Chapter 5 & 6 – Study Guide
5.3.3
Q. Why do we have to divide p’ by p?
5.3.5
Q. Where did cosine come from? Is it because B=0 in part a?
Q. Where did the 2L/(n-1) come from? How would I know that is has that period?
Q. Does that mean there are no zero eigenvalues, can ∅ = constant? Why or why
not?
5.3.8
5.3.9
Q. What does “equidimensional equation” mean?
Q. Why is the ∅ = xp used, how do we know this?
Q. Is it trivial if c and d both equal 0, so we reject it?
5.4.1
u ( x, t ) = cn e − ntn ( x )
n =1
Q. Where did this formula come from?
where 1 2 ... are the eigenvalues of the Sturm-Liouville eigenvalue problem
( K 0 ) + + c = 0,
( 0) = ( L ) = 0,
1 , 2 ,... are the corresponding eigenfunctions, and
L
f ( x ) ( x ) c ( x ) ( x ) dx
n
cn =
0
L
( x ) c ( x ) ( x ) dx
2
n
0
Q. Is this a formula I should know? What do cn, c(x), and p(x) mean?
If 1 0 then lim u ( x, t ) = 0
t →+
A)
We are seeking for the solution as a superposition of solutions u ( x, t ) = ( x ) G ( t ) with separated
variables of the differential equation that satisfy the boundary conditions.
Substituting u ( x, t ) = ( x ) G ( t ) into the given PDE gives:
( K ( x ) ( x )) G (t ) + ( x ) G (t ) = c ( x ) ( x ) ( x ) G (t )
0
( K ( x ) ( x ) ) + ( x ) = G ( t )
0
c ( x ) ( x ) ( x )
G (t )
Since the left hand side is function of x only and the right hand side is function of y only, we can deduce
that:
( K ( x ) ( x ) ) + ( x ) = G ( t ) = −
0
c ( x ) ( x ) ( x )
G (t )
where is constant. Therefore,
( K0 ) + + c = 0, G = −G
Conversely, if G and are solutions to the above pair of ODEs, then u ( x, t ) = G ( t ) ( x ) is solution of
given PDE.
Substituting u ( x, t ) = G ( t ) ( x ) into the boundary conditions get:
( L ) G (t ) = 0,
( 0) G ( t ) = 0,
Assume G ( t ) 0 , then boundary conditions become ( 0) = ( L ) = 0
Consider the regular Sturm-Liouville eigenvalue problem
( K 0 ) + + c = 0,
This problem has infinitely many eigenvalues
1 2 ...
( 0) = ( L ) = 0
Denote n to be the eigenfunction corresponding to n . According to Rayleigh quotient
L
2
2
− K 0nn + K 0 n − n dx
0
0
L
n =
L
c
2
n
dx
0
Because n ( 0) = n ( L ) = 0 , the non-integral term
L
− K 0nn = − K 0 ( L ) n ( L ) n ( L ) + K 0 ( 0 ) n ( 0 ) n ( 0 ) = 0
0
vanishes. Therefore, if 0 , = − then it follows that
2
+ n 2 dx
K
0
n
0
L
n =
L
c
n
2
0
dx
0
B)
The function G can be determined from G = −G , the general solution of this equation is
G ( t ) = c0e−t , where c0 is a constant.
Thus we obtain the following solutions:
un ( x, t ) = e−ntn ( x ) , n = 1, 2,...
n =1
n =1
u ( x, t ) = un ( x, t ) = cn e − ntn ( x )
with constants c1 , c2 ,..., cn ,... . Substituting the series into initial condition u(x,0) = f(x), we have:
f ( x ) = cnn ( x )
n =1
where
L
f ( x ) ( x ) c ( x ) ( x ) dx
n
cn =
0
L
( x ) c ( x ) ( x ) dx
2
n
0
C)
If all eigenvalues are positive then the solution u(x,t) vanishes as t → + because each term of the
corresponding series contains decaying factor e − n .
5.4.3
5.5.1 bc
5.5.2
Q. How did we get L and what is a differential operator?
Q. What was the point of simplifying the right side of the equation?
5.5.9
Q. Why and how did we expand ∅(d4∅/dx4)?
5.5.15
Define the differential operator L as following:
L=
d d
r
dr dr
Q. Why is L defined like the one above?
So the differential equation can be written as:
L ( ) + r = 0
Let the eigenfunctions n ( r ) and m ( r ) correspond to two different eigenvalues n and m .
Therefore,
n L (m ) − m L (n ) = n
d dm
d dn
r
− m r
dr dr
dr dr
=
d dm dn dm d dv dm dn
n r
m r
−
r
−
+
r
dr dr dr dr dr dr dr dr
=
d dm d dv
n r
m r
−
dr dr dr dr
=
d
d dm
r n
− m n
dr dr
dr
Integrating over the interval defined by the boundary conditions
1
d
d
0 n L (m ) − m L (n ) dr = r n drm − m drn
0
1
And substituting in L ( ) = − r gives
1
d
d
0 −nm rm + mn rn dr = r n drm − m drn 0
1
1
( n − m ) nm rdr = 1 n (1)
0
dm
d
d
d
(1) − m (1) n (1) − 0 n ( 0 ) m ( 0 ) − m ( 0 ) n ( 0 ) = 0
dr
dr
dr
dr
Since n m , n − m 0 , we have
1
r
n m
dr = 0
0
This relation tells that the eigenfunctions corresponding to different eigenvalues are orthogonal with
weighting function r.
5.6.1 ac
Q, How can I find a good trial function, it seems difficult?
Q. How can I find a trial solution that satisfies the boundary conditions?
5.6.4
(a)
The Rayleigh quotient relates an eigenfunction to the corresponding eigenvalues
1
−r 0 + r ( r ) dr
2
1
=
0
1
r (r )
2
dr
0
The boundary condition implies that:
−r 0 = −1 (1) (1) + 0 ( 0 ) ( 0 ) = 0
1
It follows that 0 .
If = 0 , then
1
r (r )
0
2
dr = 0 ( r ) = 0 ( r ) is constant function over 0 < r < 1, contradiction.
Q. Why is it a contradiction, can’t ∅(r) be a constant?
Therefore, 0
(b)
The problem
d
d d
(1) = 0 has infinitely many eigenvalues
r
+ r = 0 , ( 0 ) ,
dr
dr dr
1 2
n
Assume that the eigenfunctions are known, according to the Rayleigh quotient:
1
1
n =
−rnn + r n ( r ) dr
0
2
0
1
r (r )
n
2
dr
0
The non-integral term −rnn 0 = −1n (1) n (1) + 0n ( 0 ) n ( 0 ) = 0 vanishes
1
Then the solution is:
( r ) = cnn ( r )
n =1
where
L
( r ) ( r ) rdr
n
cn =
0
L
(r )
n
0
using orthogonality.
5.7.1
2
rdr
Q. Where did 1 < c2 < 1+a2 come from, wasn’t c2 = 1+4a2(x- ½)2?
5.7.3
(a) Assume that the solution is u ( x, t ) = ( x ) G ( t )
The given PDE becomes:
2u
2u
2
=
c
x
( ) 2
t 2
x
G ( t ) ( x ) = c 2 ( x ) G ( t ) ( x )
c2 ( x )
( x ) G ( t )
=
= −
( x ) G (t )
2
c ( x ) ( x ) + ( x ) = 0,
G ( t ) + G ( t ) = 0
This problem has infinitely many eigenvalues
1 2 ... n
The differential operator L is defined as:
L = c2 ( x )
d2
dx 2
Therefore,
L ( ) + = 0
Let the eigenfunctions n ( x ) and m ( x ) correspond to two different eigenvalues n and m .
Therefore,
d 2m
d 2n
2
n L (m ) − m L (n ) = n c ( x ) 2 − m c ( x ) 2
dx
dx
2
= c 2 ( x ) n
= c2 ( x )
d 2m
d 2n
−
m
dx 2
dx 2
d
d d m
n
− m n
dx
dx
dx
Integrating over the interval defined by the boundary conditions
b
n L (m ) − m L (n )
c2 ( x )
a
b
d
d
dx = n m − m n
dx a
dx
And substituting in L ( ) = − gives
−n mm + mnn dx = dm − dn
m
n
a
c2 ( x )
dx a
dx
b
b
1
c ( x)
2
( n − m ) nm dx = b n ( b )
b
a
dm
d
d
d
( b ) − m ( a ) n ( a ) − a n ( a ) m ( a ) − m ( a ) n ( a ) = 0
dx
dx
dx
dx
Since n m , n − m 0 , we have
b
( x ) ( x ) dx = 0
n
m
a
This relation tells that the eigenfunctions corresponding to different eigenvalues are orthogonal with
weighting function = 1.
(b) The Reyleigh quotient for the eigenfunctions states that:
d
d
−
+ c2 ( x )
dx
dx a a
dx
b
=
2
b
b
u ( x)
2
dx
a
According to the boundary conditions of the given one-dimensional wave equation, we always have
that:
b
du
du
du
−u
= −u ( b ) ( b ) + u ( a ) ( a ) = 0
dx a
dx
dx
Hence,
b
=
2
c ( x)
a
d
dx
dx
b
2
( x)
2
0
dx
a
If = 0 , then
b
c ( x) ( x)
2
2
dx = 0 ( x ) = 0 ( r ) is constant function over (a, b),
a
contradiction.
Therefore, 0
(c) Since 0 , the equation G ( t ) + G ( t ) = 0 has the following solution:
G ( t ) = c1 cos
( t ) + c sin ( t )
2
Thus we obtain the following solutions:
(
un ( x, t ) = c1 cos
( t ) + c sin ( t )) ( x ) , n = 1, 2,...
n =1
n =1
2
(
u ( x, t ) = un ( x, t ) = c1n cos
with constants c1n , c2 n ,..., .
n
( t ) ( x ) + c
n
2n
sin
( t ) ( x ))
n
5.8.3
Q. What does “transcendental equation” mean?
Q. How do I know what the horizontal axis equal and the bounds too?
5.8.8
Q. What is Green’s Formula?
Q. How did we get the ∅n?
Q. What formula is this, the An?
5.8.11
Q. Isn’t it usually
= 0, < 0, > 0. Why are there 5’s?
6.2.5
Q. How is this sum obtained, where do the 8’s come from?
Q. How did we get the magnitude of error?
Chapter 12 – Final Exam
EX. 12.2.1
EX. 12.2.2
Q. Where did the w, s, and the x+3t all come from?
Q. What was the point of integrating dx/dt and dt/ds when ultimately we integrate
dw/ds?
Q. I thought t = 3s +x, why does t = x+3t?
Q. 12.3.1
Q. 12.3.3
Q. Where’s the xx after the u?
Q. Why is this a solution?
Q. Where did that formula come from and how did we get this?
Q. 12.3.4
...

Purchase answer to see full
attachment