set 7 solutions

ESE 319-02, Spring 2014, Homework

Set 7, Due Mar. 4 !1. Zill 13.4.1. (You may use integral tables, WolframAlpha or any other method to

evaluate the integrals.) 2. Zill 13.4.5. (Hint: “simple”.) 3. Zill 13.5.1. (This requires re-solving for the new boundary conditions.) 4. Half-insulated heated rod problem (worth 20 points instead of 10). A bar of length L has an initial temperature of f(x) along its length, x. It is insulated on the “right” (meaning at x = L) and, starting at time t = 0, its left end (x = 0) is held at a constant temperature of 0.

! !By using the product u = XT, you can separate variables to find general solutions for X and T, and also express the boundary and initial conditions in terms of X and T.

! !(a) Show that the ODE for X(x) is a Sturm-Liouville problem by equating terms to equations (22) and (4) and (5) in section 12.5 of Zill. (b) Show that the eigenvalues and eigenfunctions for X are as follows.

! !

( )( )( )( ) ( )xfxu

tLutu

ktxuuuku

x

txx

=

=

=

>==

0,0,0,0

0,

( ) ( ) ( )

( ) ( )( ) ( )xfxX

LXX

CeCeTxBxAXk

CCeTBAxX

CeCeTxBxAXk

CeT

TTandXk

XTT

XX

k

tTxXtxu

tkt

t

tkt

t

=

=!=

==+=>=

==+==

==+=<−=

=

−=!=+!!⇒−=!

=!!

=

−−

−

−

−

000

sincos:0

:0

sinhcosh:0

0

,

2

2

2

2

αλ

λ

αλ

λ

αααλ

λ

αααλ

λλ

λ

( ) ( )!,3,2,1

212sin2 =

−=== n

Ln

xBxXk nnnnnπ

αααλ


(c) Since the ODE for X is a Sturm-Liouville problem, its eigenfunctions can be used to express any continuous function, f(x), on the interval [0,L] as this sum.

! !By multiplying both sides by ! (for integer m = 1, 2, … ) and integrating from 0 to L, show that the formula for cn is as follows. (See pages 632-33 of Zill and Lecture 5. Use orthogonality.)

! !(d) Use the above results to write the complete solution, u(x,t), to the original half-insulated heated rod problem. !(e) Write the solution for the specific case when u(x,0) = 100, L = π and k = 1. !!!!

( ) ( )!"

#$%

& −=∑

∞

= Lxn

cxfn

n 212sin

1

π

( )!"

#$%

& −

Lxm

212sin π

( ) ( )∫ "#

$%&

' −=

L

n dxLxn

xfL

c0 2

12sin2 π


1. Zill 13.4.1. The boundary conditions match that of the case solved in the text, so use those results. (You may use tables, WolframAlpha or any other method to evaluate the integrals. I use tables here.) !

! !

( ) ( )

( ) ( )

( )

( )

( )

( )

( ) ( )[ ]

[ ] ( )[ ]

( ) ( )[ ] ( )∑∑

∫∫∫∫

∫∫

∫∫

∫∫∫

∫

∑

∞

=

∞

=

−

∞

=

=

%%&

'(()

* −−=%

%&

'(()

* −−=

−−=−++−=

−+−%&

'()

*−−%&

'()

*=

++,

-

.

./

0%&

'()

* ++%&

'()

*−%&

'()

*−

+,

-./

0 −%&

'()

*=

+++−=+−=

+−=

++=

+−=

%&

'()

*%&

'()

*−%&

'()

*=

−=−=

==

%&

'()

* +=

=∂

∂−=

==

133

2

133

2

33

2

33

222

232

0

23

0

2

222

1

0

23

0

2

0

2

0041

0

1

041

sincos11

sincos11

,

11cos1cos

2cos

2

2cos2cos21

cos21

sincos2cos21

cossin21

sincos2coscos2cossin

coscossin

sincoscos

cossinsin

sin21

sin21

sin21

sin21

sin2

0sin2

sinsincos,

00,

0,0,0

n

n

n

n

n

L

L

n

nnn

LL

LLL

n

L

n

nnn

t

xLn

tLan

nL

xLn

tLan

nL

txu

nL

nnL

nnL

nnL

nnnnL

Lnn

nL

xLn

xLn

xLn

xLn

xLn

nL

L

xLn

xLn

xLn

nL

A

Cuuuuuuduuuuuduu

uduunuuuduu

Cuuuuduu

Cuuuuduu

dxLnx

Ln

xLn

nL

Ldx

Lnx

Ln

xLn

nL

xdxLn

xL

xdxLn

xxdxLn

xLxL

A

xdxLn

xgan

B

xLn

tLan

BtLan

Atxu

tu

xLxxu

tLutu

ππ

π

ππ

π

ππ

ππ

ππ

π

ππππ

πππ

πππππ

π

πππ

π

πππ

π

πππ

π

πππ

π

π

πππ


!2. Zill 13.4.5. The boundary conditions match that of the case solved in the text, so use

those results. Since g(x) is the sine function itself, just match coefficients rather than evaluate the integral: the “simple method”. !

! !!!

( ) ( )

( ) ( ) ( )

( ) ( )

( )

( ) ( )

( ) xata

txu

nna

Bnn

aB

nxnaBxxg

xdxLn

xfL

A

nxnatBnatAtxu

xgxtu

xfxu

Ltutu

nn

nn

L

n

nnn

t

sinsin1

,

101/1

1011

sinsin

0sin2

sinsincos,

sin00,

0,0,0

1

0

1

0

=

!"#

≠

==⇒

!"#

≠

==

==

==

+=

==∂

∂==

=⇒==

∑

∫

∑

∞

=

∞

=

=

π

ππ


3. Zill 13.5.1. The boundary conditions do not match that of the case solved in the text, so you need to go through the possibilities. !

! !

( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) ( )

( ) ( )

( ) ( )∑ ∫

∫∫

∑ ∫∑

∑

∞

=

∞

=

∞

=

∞

=

$$$$

%

&

''''

(

)

=

==

$$%

&''(

)===

=

===+=

==⇒

=⇒====+==

=⇒====+=−=

=⇒====+==

===−,,

===+,,

====

1 0

00

1 01

1

2

2

sinsinhsinsinh

12,

sinsinh

2sin

2sinh

sinsin2

sinsinh,

sinsinh,

sinh00sinhcosh

sin

0sin00sincos:

0)(0sinh00sinhcosh:

0000:0

0000000

,00,0,0,0

n

a

a

n

a

n

n

a

nn

nn

n

xan

yan

xdxan

xf

abna

yxu

xdxan

xf

abn

aAxdx

an

xfaa

bnA

xan

xdxan

xfa

xfxan

abn

Abxu

xan

yan

Ayxu

yan

DbYCYyan

Dyan

CY

an

xan

BxX

an

aBaXAXxBxAX

xXaBaXAXxBxAX

xXAaaXBXBAxX

xfbYYYYaXXXX

xfbxuxuyauyu

ππππ

ππ

ππ

ππππ

ππ

πππ

πλ

π

παααααλ

ααααλ

λ

λ

λ


4. Half-insulated heated rod (20 points). !(a) Show that the ODE for X(x) is a Sturm-Liouville problem by equating terms to equations (22) and (4) and (5) in section 12.5 of Zill.

! !(b) Show that the eigenvalues and eigenfunctions for X are as follows.

!

! (c) Since the ODE for X is a Sturm-Liouville problem, its eigenfunctions can be used to express any continuous function, f(x), on the interval [0,L] as this sum.

! !By multiplying both sides by ! (for integer m = 1, 2, … ) and integrating from 0 to L, show that the formula for cn is as follows. (See pages 632-33 of Zill and Lecture 5.)

!

( ) ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) 000100:50:4

0010:22

1221

22

11

=!=⇒=======

=!+

=!+

=+!!⇒======

=++!+!!

LXXXyBABALbabyBbyAayBayA

XXXyxcxbxdxayxdxcyxbyxa

kkλλλ

λ

( ) ( )!,3,2,1

212sin2 =

−=== n

Ln

xBxXk nnnnnπ

αααλ

( )

( ) ( ) ( )

( )( ) ( ) ( )

( )

( ) ( ) ( )( )

( ) xBxXk

nL

nL

trivialnonxBxXLLBLXAX

xBxAXxBxAXk

rejecttrivialxXALXBXAXBAxX

rejecttrivialxXBLBLXAX

xBxAXxBxAXk

nnnnn

n

ααλ

πα

πππα

αααα

αααααααλ

λ

αα

αααααααλ

sin

,3,2,1212

,25,

23,2

sin0cos0cos00

cossinsincos:03

,0000:02

,000cosh00

coshsinhsinhcosh:01

2

2

2

==

=−

=⇒=

−=⇒=⇒==&⇒==

+−=&+=>=

=⇒==&⇒==

=&+==

=⇒=⇒==&⇒==

+=&+=<−=

!!

( ) ( )!"

#$%

& −=∑

∞

= Lxn

cxfn

n 212sin

1

π

( )!"

#$%

& −

Lxm

212sin π

( ) ( )∫ "#

$%&

' −=

L

n dxLxn

xfL

c0 2

12sin2 π


!

! !(d) Use the above results to write the complete solution, u(x,t), to the original half-insulated heated rod problem.

! !(e) Write the solution for the specific case when u(x,0) = 100, L = π and k = 1.

( ) ( ) ( ) ( )

( )

( )( ) ( )

( )( ) ( )

( )( )

( ) ( )∫

∫

∫

∫∑∫

#$

%&'

( −=

=#$

%&'

( −

−=

#$

%&'

( −−

−

−=

−#$

%&'

( −

−=

#$

%&'

( −=

#$

%&'

( −#$

%&'

( −=#$

%&'

( − ∞

=

L

n

nn

L

n

L

n

L

n

L

nn

L

dxLxn

xfL

c

Lc

nnL

c

Lxn

Lxn

nL

c

dxL

nLxn

nL

c

ityorthogonaldxLxn

c

dxLxn

Lxm

cdxxfL

xm

0

0

0

2

0

2

0 10

212sin2

2412

122

12sin41

212

21

122

212

212sin

122

212sin

212sin

212sin

212sin

π

π

π

ππ

π

ππ

π

π

πππ

∵

( ) ( )

( )( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )!"

#$%

& −

()*

+,-

!"

#$%

& −=

!"

#$%

& −=

!"

#$%

& −=

=−

===

∑ ∫

∫

∑

∞

=

!"

#$%

& −−

∞

=

!"

#$%

& −−

−

Lxn

edxLxn

xfL

txu

or

dxLxn

xfL

B

Lxn

eBtxu

nL

nxeBTXtxu

n

tL

nkL

L

n

n

tL

nk

n

nntk

nnnn

tn

212sin

212sin2,

212sin2

212sin,

,2,1212sin,

1

212

0

0

1

212

2

2

ππ

π

π

παα

π

π

α !


!

( )( ) ( ) ( ) ( )

( )( )

( ) ( )

( )( )

( )

( )( )

( ) ( )∑

∫∫

∑∑

∞

=

$%

&'(

) −−

∞

=

$%

&'(

) −−∞

=

$%

&'(

) −−

$%

&'(

) −

−=

−=$%

&'(

) −−

−=

−$%

&'(

) −

−=$%

&'(

) −=

$%

&'(

) −=$%

&'(

) −=

1

212

0

00

1

212

1

212

212

sin12

1400,

121400

212

cos12

2200

212

212

sin12

2200212

sin1002

212

sin212

sin,

2

22

n

tn

n

n

tn

nn

tL

nk

n

xne

ntxu

orn

xnn

dxnxn

ndx

xnB

xneB

Lxn

eBtxu

π

ππ

ππ

π

π

ππ

π

set 7 solutions

Documents