y MY 0=

=

SLP⇒

Supplemental Sturm-Liouville Problems

X X 0µ′′ − =

2n nµ λ= −

2n n nX Xλ′′ = −

( )nX x

[ ]x 0X 0=

=

[ ]x LX 0=

=

SLP⇒

( ) 5 6u x, y u u= +


587

VIII.2.4.5 Laplace Equation in Cylindrical Coordinates

Laplacian in cylindrical coordinates:

2 2

22 2 2

1 u 1 u uu rr r r r zθ

∂ ∂ ∂ ∂ ∇ ≡ + + ∂ ∂ ∂ ∂

2 2 2

2 2 2 2

u 1 u 1 u ur rr r zθ

∂ ∂ ∂ ∂= + + +

∂∂ ∂ ∂

The Steady State Heat Equation with volumetric heat generation g :

2 2

2 2 2

1 u 1 u u gr 0r r r kr zθ

∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂

2 2 2

2 2 2 2

u 1 u 1 u u g 0r r kr r zθ

∂ ∂ ∂ ∂+ + + + =

∂∂ ∂ ∂

1-D Long cylinder with angular symmetry ( )u r

Long cylindrical surface with angual symmetry ( )u θ

Elongated solid (fins), flow in pipes ( )u z

2-D Long cylinder u 0z

∂=

∂ ( )u r,θ

Cylinder with angular symmetry u 0θ

∂=

∂ ( )u r,z

Domain: circular segment (cake slice) annular outer

r

0

2r1r

r

1r

0

r

z

y

x

r

0θ

z( )r, ,zθ


Short Cylinder with Angular Symmetry The Heat Equation – Poisson’s equation

2 gu 0k

∇ + =

( )2

2

g r,z1 u ur 0r r r kz

∂ ∂ ∂ + + = ∂ ∂ ∂ 10 r r≤ < 0 z L< <

Boundary conditions: z 0= [ ] ( )1u f r= 10 r r≤ < z L= [ ] ( )2u f r= 10 r r≤ < 1r r= [ ] ( )3u f z= 0 z L< < r 0= ( )u 0,z < ∞ 0 z L< < Basic cases for LE: Basic Case for PE: Superposition Principle: ( ) 1 2 3 4u r,z u u u u= + + +

r

z

1r

z 0=

z L=

( )r, ,zθ

21u 0∇ = [ ]

11 r ru 0

==

[ ]1 1z 0u f

==

[ ]1 z Lu 0

==

22u 0∇ = [ ]

12 r ru 0

==

[ ]2 z 0u 0

==

[ ]2 2z Lu f

==

23u 0∇ =

13 3r ru f

==

3 z 0u 0

==

3 z Lu 0

==

24

gu 0k

∇ + =14 r r

u 0=

=

4 z 0u 0

==

4 z Lu 0

==


589

Separation of Variables: ( ) ( ) ( )u r,z R r Z z=

Laplace’s equation: 2 2

2 2

u 1 u u 0r rr z

∂ ∂ ∂+ + =

∂∂ ∂

1R Z R Z Z R 0r

′′ ′ ′′+ + =

R 1 R Z 0R r R Z′′ ′ ′′

+ + = separable equation

Example (Exercises, Problem #5)

2 2

2 2

u 1 u u 0r rr z

∂ ∂ ∂+ + =

∂∂ ∂ 10 r r≤ < 0 z L< <

Boundary conditions: z 0= [ ]u 0= 10 r r≤ < z L= [ ] ( )2u f r= 10 r r≤ < 1r r= [ ]u 0= 0 z L< < r 0= ( )u 0,z < ∞ 0 z L< < finite Homogeneous boundary conditions: z 0= [ ] ( ) ( )z 0u R r Z 0 0

== = ⇒ ( )Z 0 0=

1r r= [ ] ( ) ( )

11r ru R r Z z 0

== = ⇒ ( )1R r 0=

r 0= ( ) ( ) ( )u 0,z R 0 Z z= < ∞ ⇒ ( )R 0 < ∞

Separate variables: R 1 R ZR r R Z

µ′′ ′ ′′

+ = − =

Equation for R : R 1 RR r R

µ′′ ′

+ =

Write in self-adjoint form: rR R rR 0µ′′ ′+ − =

( ) ( )rR 0 r R 0µ′′ + + − = SLT⇒ ( )2 , p r rµ λ= − =

Rewrite as the Bessel Equation

′′ ′+ + − =

2v2 2 2r R rR r 0 R 0λ BE of order 0

General solution (Table 18, Bessel Functions): ( ) ( ) ( )1 0 2 0R r c J r c Y rλ λ= +

( )R 0 < ∞ ⇒ 2c 0=

( ) ( )1 1 0 1R r 0 c J rλ= = ⇒ ( )0 1J r 0λ =

22u 0∇ = [ ]

12 r ru 0

==

[ ]2 z 0u 0

==

[ ]2 2z Lu f

==


Solution of SLP: ( )2 2 2r R rR r 0 R 0λ′′ ′+ + − =

( )1R r 0=

( )R 0 < ∞ Eigenfunctions: ( ) ( )n 0 nR r J rλ= Eigenvalues: nλ are positive roots of characteristic equation: ( )0 1J r 0λ =

Norm: ( ) ( ) ( )1r 2

2 2 21n 0 n 1 n 1

0

rR r J r rdr J r2

λ λ= =∫

Equation for Z ZZ

µ′′

− =

2n

ZZ

λ′′

− = −

2

nZ Z 0λ′′ − = ( )Z 0 0= ( ) ( ) ( )1 n 2 nZ z c cosh z c sinh zλ λ= + ( ) ( ) ( )1 2Z 0 0 c cosh 0 c sinh 0= = + 1c= ( ) ( )n nZ z sinh zλ=

Solution in the form of series: ( ) ( ) ( )n n nn 1

u r,z c R r Z z∞

=

= ∑

( ) ( ) ( )n 0 n nn 1

u r,z c J r sinh zλ λ∞

=

= ∑

B.c. at z L= [ ] ( )2u f r= ( ) ( )n 0 n nn 1

c J r sinh Lλ λ∞

=

= ∑

Fourier-Bessel Series: ( )2f r ( ) ( )n n 0 nn 1

Fourier coefficients

c sinh L J rλ λ∞

=

= ∑

( )n nc sinh Lλ ( ) ( )

( )

1

1

r

3 0 n0

r20 n

0

f r J r rdr

J r rdr

λ

λ=

∫

∫

( ) ( )

( )

1r

3 0 n0

2211 n 1

f r J r rdr

r J r2

λ

λ=

∫

Solution: ( ) ( ) ( )n 0 n nn 1

u r,z c J r sinh zλ λ∞

=

= ∑ , nc( ) ( )

( ) ( )

1r

3 0 n0

2 21 1 n 1 n

2 f r J r rdr

r J r sinh L

λ

λ λ=

∫


591

5-LE-CYL.mws 5.5 SS conduction in cylindrical coordinates Example LE > restart; > with(plots): > r1:=2;z1:=5;

:= r1 2

:= z1 5

> f(r):=(r-r1)^2; := ( )f r ( ) − r 2 2

> plot({f(r)},r=0..r1,axes=boxed,filled=true);

Characteristic Equation: > w(x):=BesselJ(0,x*r1);

:= ( )w x ( )BesselJ ,0 2 x

> plot(w(x),x=0..5);

Eigenvalues: > n:=1: for m from 1 to 30 do z:=fsolve(w(x)=0,x=m*(1)..(m+1)*(1)): if type(z,float) then lambda[n]:=z: n:=n+1 fi od: > for i to 4 do lambda[i] od;

1.202412779

2.760039055

4.326863956

5.895767220

> N:=n-1; := N 19

> n:='n':m:='m':z:='z': Eigenfunctions: > R[n]:=BesselJ(0,lambda[n]*r);

:= Rn ( )BesselJ ,0 λn r

squared norm: > NR2[n]:=int(R[n]^2*r,r=0..r1);

:= NR2n

+ 2 π λn ( )BesselJ ,0 2 λn2 2 π λn ( )BesselJ ,1 2 λn

2

π λn

> NR2[n]:=r1^2/2*BesselJ(1,lambda[n]*r1)^2; := NR2n 2 ( )BesselJ ,1 2 λn

2

Fourier Coefficients: > a[n]:=int(f(r)*R[n]*r,r=0..r1)/NR2[n]/sinh(lambda[n]*z1): Solution: > u(r,z):=sum(a[n]*R[n]*sinh(lambda[n]*z),n=1..12): > plot3d(u(r,z),r=0..r1,z=0..z1,axes=boxed);


> un(r,z):=subs(r=-r,u(r,z)): > u0:=subs(z=0,u(r,z)):n0:=subs(r=-r,u0): > u1:=subs(z=1,u(r,z)):n1:=subs(r=-r,u0): > u2:=subs(z=2,u(r,z)):n2:=subs(r=-r,u0): > u3:=subs(z=3,u(r,z)):n3:=subs(r=-r,u0): > u4:=subs(z=4,u(r,z)):n4:=subs(r=-r,u0): > u5:=subs(z=5,u(r,z)):n5:=subs(r=-r,u0): >cylinderplot({[r,theta,u5/1+5],[r,theta,u4/1+4],[r,theta,u3/1+3],[r,theta,u2/1+2],[r,theta,u1/1+1],[r,theta,u0/1+0]},r=0..r1,theta=0..2*Pi);

plot({0,u0,n0,1,u1+1,n1+1,2,u2+2,n2+2,3,u3+3,n3+3,4,u4+4,n4+4,5,u5+5,n5+5,5,u5+5,n5+5} ,r=-r1..r1,color=black,view=[-r1..r1,0..9]);

( )u r,z

=z 5

=z 4

=z 3

=z 2

=z 1

=z 0

=z 5

=z 4

=z 3

=z 2

=z 1

=z 0

( )u r,z


593

> plot({u0,n0,u1,n1,u2,n2,u3,n3,u4,n4,u5,n5,u5,n5},r=-r1..r1, axes=boxed,color=black,view=[-r1..r1,0..4]);

> animate({un(r,z)},r=-r1..r1,z=0..z1,frames=200,axes=boxed,numpoints=200);


Example Long Cylinder (Tree Trunk)

Laplace’s Equation:

2 2

2 2 2

u 1 u 1 u 0r rr r θ

∂ ∂ ∂+ + =

∂∂ ∂ 10 r r≤ < , πθ 20 ≤≤

Boundary condition: ( ) ( )1 1u r , uθ θ= πθ 20 ≤≤ ( )u 0,θ < ∞ bounded solution Periodicity: ( ) ( )u r, 2 u r,θ π θ+ = 10 r r≤ < Separation of variables We assume that function ( )θ,rus can be represented as a product of

two functions each of a single variable ( ) ( ) ( )θΘθ rR,rus =

Substitution into Laplace’s equation leads to

0Rr1R

r1R 2 =′′+′+′′ ΘΘΘ

Multiplying equation by 2r and separating the variables, we receive:

µΘΘ

=′′

−=′

+′′

RRr

RRr 2

Solution Consider first the equation

µΘΘ

=′′

−

0=+′′ ΘµΘ

Solution of this equation can be

( )

21 2

1 22

1 2

c cosh c sinh 0 c c 0 c cos c sin 0

λθ λθ µ λΘ θ θ µ

λθ λθ µ λ

+ = − <

= + = + = >

According to the geometry of the problem, the function ( )θΘ is a periodic function with a period p 2π= . Any function which is periodic with the fraction of 2π

n2pnπ

= n 1,2,...=

is also 2π periodic. A constant function also is p 2π= periodic.

n2 2p

nπ πλ

= = ⇒ n nλ = n 1,2,...=

r

1r

0

( )1u θ

x

θ

( )u r,θ


595

We can see, that in the solution, such functions and corresponding to them values of separation constant can be only:

0 0µ = ( )0 2cΘ θ =

2 2

n n nµ λ= = ( )n n nc cos n b sin nΘ θ θ θ= + With the determined eigenvalues, equation for ( )rR becomes:

0 0µ = 2r R rR 0′′ ′+ = ( ) 1 2R r c c ln r= + ⇒ 2c 0= bounded ( )0R r 1=

2 2n n nµ λ= = 0RnRrRr 22 =+′+′′

which is a Cauchy-Euler equation with a general solution

( ) n2

n1 rcrcrR −+=

For positive n , the second term in this solution approaches infinity when r goes to 0 . But 0r = is an interior point in our domain where a solution from the physical considerations should be finite. Therefore, we have to assume coefficient 2c to be equal zero, and the solutions becomes ( ) n

nR r r= We construct now a steady state solution in a form of the series:

( ) ( )n0 n n

n 1u r, c r c cos n d sin nθ θ θ

∞

=

= + +∑

Coefficients are determined from the boundary condition:

( ) ( ) ( )n n1 0 n 1 n 1 1

n 1u r , c c r cos n b r sin n uθ θ θ θ

∞

=

= + + =∑

which, if it is treated as a standard Fourier series expansion of the function ( )θ0u over the interval ( )0,2π , are

( )∫=π

θθπ

2

000 du

21c

( )2

nn 1 1

0

1c r u cos n dπ

θ θ θπ

= ∫ ⇒ ( )2

n 1n01

1c u cos n dr

π

θ θ θπ

= ∫

( )2

nn 1 1

0

1b r u sin n dπ

θ θ θπ

= ∫ ⇒ ( )2

n 1n01

1b u sin n dr

π

θ θ θπ

= ∫

Solution: ( ) ( )n0 n n

n 1u r, c r c cos n d sin nθ θ θ

∞

=

= + +∑


> r1:=0.5; := r1 0.5

> u1:=Heaviside(t-5*Pi/4); := u1

Heaviside − t 5 π

4

> plot(u1,t=0..2*Pi);

Fourier Coefficients: > c[0]:=int(u1,t=0..2*Pi)/Pi/2.0;

:= c0 0.3750000000

> c[n]:=int(u1*cos(n*t),t=0..2*Pi)/Pi/r1^n;

:= cn

− 2 ( )sin π n ( )cos π n

sin 5 π n

4n π 0.5n

> b[n]:=int(u1*sin(n*t),t=0..2*Pi)/Pi/r1^n;

:= bn − − − 2 ( )cos π n 2 1

cos 5 π n

4n π 0.5n

Solution: > u:=c[0]+sum((c[n]*cos(n*t)+b[n]*sin(n*t))*(r^n),n=1..15): > cylinderplot([r,t,u],r=0..r1,t=0..2*Pi,axes=boxed,grid=[50,50]);

> uu1:=subs(r=r1,u): > plot({u1,uu1},t=0..2*Pi);

> spacecurve({[r1*cos(t),r1*sin(t),uu1],[r1*cos(t),r1*sin(t),u1], [r1*cos(t),r1*sin(t),0]},t=0..2*Pi,axes=boxed);


597


REVIEW QUESTIONS 1. What is the main assumption in the method of separation of variables? 2. What is a separation constant?

3. How does the Sturm-Liouville Problem appear in the method of separation of variables?

4. What is the form of the solution of the BVP in the method of separation of variables?

5. How the methods of separation of variables and the eigenfunction expansion methods are related?

EXERCISES 1. 1) Find the solution of the Laplace Equation

2 2

2 2

u u 0x y

∂ ∂+ =

∂ ∂

in the domain ( ) ( )D 0,L 0,M= × subject to boundary conditions:

x 0

u 0x =

∂=

∂,

x Lu 0

== , ( )1y 0

u f x=

= ,y M

u 0y

=

∂=

∂

2) Sketch the graph of solution for L 2= , M 3= and ( )1f x x( L x )= −

3) What is the solution for ( ) xf x cosL

π= ? Sketch the graph.

2. 1) Find the solution of the Laplace Equation

2 2

2 2

u u 0x y

∂ ∂+ =

∂ ∂

in the domain ( ) ( )D 0,L 0,= × ∞ subject to boundary conditions:

x 0 x L

u u 0x x= =

∂ ∂= =

∂ ∂ and ( )y 0

u f x=

= , y

u→∞

< ∞

2) Find the solution and sketch the graph for L 5= and a) ( )f x 10=

b) ( )f x sin x2Lπ

=

x

[ ]x Lu 0=

=[ ]x 0u 0=

′ =

y

M

L0

[ ]y Mu 0=

′ =

[ ] ( )1y 0u f x=

=

2u 0∇ =


599

3. 1) Find the solution of the Poisson Equation

( )2 2

2 2

u u F x, yx y

∂ ∂+ =

∂ ∂


x 0

u 0x =

∂=

∂,

x L

u 0x =

∂=

∂, ( )y 0

u f x=

= , y M

u 0=

=

2) Sketch the graph of solution for L 4= , M 2= and a) ( )f x 5= and ( ) ( )( )F x, y xy L x M y= − −

b) ( )f x x= and ( )F x, y of your choice

4. Catalytic Reactor. Steady State Solution.

a) Find the solution of the Poisson Equation

( )⊗∂ ∂

+ + ⋅ − =∂ ∂

2 2

2 2u u S u T 0

x y


x 0

u 0x =

∂=

∂,

=

∂=

∂ x L

u 0x =

=y 0

u 1 , =

=y M

u 0

( )⊗= ⋅ −c cq S u T Volumetric rate of energy produced by chemical reaction cq is in general a complicated function of temperature, pressure, composition, and catalyst activity. For simplicity, it can be approximated as a linear function of temperature

=c 3WS 10

m K, ⊗ = oT 20 C empirical constants for the given reactor (T⊗ can be

0T T⊗ ≤ for energy sink, 0T T⊗ ≥ for energy source)

b) Sketch the graph of solution for L 4= , M 2= and S 10= , T 20⊗ = c) Find the solution of the Poisson Equation

( )2 2

2 2

u u uv S u T 0xx y ⊗

∂ ∂ ∂+ + + − =

∂∂ ∂


x 0

u 0x =

∂=

∂,

x L

u 0x =

∂=

∂,

y 0u 1

== ,

y Mu 0

==

d) Sketch the graph of solution for L 4= , M 2= , S 10= , T 20⊗ = , v 1=

x

=

∂ = ∂ x 0

u 0x

y

M

L0

( ) =u x,M 0

2u 0∇ ==

∂ = ∂ x L

u 0x

( ) =u x,0 1


5. 1) Solve the Laplace Equation in the cylindrical domain

0zu

ru

r1

ru

2

2

2

2

=∂∂

+∂∂

+∂∂ ( )z,ru : 1rr0 <≤ , 1zz0 <<

Boundary conditions: ( ) 0z,ru 1 = 1zz0 << ( ) 00,ru = 1rr0 <≤ ( ) ( )rfz,ru 1 = 1rr0 <≤ 2) Display some creativity in visualization of the solution for

( ) ( )

1

12

1

r 2z 5

f r r r

==

= −

6. 1) Solve the Laplace Equation in the cylindrical domain

0zu

ru

r1

ru

2

2

2

2

=∂∂

+∂∂

+∂∂ ( )z,ru : 1rr0 <≤ , 1zz0 <<

Boundary conditions: ( )1u r ,z 2= 1zz0 <<

( )u r,0 0= 1rr0 <≤

( )1u r,z 3= 1rr0 <≤ 2) Display some creativity in visualization of the solution for

1

1

r 4z 10

==

7. 1) Short Cylinder with angular symmetry, ∂=

∂u 0θ

,

with prescribed temperature on the lateral surface ∇ =2u 0

∂ ∂ ∂ + = ∂ ∂ ∂

2

21 u ur 0r r r z

10 r r≤ < 0 z L< <

Boundary conditions: z 0= [ ] =u 0 10 r r≤ < z L= [ ] =u 0 10 r r≤ < 1r r= [ ] ( )3u f z= 0 z L< < r 0= ( )u 0,z < ∞ 0 z L< < finite 2) Display some creativity in visualization of the solution for

( ) ( )≤ <= = = = − ≤ ≤

1

0 0 z 6r 4, L 10, f z H z 6

1 6 z 10

r

z

1r

z 0=

z L=

( ) ( )=1u r ,z f z


601

4.6 Laplace Equation in Spherical Coordinates

ALLEGORY OF GEOMETRY

(Museum of Louvre)


Universite Rene Descartes, Paris

viii.2 method of separation of variables – stationary ...vps/me505/iem/08 02.pdf · chapter viii...

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