Announcements

Lecture 10

Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-31

Math objects made using MathType; graphs made using Winplot.

Announcements

Exam:

• 20 questions

• 4 SA (6 marks each), including 1 proof • 16 MC (3 marks each)

Highlights:

• Integration • PDE

• More weight on new stuff

• Surface integrals (especially Stewart’s stuff) No calculator, formula sheet

Chapters: 12.1 – 12.4, 9.1 – 9.17, 13.1 – 13.6 Office hours Friday and Monday! 

3 hours

1 huge test room

Chapter 13.5 – Laplace’s Equation

There’s no time, so there won’t be initial conditions.

Steady-state: doesn’t depend on time

2 2

2 2 0

0 , 0

u v

x y

x a y b

∂ _⋅∂ ₌

∂ ∂

≤ < < <

Boundary Conditions (BCs): two groups – one for boundaries along x-axis; one for boundaries along y-axis

Insulation:

0

0

x x a

u u

x ₌ x ₌

∂ _{= =}∂

∂ ∂

( )

, 0 0,

( )

,

( )

, 0

u x = u x b = f x < <x a

( )

,

( ) ( )

u x y = X x Y y

2 2 2

2

0 u

X Y x

u XY y

X Y

XY X Y

X Y λ

∂ _{= ⋅} ∂

∂ _{= ′′} ∂

′′ ′′ ′′+ ′′ = ⇔ = − = −

*Breaks into 2 ODE’s* 0

X′′ +λX =

( )

0 0

( )

X′ = =X a′

2

2

0 0 0 λ

λ α

λ α = = − < = >

0 Y′′ −λY =

Note the negative!

( )

0 0

Y =

(use homogeneous BCs)

3 cases:

i)λ =0,X x

( )

=1

ii) 2

0 λ= −α <

gives trivial solution iii) 2

0 λ α= >

Gives non-trivial solution Homogeneous BCs

Eigenvalues:

2

, 1, 2, 3,... n

n n a

π

λ =_ _ =

 

Eigenfunctions: cos n x ,n 1, 2, 3,.. a

π

 _{ =}

 

 

( )

0, 0 0 Y′′ −λY = Y =

λ = 0

( )

1 2

0 Y

c y c Y y ′′ =

⇒ + =

Plug in BCs

( )

0 1

( )

0 2 0 2 0

Y =c + = ⇒c c =

( ) (

constant

)

Y y = y

( )

2

2

, 1, 2, 3,...

0, 0 0 n

n a n

Y Y Y

a π λ

π

 

=_{ } =

 

 

′′ −_{ } = =

 

( )

1cosh 2sinh

n n

Y y c y c y

a a

π π

   

= _{ }+ _{ }

   

( )

0 0 1

( )

1 1 0

Y = =c ⇒ =c

So it only contains sinh terms.

( )

2sinh

n

Y y c y

a π

 

= _{ }

 

0 0

0

, 0

, 1

, 0

cos sinh , 1

n

n n

n X Y n u

X Y n

A y n

n x n y

A n

a a

π π

= 

=  _≥



= 



=      _≥

   

 _{   }



Although the solutions satisfy PDE and homogeneous BCs, we still need to satisfy the inhomogeneous BCs. How? Fourier Series.

Inhomogeneous BCs

( )

0

1

, _ncos sinh

n

n x n y

u x y A y A

a a

π π

∞ =

   

= + _{   }

   

∑

Apply inhomogeneous BCs: (plug in y = b)

( )

( )

0

1

, _ncos sinh

n

n x n b

f x u x b A b A

a a

π π

∞ =

   

= = + _{   }

   

∑

Now, make this a Fourier series. This currently looks like the half-range expansion of f in a cosine series.

( )

( )

0

0 0

0

0

2

, d

2

2

sinh , cos d

a

a

n n n

a

A b a f x x

a

n b n x

A a a f x x

a a a

π π

= =

 _{= =}  

   

   

∫

∫

( )

( )

( )

0 ₀

0

0 1

1

d 2

cos d

sinh

, cos sinh ,

a

a n

n n

A f x x

ab

n x

A f x x

n b a

a

a

n x n y

u x y A y A

a a

π π

π π

∞ =

⇔ =

 

= _{ }

   

 

 

   

= + _{   }

   

∫

∫

∑

Where A0, An, are as above.

Dirichlet Problem

A boundary-value problem, where we seek a solution to an elliptic PDE (such as Laplace’s) within a region, R, such that u takes on prescribed values on the entire boundary of R is called the Dirichlet Problem.

The Dirichlet Problem can be solved on a rectangle via separation of variables, when

homogeneous BCs are given on two parallel boundaries. However, separation of variables will not work when the BCs on all four sides are inhomogeneous.

You cannot directly use separation variables. The approach is to use superposition.

( )

( ) (

)

( )

( )

( ) ( )

( )

2 2

2 2 0, 0 , 0

0, , ,

, 0 , ,

u u

x a y b

x y

u y F y u a y G y u x f x u x b g x

∂ ₊∂ _{= < < < <}

∂ ∂

= =

= =

Break the problem into 2 pieces, 2 separate problems.

Problem 1 Problem 2

( )

( )

( )

( ) ( )

( )

2 2

1 1

2 2

1 1

1 1

0

0, 0 0,

, 0 , ,

u u

x y

u y u y

u x f x u x b g x

∂ ₊∂ ₌

∂ ∂

= =

= =

Compared to the previous problem, instead of having 4 inhomogeneous BCs, there are now only 2

( )

( )

(

)

( )

( )

( )

2 2

2 2

2 2

2

2

2 2

0 0,

,

, 0 0 ,

u u

x y

u y F y

u a y G y

u x u x b

∂ ₊∂ ₌

∂ ∂

= = = =

Principle of superposition lets us break up the problem: ifu x y1

( )

, solves problem 1 andu2

( )

x y,

solves problem 2, thenu₁+u₂solves the original problem. Setu= +u₁ u₂to test it.

( )

( )

( )

( )

1 2

0, 0, 0,

0

u y u y u y

F y

= +

= +

( )

( )

( )

( )

1 2

, , ,

0

u x b u x b u x b g x

= +

= +

13.6 – Non-homegeneous BVPs

Warning: difficult 

A BVP is non-homogeneous when either the PDE or the BCs are non-homogeneous. Cannot always perform separation of variables if the problem is inhomogeneous. e.g. 1)

Heat generated in a rod:

2

u u

k r

x t

∂ ∂

+ =

∂ ∂

Not separable 

Use the principle of superposition.

Change the dependent variable, u, foru= +v ψ Case 1

Inhomogeneous term in PDE AND BCs are time-independent.

( )

( )

( )

( )

( )

2 2

0 1

0, , ,

, 0

u u

k F x

x t

u t u u L t u

u x f x

∂ ∂

+ =

∂ ∂

= =

=

( ) ( )

, ,

( )

u x t =v x t +ψ x ←time independent case

Choose v to satisfy the associated homogeneous problem and the ψ to solve another equation that accounts for the inhomogeneous BCs.

( )

2 2

2 2

0

u v

x

x x

u v

t t ψ

∂ ∂ _′′

= +

∂ ∂

∂ ₌∂ ₊

∂ ∂

( )

( )

2 2

v v

k x F x

x ψ t

∂ _{+ ′′ + =}∂

∂ ∂

Separate into 2 equations:

( )

( )

0

kψ ′′ x +F x =

( )

0 u0,

( )

L u1

ψ = ψ =

Solve with integration.

2 2

v v

k

x t

∂ ₌∂

∂ ∂

( )

( )

( )

( )

( )

0, 0 ,

, 0

v t v L t

v x f x ψ x = =

= −

Finish like before.

e.g. 2) Heat stuff:

2 2

u u

k r

x t

∂ ∂

+ =

∂ ∂

Solve subject to following boundary conditions:

( )

( )

( )

0

0, 1,

, 0 0

u t u u t

u x

= =

=

Then,

( )

2 2

v v

k k x r

x ψ t

∂ _{+ ′′ + =}∂

∂ ∂

( )

0 kψ ′′ x + =r

( )

0 u0

( )

1

ψ = =ψ

2 2

v v

k

x t

∂ ₌∂

∂ ∂

( )

( )

( )

( )

, 0

0, 0 1,

v x x

v t v t

ψ = − = = Solve for ψ first, using integration.

( )

( )

( )

1

2

1 2

2 r x

k rx

x c

k rx

x c x c

k ψ

ψ

ψ

− ′′ =

′ = − +

= − + +

Solve for c1, c2, using BCs

( )

( )

2 0

1 0

0 0 0 1

2

c u r

c u k

ψ

ψ

= + + = −

= + + = u₀

( )

(

)

1

2

0

0

2

2 2

1 2

r c

k

r r

x x x u

k k

r

u x x

k ψ

⇒ =

−

= + +

= − −

( )

1

2

2

0

2 1

2 2

r c

k c

r r c

k k

u ψ =

=

= − + +

=

You know u0 is not 0 because it’s homogeneous.

( )

( )

( )

( )

( )(

)

2 2

0

0, 0 1,

, 0 1

2

v v

k

x t

v t v t

r

v x x x x u

k ψ

∂ ∂

=

∂ ∂

= =

= − = − −

Use of separation of variables 0

X′′ +λX = Dirichlet BCs

( )

sin

(

)

,

1, 2, 3,...

X x c n x

n

π

⇒ =

=

( )

( )2 2

0 k n t T k T

T t e π

λ λ − ′ + =

=

(

)

( )

(

)

2 2 2 2 1 sin , sin kn t n n kn t n n

v A e n x

v x t A e n x

π π π π − ∞ − = = =

∑

( )

(

)

(

)

0 1

, 0 1

2

sin n n

r

v x x x u

k

A n xπ

∞ = = − − =

∑

Choose

(

)

(

)

( )

(

)

1 0 0 0 2 2

2 1 sin d

2

2 1 1

n

n r

A x x x n x x

k u r n kn π π π   = _ − − _     = _ + _ − −  

∫

( )

(

)

2 2

(

)

0 1 , 1 sin 2 kn t n n

u x t v

r

u x x A e n x

k π ψ π ∞ − = = + = − − +

∑

When 0

(

( )

)

2 2

2 1 n 1

n

u r

A

nπ kn π

 

= _ + _ − −

 

Ast→ ∞, what happens to u.v x t

( )

, →0(transient solution)

( )

( )

lim ,

t→∞u x t =ψ x ← steady-state solution

Another possibility: Inhomogeneous terms and BCs that depend on time.

( )

( )

( )

( )

( )

2 2 0 1 , 0, , , 0 u t

k F x t

x t

u t u t u L t u t t ∂ _{+ =} ∂ ∂ ∂ = = >

( )

, 0

( )

u x = f x

Want to tryu x t

( ) ( )

, =v x t, +ψ

( )

x t, , where v satisfies associated homogeneous equation.

( )

2 2

2 2 ,

v v

k k F x t

x x t t

ψ ψ

∂ ₊ ∂ _{+ =}∂ ₊∂

∂ ∂ ∂ ∂

Split into ψ pieces and v pieces.

( )

( )

( )

( )

( )

2 2 0 1 , 0, ,

k F x t

x t

t u t L t u t

ψ ψ ψ ψ ∂ _{+ =}∂ ∂ ∂ = =

( )

( )

( )

( )

( )

2 2

0, 0 ,

, 0 , 0

v v

k

x t

v t v L t

v x f x ψ x ∂ ₌∂

∂ ∂

= =

= −

You can’t solve for ψ, so this approach fails. We can’t hope that v will solve a homogeneous BVP.

Instead, take

( )

, 0

( )

1

( )

0

( )

x

x t u t u t u t L

ψ = + _ − _

2

2 0

x ψ ∂

= ∂

Plug this into PDE:

( )

( )

( )

( )

( )

( )

2

2 ,

0, 0 ,

, 0 , 0

v v

k G x t

x t

v t v L t

v x f x ψ x

∂ _{+ =}∂

∂ ∂

= =

= −

( )

,

( )

, _t

G x t =F x t −ψ To solve for v:

1. Form associated homogeneous problem to get

2 2

v v

k

x t

∂ ₌∂

∂ ∂

2. Use separation of variables to get eigenvalues/functions of Sturm-Liouville problem 0

X′′ +λX =

3. Writev x t

( )

, andG x t

( )

, as Fourier series in eigenfunctions you found above. However, coefficients will depend on t.

4. Explicitly find coefficients forG x t

( )

, 5. Use that and v x t

( )

, in PDE in v

6. Equate coefficients to find Fourier series forv x t

( )

, e.g. 3)

Example 2 AEM

Solve:

( )

( ) ( )

( )

2

2 , 0 1, 0

0, cos , 1, 0 , 0 0

u u

x t

x t

u t t u t

u x

∂ ∂

= < < >

∂ ∂

= =

=

( ) ( )

, ,

u=ψ x t +v x t

( )

( )

( )

( )

( )

( )

( ) (

) ( )

0 1 0

,

cos 0 cos

, 1 cos

x

x t u t u t u t L

t x t

x t x t

ψ

ψ

= + _ − _

= + _ − _ = −

(

) ( )

(

)

( )

( )

( )

( )

2 2

2 2

1 sin

1 sinh

0, 0 1,

, 0 1

v v

x t

x t

v v

x t

x t

v t v t

v x x

∂ ₌∂ _{+ −}

∂ ∂

∂ ∂

⇔ + − =

∂ ∂

= = = −

Related homogeneous problem:

( )

( )

( )

( ) ( )

2 2

1

1

1

0, 0 1,

, 0 1

v v

x t

v t v t

v x x

v X x T t ∂ ₌∂

∂ ∂

= = = − =

( )

( )

( )

( )

0

0 0 1

X x X x

X X

λ

′′ + =

= =

=>Eigenvalues:λ_n =n2π2

Eigenfunctions:sin

(

n x nπ

)

, =1, 2, 3,...

( )

,

( )

, t

(

1

) ( )

sin

G x t =F x t −ψ = −x t

( )

( ) (

)

1

, _n sin

n

v x t v t n xπ

∞ =

=

∑

( )

( ) (

)

(

1

)

, sin

1 n n

G x t G t n x

x

π

∞ =

=

= −

∑

Solve for G t_n

( )

.

( )

(

) ( ) (

)

( )

1 0

2 1 sin sin d

2 sin n

G t x t n x x

t n

π

π

= −

=

∫

Therefore,

(

) ( )

( ) (

)

1

2

1 sin sin sin

n

x t t n x

nπ π

∞ =

− =

∑

Want Fourier coefficients ofv t_n

( )

. Get them by plugging Fourier series ofG x t

( )

, and

( )

( ) (

)

1

, _n sin

n

v x t v t n xπ

∞ =

=

∑

in the PDE

( ) (

)

( ) (

)

( ) (

)

2 2

1 1

2

sin sin sin sin

n n

n n

n v t n x t n x v t n x

n

π π π π

π

∞ ∞

= =

′

− + =

∑

∑

∑

Equate coefficients

( )

( )

( )

( )

( )

( )

2 2 2 sin 2 2 2 sin

n n n n

t t

n v t v t v t n v t

n n

π π

π ′ ′ π

− + = ⇔ + =

Using 2Z03’s undetermined coefficients, you obtain

( )

( )

( )

2 2

2 2

4 4

sin cos 2

1

n t

n n

n t t

v t c e

n n

π

π

π π

−

 − 

=  +

+

 

( )

₍

( )

₎

( )

2 2

(

)

2 2

4 4 1

sin cos

, 2 sin

1

n t n n

n t t

v x t c e n x

n n

π

π

π

π π

∞

− =

 ₋ 

 

= +

+

 

 

∑

Solve for cn by applying initial condition:

4 4

2 2

1 n

c

n

nπ n π π

= −

 + 

 

( )

( )

₍

₎

(

)

2 2 2 2

2 2

4 4 1

sin cos 2

, sin

1

t n t n t n

n t e e

v x t n x

n n n

π π

π

π

π π

− − ∞

=

 _{− +} 

 

= −

+

 

 

∑

( )

( ) ( )

(

) ( ) ( )

, , ,

1 cos ,

u x t x t v x t x t v x t ψ

= +

= − +

( )

, 0

( )

1

( )

0

( )

x

x t u t u t u t L

ψ = + _ − _

( )

( )

( )

2

2 , , , , t

v v

k G x t G x t F x t

x t ψ

∂ _{+ =}∂ _{= −}

∂ ∂

( )

( )

( )

( )

( )

0, 0 ,

, 0 , 0

v t v L t

v x f x ψ x = =

= −

Form associated homogeneous BVP:

( )

( )

( )

( )

( )

2

1 1

2

1 0, 0 1 ,

, 0 , 0

v v

k

x t

v t v L t

v x f x ψ x ∂ ₌∂

∂ ∂

= =

= −

Solve associated Sturn-Liouville problem: 0

X′′ +λX =

BCs in X get eigenfunctions

Then, assume v x t

( )

, andG x t

( )

, may be written as Fourier series in these eigenfunctions with coefficients that depend upon t.

Solve for Fourier Coefficients of G.

Plug G, v’s Fourier expansions into PDE, equate coefficients. This usually results in 1st order ODE in t. Yields Fourier coefficients forv x t

( )

,

Solve for remaining constants, using initial conditions, which gives you v. u= +v ψ

Review

Math 2MM3 Exam 2009 #7 1. Parametrize surface

2 2

2 2

4 4

z x y

z x y

= ± − −

= − −

2 2

2 2

0, 4

1, 3

z x y

z x y

= + =

= + =

Note: Cylindrical coordinates

( )

{

, | 3 2, 0 2

}

D= r θ ≤ ≤r ≤ ≤θ π

( )

2

2

1

, cos , sin , 1

z r

r θ r θ r θ r

= −

= −

r

Find surface area of parametric surface in terms of r,θ.

( )

(

)

1 2 2

2

d 1

cos , sin , 1 2 2

cos , sin , 1 sin , cos , 0 r

D r

A s A

r r

r r

r r

θ

θ

θ θ

θ θ

θ θ

−

= ×

= − −

− =

− = −

∫∫

r r

r

r

(

)

1 2

2

2 2

2 2

2 2

2 2

2 2

4 4

2 2 2

2 2

4 2 2

4 2 4

2

2 2

2

ˆ ˆ ˆ

cos sin

1

sin cos 0

cos sin

ˆ ˆ ˆ _{cos sin}

1 1

cos sin

, ,

1 1

cos sin

1 1

1

1

1 1 r

r r r

r r

r r

r r

r r

r r

r

r r

r r

r

r r

r r r r r r

r r

r

r r

θ

θ

θ θ

θ θ

θ θ _{θ θ}

θ θ

θ θ

−

− × =

− −

  −  _{ }

= _{ }− _{ }+ _ + _

− −

   

= ×

− −

= ⋅ + +

− −

= +

− + − =

−

= −

= −

i j k

r r

i j k

r r

(

)

1 2 2

2 2 2

2

0 3

1 d

d d 1

D

r r A

r r r

π

θ

−

= −

=

−

∫∫

∫ ∫

No Chapter 13.7!