Then the second boundary condition leads to c 1 μ

Then the second boundary condition leads to
c
1
μ cosh( μ ) + μ
2
c
1
sinh( μ ) = 0
or (since c
1
and μ are both not zero),
cosh( μ ) + μ sinh( μ ) = 0 .
But, without any lose in generality, we may always assume that μ > 0 , and
so
cosh( μ ) + μ sinh( μ ) > 1
showing that it can never be zero. Therefore c = μ
2
> 0 leads to no
solutions. Next we assume that c = 0 . Then we get
ϕ
00
( x ) = 0
with
ϕ (0) = 0 and ϕ
0
(1) = 0
which leads to ϕ ( x ) = 0 . Finally we assume that c = − λ 2 < 0 . This leads
to
ϕ
00
( x ) + λϕ ( x ) = 0
with
ϕ (0) = 0 and ϕ
0
(1) − λ
2
ϕ (1) = 0 .
This problem was the subject of Problem #2 above and has the eigenfunc-
tions
ϕ
n
( x ) = sin( λ
n
x )
with cos( λ
n
) = λ
n
sin( λ
n
) .
17

b.) Continuing with the separation of variables method, then we have
γ
0
n
( t ) + λ
2
n
γ
n
( t ) = 0
so that
γ
n
( t ) = e
− λ
2
n
t
for n = 1 , 2 , 3 , . . . . Then we see that
u ( x, t ) =
∞
X
n =1
a
n
e
− λ
2
n
t
sin( λ
n
x ) .
The initial condition u ( x, 0) then requires that
u ( x, 0) = f ( x ) =
∞
X
n =1
a
n
sin( λ
n
x ) =
∞
X
n =1
a
n
ϕ
n
( x ) .
To solve for a
n
, let us de fi ne an inner product by
h F ( x ) , G ( x ) i =
Z
1
0
F ( x ) G ( x ) dx
and compute
h f ( x ) , ϕ
m
( x ) i =
∞
X
n =1
h ϕ
n
( x ) , ϕ
m
( x ) i a
n
.
This is a system of in fi nity equation in the in fi nite number of unknowns
{ a
1
, a
2
, a
3
, . . . , a
n
, . . . } .
The fact that ϕ
n
( x ) and ϕ
m
( x ) are non-orthogonal makes the remain un-
coupled and therefore much more di ffi cult to solve. Instead of solving this
system, we may get around this problem by converting the set of functions
ϕ
n
( x ) = sin( λ
n
x )
into an orthogonal set ψ
n
( x ) . Toward this end we use the Gram-Schmidt
process and set
ψ
1
( x ) = ϕ
1
( x ) = sin( λ
1
x )
18

and then compute ψ
2
( x ) using,
ψ
2
( x ) = ϕ
2
( x ) −
h ϕ
2
( x ) , ψ
1
( x ) i
h ψ
1
( x ) , ψ
1
( x ) i
ψ
1
( x )
= sin( λ
2
x ) −
R
1
0
sin( λ
2
x ) sin( λ
1
x ) dx
R
1
0
sin( λ
1
x ) sin( λ
1
x ) dx
sin( λ
1
x )
= sin( λ
2
x ) −
− sin( λ
2
) sin( λ
1
)
1
2
λ 2
1
sin
2
( λ
1
)
sin( λ
1
x )
and so
ψ
2
( x ) = sin( λ
2
x ) +
2 sin( λ
2
)
λ 2
1
sin( λ
1
)
sin( λ
1
x ) .
To compute ψ
3
( x ) , we use
ψ
3
( x ) = ϕ
3
( x ) −
h ϕ
3
( x ) , ψ
1
( x ) i
h ψ
1
( x ) , ψ
1
( x ) i
ψ
1
( x ) −
h ϕ
3
( x ) , ψ
2
( x ) i
h ψ
2
( x ) , ψ
2
( x ) i
ψ
2
( x ) .
and so on. Once we have a su ffi cient number of these, then we may expand
u ( x, t ) as
u ( x, t ) =
∞
X
n =1
b
n
e
− λ
2
n
t
ψ
n
( x ) .
The initial condition u ( x, 0) then requires that
u ( x, 0) = f ( x ) =
∞
X
n =1
b
n
ψ
n
( x ) .
To solve for b
n
, we then use
b
n
=
h f ( x ) , ψ
n
( x ) i
h ψ
n
( x ) , ψ
n
( x ) i
.
Of course, fi nding a general simple expression for ψ
n
( x ) may not be easy.
So either way, fi tting the initial condition is not easy when the original set
of eigenfunctions is not orthogonal.
––––––––––––––––––––––––––––––––––––
19

––––––––––––––––––––––––––––––––––––
Solution to Problem #6
First we solve for u
p
( x ) so that it satis fi es xu
00
p
( x ) − u
0
p
( x ) = 0 , and u
p
(0) = 1
and u
p
(1) = 0 . Solving the di ff erential equation for u
p
( x ) , we get
du 0
p
u 0
p
=
dx
x
or ln( u
0
p
) = ln( x ) + c
1
which leads to u
0
p
( x ) = Ax , which then results in the general solution
u
p
( x ) = A
Z
xdx + B =
1
2
Ax
2
+ B.