5. Boundary Value Problems (using separation of variables).
Seven steps of the approach of separation of Variables:
1) Separate the variables:
(by writing e.g. u(x, t) = X(x)T (t) etc..
2) Find the ODE for each “variable”.
3) Determine homogenous boundary values to stet up a Sturm- Liouville
problem.
4) Find the eigenvalues and eigenfunctions.
5) Solve the ODE for the other variables for all different eigenvalues.
6) Superpose the obtained solutions
7) Determine the constants to satisfy the boundary condition.
Second example: Initial boundary value problem for the wave equation with
periodic boundary conditions on D = (−π, π) × (0, ∞)

utt − c2 uxx = 0



 u(x, 0) = f (x)
(IBVP) ut (x, 0) = g(x)



 u(−π, t) = u(π, t)
ux (−π, t) = ux (π, t)
(Hyperb. PDE)
IC for u ,
IC for u′ ,
BC for u
BC for ux
Step 1.
u(x, t) = X(x)T (t)
uxx (x, t) = X ′′ (x)T (t)
and
utt (x, t) = X(x)T ′′ (t)
1
Step 2.
Substituting that into the equation and and dividing the result by
c2 X(x)T (t)
yields the ODEs
X ′′ + λX = 0 and T ′′ + λc2 T = 0 .
Step 3.
We obtain the B.C:
X(−π) = X(π) , and X ′ (−π) = X ′ (π)
Step 4.
The eigenvalues and eigenfunctions of
 ′′
 X + λX = 0 ,
X(−π) = X(π) ,
 ′
X (−π) = X ′ (π)
are
λ0 = 0 with eigenfunction φ(x) ≡ 1
and
λn = n2 with eigenfunctions φn (x) = cos nx and ψn (x) = sin nx .
Step 5.
1 ) For λ0 = 0 , the ODE for T0 reads:
T0′′ = 0;
with general solution T0 = A0 + B0 t .
2 ) For λn = n2 , the ODE for Tn reads:
2
Tn′′ + n2 c2 Tn = 0;
with general solution τn (t) = An cos nct + Bn sin nct .
Step 6.
We get the following solution of the PDE satisfying the boundary conditions:
u0 (x, t) = (Ã0 + B̃0 t)C
and
un (x, t) = (Ãn cos nct + B̃n sin nct)(C̃n cos nx + D̃n sin nx)
“Streamlining” the constants, we write these as
u0 (x, t) = (A0 + B0 t)
and
un (x, t) =
(An cos nct + Bn sin nct) cos nx + (Cn cos nct + Dn sin nct) sin nx .
Superposing these functions we get
u(x, t) = (A0 + B0 t)+
∞
X
[(An cos nct + Cn sin nct) cos nx + (Bn cos nct + Dn sin nct) sin nx]
n=1
which is a solution of the PDE satisfying the homogeneous boundary condition, provided the coefficients allow for an appropriate convergence of the
sequence.
Step 7.
Determine the coefficients so that u(x, t) satisfies the initial conditions.
1 ) First initial condition gives:
∞
X
f (x) = A0 +
[An cos nx + Bn sin nx]
n=1
3
If we choose these constants to be the coefficients of the Fourier Series
of f , then the first IC is satisfied:
Z π
Z
1
1 π
A0 =
f dx , An =
f cos nxdx and
2π −π
π −π
1
Bn =
π
Z
π
f sin nxdx
−π
2 ) Second initial condition gives:
g(x) = ut (x, 0) = B0 +
∞
X
n=1
[(An − nc sin nct + Cn nc cos nct) cos nx+
(Bn − nc sin nct + Dn nc cos nct) sin nx]
= B0 +
∞
X
t=0
[ncCn cos nx + ncDn sin nx]
n=1
If we choose the constants B0 , ncCn
and
ncDn , to be the coeffi-
cients of the Fourier Series of g then the second IC is satisfied: so
Z π
Z π
1
1
gdx , Cn =
g cos nxdx and
B0 =
2π −π
ncπ −π
1
Dn =
ncπ
Z
π
g sin nxdx
−π
With the above choice of constants u(x, t) is the solution of the problem.
Example 3.
Now we deal with a mixed boundary value problem for the Laplace equation.:
4

uxx + uyy = 0



 uy (x, 0) = 0
(IBVP) u(x, m) = 0



 u(0, y) = g(y)
u(l, y) = 0
Laplace EQ. in (0, l) × (0, m)
NC at y = 0 ,
DC at y = m ,
nonhomogeneous DC at x = 0
DC at x = l
Step 1.
u(x, y) = X(x)Y (y)
uxx = X ′′ Y
and
uyy = XY ′′
Step 2.
Substituting that into the equation and and dividing the result by XY
yields the ODEs
X ′′ ± λX = 0 and Y ′′ ∓ λY = 0 .
Step 3.
We obtain the B.C:
Y ′ (0) = Y (m) = 0 , and X(l) = 0 , X(0)Y (y) = g(y) .
So we have Y as the unkown in the Sturm- Liouville problem.
Step 4.
We have to find the eigenvalues and eigenfunctions of
′′
Y + λY = 0 ,
Y ′ (0) = Y (m) = 0
i) λ<0
No negative eigen values since B = 0 .
5
ii ) λ = 0
The general solution is φ(y) = A + By , and we get from the first BC
0 = φ′ (0) = B and then from the second
0 = φ(m) = A
hence zero is not an eigenvalue.
iii ) λ > 0 The general solution is φ(y) = A cos νy + B sin νy .
√
ν = λ and we get from the first B.C
0 = ν(A(− sin νy) + B cos νy)
y=0
with
= νB, ⇒ B = 0 ,
and then from the second BC
0 = A cos νm ,
1
which allows for eigenvalues if cos νm = 0 or νm = (n − )π , that is
2
(2n − 1)π 2
λn = (
) , n = 1, 2, . . . ;
2m
with eigenfunction
φn (y) = cos(
(2n − 1)πy
).
2m
Step 5.
With λn = (
(2n − 1)π 2
) = : νn2 ,
2m
the ODE
Xn′′ − λXn = 0;
has the general solution ψn (x) = An cosh νn x + Bn sinh νn x .
Step 6.
So the solutions
un (x, t) = (An cosh νn x + Bn sinh νn x)(cos νn y)
6
of the PDE also satisfy the boundery conditions uy (x, 0) = u(x, m) = 0 .
Superposing these functions we get
∞
X
u(x, y) =
(An cosh νn x + Bn sinh νn x)(cos νn y)
n=1
which is a solution of the PDE satisfying the homogeneous boundary condition for y , provided the coeffcients allow for an appropriate convergence
of the sequence.
Step 7.
Determine the coefficients so that the PDE satisfies the other boundary
conditions:
In order to deal firstly with the homogeneous boundary condition we write
∞
X
u(x, y) =
(An cosh νn (x − l) + Bn sinh νn (x − l)(cos νy) .
n=1
( Recall: Solutions of ODEs with constant coefficients are translation invariant.) We get from the second B.C:
∞
X
0 = u(l, y) =
An
n=1
Hence An = 0, n = 1, 2, . . . , and the first BC requires
∞
X
g(y) =
Bn sinh νn (−l)](cos(νn y)
n=1
So if we choose the constants Bn sinh νn (−l) to be the coefficients of the
generalized Fourier Series of g , then the BC for x = 0 is satisfied.
Consequently:
2
Bn =
m sinh νn (−l)
Z
m
g(y) cos
0
(2n − 1)πy
dy .
2m
With the above choice of constants Bn the solution of the above boundary
7
value problem for the Laplace equation is:
u(x, y) =
∞
X
n=1
sinh(
(2n − 1)π
(2n − 1)π
(x − l)) cos(
y) .
2m
2m
Remark:
If the boundary conditions are inhomogeneous at more than one side of the
rectangle (0, l) × (0, m) then we separate the problem into problems with
inhomogeneous BC given at one side only, and we obtain the solution by
superposing the solution of the separated problems.
Inomogeneous problems and higher dimensional problems
As a first example let us consider

u − k uxx = f (x, t) (inhom. PDE)

 t
u(x, 0) = 0
Initial condition for u ,
BC at x = 0

 u(0, t) = 0
u(π, t) = 0
BC at x = π
Our first approach is to find solution of the kind
∞
X
u(x, t) =
An (t)ϕn (x) ,
n=1
Where (ϕn ) is a sequence of eigenfunctions of Sturm- Liouville problem of
the associated homogeneous problem.
In our case that is the sequence of eigenfunctions of the Fourier Sine Series.
So here we have:
u(x, t) =
∞
X
An (t) sin nx .
n=1
Assuming that we can interchange the differentiation with the infinite sum,
we calculate
8
ut (x, t) =
∞
X
A′n (t) sin nx .
n=1
uxx (x, t) =
∞
X
An (t)(−n2 sin nx) .
n=1
With those derivatives the PDE reads
∞
∞
X
X
′
An (t) sin nx + k
An (t)n2 sin nx = f (x, t)
n=1
n=1
Now we multiply the inhomogeneous PDE with an eigenfunction ϕk =
sin(kx) , say, and integrate over the domain (0, π) . Because of the orthogonality of the eigenfunction we get
Z
Z π
2
2
′
sin nxdx + kn An (t)
An (t)
=
Dividing by
say,
sin2 nxdx
0
0
Z
π
b
f (x, t) sin nx .
a
Z
π
sin2 nxdx we get on the right hand side a function in t
0
ψn (t) = (
Z
π
2
sin nxdx)
0
−1
Z
π
f (x, t) sin nxdx .
0
and we get the first order ODE for An (t) as
A′n (t) + kn2 An (t) = ψn (t) .
(*)
The initial condition
u(x, 0) = 0 ,
is satisfies if we add the initial condition
An (0) = 0
to the ODE. This is a first order ODE for which we can write down the
9
solution in the form
An (t) = e
−n2 kt
Z
t
ψn (τ )en
2
kτ
dτ ,
0
which in turn we can write as
Z t
2
ψn (τ )e−n k(t−τ ) dτ .
An (t) =
0
Example: f (x, t) = t , so
(
Z π
0
if n , is even
2
4t
ψn (t) =
.
t sin nxdx =
if n , is odd
π 0
nπ
In that case a more direct approach to solve the ODE’s is possible. We have
that An (t) = Cn e−kn
2
t
and is a solution of the homogeneous equation for
(*) . That is the case if n is even.
If n is odd, we use as trial solutions for a particular solution of (*)
(An )p (t) = αt + β , with (An )′p (t) = α
From the Ode we get
α + kn2 (αt + β) =
4t
nπ
Comparing coefficients we get
kn2 αt =
4t
, and α + kn2 β = 0
nπ
or
α=
4
4
,
and
β
=
−
.
πkn3
πk 2 n5
Finally the initial condition An (0) = 0 , gives
Cn = 0 if n even and Cn = −βn =
Altogether, with 2m − 1 = n we have
10
4
, if n is odd.
πk 2 n5
u(x, t) =
∞
X
m=1
with
A(t) = (
An (t) sin(2m − 1)x .
4
4
4
−k(2m−1)2 t
e
+
t
−
).
πk 2 (2m − 1)5
πk(2m − 1)3
πk 2 (2m − 1)5
Inhomogeneous boundary conditions
As an example let us deal here with the one-dimensionnal wave equation
(vibrating string equation)

utt − c2 uxx = 0



 u(x, 0) = 0
ut (x, 0) = 0



 (u(0, t) = g(t)
u(l, t) = h(t)
on (0, l) × (0, ∞) :
(inhom. PDE)
Initial cond. for u ,
Initial cond. for ut ,
BC at x = 0
BC at x = l
The most straight forward approach here is to write u(x, t) = w(x, t) +
v(x, t) where v(x, t) is a “simple function ” satisfying the boundary conditions.
Then w(x, t) = u(x, t) − v(x, t) satisfies the inhomogeneous PDE with homogeneous B.C.:
Usually we set
1
1
v(x, t) = g(t)(1 − x ) + h(t)(x ) .
l
l
We calculate
1
1
vtt = g ′′ (t)(1 − x ) + h′′ (t)(x ) ,
l
l
and
vxx = 0 .
So we need w to be a solution of
11

1
1


wtt − c2 wxx = −(g ′′ (t)(1 − x) + h′′ (t)( x);


l
l


1
1

w(x, 0) = −(g(0)(1 − x ) + h(0)(x )),
l
l

1
1



wt (x, 0) = −(g ′ (0)(1 − x ) + h′ (0)(x )),


l
l

(w(0, t) = w(l, t) = 0 .
which we can solve using the discussed methods.
Elliptic equations on spherical domains.
(1)
uxx + uyy = 0
(x, y) = f (x, y)
on
for
p
2
2
p x +y <R
x2 + y 2 = R
We introduce a change of variables setting
U (r, ϕ) = u(x, y) x=r cos ϕ
y=r sin ϕ
p
x2 + y 2 < R , imply that
then uxx + uyy = 0 for
Urr + 1r Ur + r12 Uϕϕ = 0
for 0 < r < R , −π < ϕ < π ,
(2)
U (R, ϕ) = F (ϕ),
with F (ϕ) = f (R cos ϕ, R sin ϕ)
Somehow the set of auxiliary conditions does not seem to be complete but
if we separate the variables first, this will give us a better picture of what
will be needed.
Step 1:We set
U (r, ϕ) = R(r) Φ(ϕ) ,
Ur = R′ Φ ,
and get
Urr = R′′ Φ
and
The PDE of (2) implies
1
1
R′′ Φ + RΦ + 2 RΦ′′ = 0 , ;
r
r
r2
Step 2: Multiplying with
. yields
RΦ
12
Uϕϕ = Rφ′′ .
Φ′′
r2 R′′ + rR′
+
= 0,
R
φ
which gives us the ODE’s
(3)
r2 R′′ + rR′ ∓ λR = 0, , ,
(2)
Φ′′ ± λΦ = 0 .
Step 3. We would like to work with Φ as the ODE for a Sturm-Lionville
problem. In order to do this we observe that the “reasonable” requirement
that the solution is continuously differentiable implies
U (r, −π) = U (r, π)
and
Uϕ (r, −π) = Uϕ (r, π) .
Step 4 For Φ, we get the Sturm-Liouville problem
 ′′
 Φ + λΦ = 0
ST.L. Φ(−π) = Φ(π)
 ′
Φ (−π) = Φ′ (π)
with the eigen values λ0 = 0,
λn = n2 ,
n = 1, 2, . . . and the eigen
function 1, cos x, sin x, cos 2 x sin 2 x . . ..
Step 5 We have to consider the ODE’s
r2 R′′ + rR′ = 0
(for λ = 0)
and
r2 R + rR′ − n2 R = 0
(for λn = n2 ) .
For λ = 0 we get with w = R′
w′ (π) + 1r w(r) = 0 .
The solution can be found by a “separation of variables approach”. We get
1
w′
=−
w
r
ln w = ln(−r) + C
1
w = C1 ·
r
13
or
ϕ0 (r) = A0 + B0 ln(r)
The second ODE an Euler equation has the solutions
ϕn (r) = An rn + Bn r−n
Step 6
Now superposing the solution of (1) and (2) we get
U (r, φ) = A0 +
+C0 (ln(r)) +
∞
X
rn (An cos nϕ + Bn sin nϕ)
n=1
∞
X
1
( )n (Cn cos nϕ + Dn sin nϕ)
r
n=1
Step 7
Since ln(r) and
1
rn
are unbounded the functions un (r, θ) which need to
superpose are not solutions of the PDE if those terms a part of un (r, θ) . So
we set the coefficients Cn , n = 0, 1, 2, . . . and Dn , n = 1, 2, . . . zero. Then
u(r, ϕ) = A0 +
∞
P
rn (An cos nϕ + Bn sin nϕ)
n=1
Is a solution of the PDE in the disk and we get we get from the initial
condition
U (R, ϕ) = F (ϕ) = A0 +
∞
P
Rn An cos nϕ +
n=1
∞
P
Rn Bn sin nϕ
n=1
which are satisfied if A0 , Rn An , Rn Bn are the Fourier coefficients of F ,
i.e.:
1
A0 =
2π
Zπ
F (ϕ)dϕ
−π
14
An =
1
πRn
Z+π
F (ϕ) cos nϕdϕ
−π
1
Bn =
πRn
Z+π
F (ϕ) sin nϕ dϕ
−π
Note, that we can not set the coefficients Cn , n = 0, 1, 2, . . . and Dn , n =
1, 2, . . . the be zero in case the domain is a “washer”, say
D = {(x, y) 0 < a2 < x2 + y 2 < R2 } .
In that case there will be an additional boundary condition for x2 +y 2 = a2 .
and we get from the BC’s two equation for each pair of coeffiecents An , Cn
and two equation for the pair of coeffiecents Bn , Dn , respectively, c.f. the
related Homework problem.
We want to consider the solution for the Disk somewhat closer:
We have
1
u(r, φ) =
2π
Zπ
F (θ)dθ
−pi
Z+π
∞
X
rn
+
F (θ)(cos nφ cos nθ + sin φ sin nθ)dθ
n
πR
n=1
−π
1
=
2π
Zπ
−pi
1
=
2π
Zπ
−pi
∞
X
r
F (θ)(1 + 2
( )n cos n(φ − θ))dθ
R
n=1
∞
X
r
F (θ)(1 +
( )n (ein(φ−θ) + e−in(φ−θ) ))dθ
R
n=1
15
1
=
2π
Zπ
−pi
rei(φ−θ)
re−i(φ−θ)
F (θ)(1 + (
+
))dθ
R − rei(φ−θ)
R − re−i(φ−θ)
and so we get Poissons integral formula
1
u(r, φ) =
2π
Zπ
F (θ)
−π
R2 − r 2
dθ
R2 + r2 − 2rR cos(φ − θ)
For r = 0 we get
1
u(0, φ) =
2πR
Zπ
U (R, θ)R dθ ,
−π
That is the vaule of u at the center is the average of the values of u at the
boundary of the disk.
Remark:
Consequences: Mean Value Theorem and Maximum Principle for solutions
of the Laplace equations.
Higher dimensional problems
Cooling of the sphere:
Under the assumption of a rotational symmetric initial temperature distribution and an constant outer temperature distribution (normalized to zero
the inital boundary value problem for heat distribution in a sphere

t > 0, k x k < π ,
 ut − k∆u = 0 ,
u(x, 0) = f (k x k) ,

if k x k = π ,
u(x, t) = 0 ,
becomes
16

2

 ut − k(uρρ + uρ ) = 0 ,
ρ
u(ρ,
0)
=
f
(ρ)
,


u(π, t) = 0 .
t > 0, ρ < π,
(Assuming that the heat distribution is only a matter of the distance to the
center the derivatives with respect to φ are vanishing.)
Separating the variables u(ρ, t) = RT we get
2
T ′ R + k(R′′ T + (R′ T )) = 0 , and so
ρ
R′′ + ρ2 (R′ )
T′
=−
=λ
kT
R
Since we have now choice for which variable we might use for a Sturm Liouville problem, we need to consider
2
R′′ + R′ + λR = 0,
ρ
0<ρ<π
But this is not in the form
(pX ′ )′ − qX + λmX = 0 .
Further more the boundary conditions 0 = u(π, t) = R(π)T (t) , provides
one homogeneous BC
R(π) = 0
but not a second one.
so we are not dealing with a regular Sturm - Liouville problem, here.
Introducing the functions X(ρ) = ρR(ρ) we get for X the ODE
X ′′ + λX = 0
and so for ν 2 = λ > 0 we have the solutions
17
R(ρ) =
1
1
C1 cos(νρ) + C1 sin(νρ) .
ρ
ρ
Since we can use only bounded solutions we would like to disregard the unbounded functions in this sum. So replacing the second boundary condition
by the condition that the solution need to be bounded might replace the
second boundary condition. Then multipying the ODE with ρ2 gives the
singular Sturm-Liouville problem

 ρ2 R′′ + 2ρR′ + ρ2 λR = 0,
R(π) = 0 ,

R a bounded function in (0, π)
0 < x < π,
,
whose eigenfunctions have the properties discussed in the last chapter with
respect to the weigt function m(ρ) = ρ2 .
Is it an exercise using the above transformation to show that all eigen values
are λn = n2 , n = 1, 2, 3, . . . with eigenfunctions
φn (ρ) =
sin(nρ)
.
ρ
The solutions of the ODE for T are e−n
un (ρ, t) = An e−n
2
kt sin(nρ)
ρ
2
kt
so that we have
.
are bounded solutions of the PDE with un (π, t) = 0 .
For u(ρ, t) =
∞
X
un (ρ, t) , we determine the coefficents from the remaing
n=1
auxilliary condition:
f (ρ) = u(ρ, 0) =
∞
X
n=1
An
sin(nρ)
ρ
With the generalized Fourier cofficient of the eigenfunctions φn (ρ) that is
Z π
Z π
sin(ρ) 2 2
sin ρ 2
−1
(
An = (
) ρ dρ)
ρ dρ .
f (ρ)
ρ
ρ
0
0
18
=(
Z
π
2
sin (ρ) dρ
0
Z
π
f (ρ)ρ sin ρ dρ .
0
Note that that An are the Fourier Sine coefficients of ρf ,
which we would get if we work the entire problem with the transformed
functions.
Two dimensional Wave equation
Example

utt − c2 ∆ w = 0 ,
(x, y, t) ∈ (0, L) × (0, M ) × R ,




 u(x, y, 0) = f (x, y) ,
ut (x, 0) = g(x, y) ,


u(0, y, t) = u(L, y, t) = 0 ,



uy (x, 0, t) = uy (0, M, t) = 0 ,
Separation of variables:
Substiuting u(x, y, t) = X(x)Y (y)T (t) into the PDE we get
X(x)Y (y)T ′′ (t) − c2 (X ′′ (x)Y (y)T (t) + X(x)Y ′′ (y)T (t)) = 0 ,
or
X ′′
Y ′′
T ′′
=
+
= −λ
c2 T
X
Y
this gives us first the ODE’s for T, X and Y :
T ′′ + c2 λT = 0
X ′′ + µX = 0
Y ′′ + κY = 0 with λ = µ + κ .
From the boundary condition we get the Sturm-Liuoville problems
′′
X + µX = 0
X(0) = X(L) = 0 .
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and
Y ′′ + κY = 0
Y ′ (0) = Y ′ (M ) = 0 .
with are the Sturm-Liuoville problems for the Fourier Sine and Fourier
mπ 2
Cosine Series, respectively. So we have the eigen values µm = (
) with
L
eigen functions
√
φm = sin( µm x , m = 1, 2, 3, 4, . . . ,
for the first Sturm-Liuoville problem.
For the second Sturm-Liuoville problem we have the eigenvalues κk = (
with eigen functions
√
ψk = cos( κk y), , k = 1, 2, 3, 4, . . . ,
and
ψ0 ≡ 1 .
For each combination λmk = µm + κk we get the solution
√
√
τmk = Amk cos(c µm + κk t) + Bm,k sin(c µm + κk t)
of the ODE’s for T
T ′′ + c2 λmk T = 0 .
(Step 6 ) So the functions
√
√
√
um,k (x, y, t) = Am,k cos(c µm + κk t) sin( µm x) cos( κk y)
√
√
√
+Bm,k sin(c µm + κk t) sin( µm x) cos( κk y)
for m, k = 1, 2, 3, . . . and
20
kπ 2
)
M
√
√
√
um,0 = Am,0 cos(cµm t) sin( µm x) + Bm,0 sin(c µm t) sin( µm x)
for m = 1, 2, 3, . . . are solutions of the PDE together with the homogeneous
boundary condition. We superpose those function to get
u(x, y, t)
=
∞
X
√
√
[Am,0 cos(c µm t) sin( µm x)+
m=1
√
√
Bm,0 sin(c µm t) sin( µm x)]
+
∞
X
m=1
∞
X
√
√
√
[Am,k cos(c µm + κk t) sin( µm µm x) cos( κk y)
k=1
√
√
√
+Bm,k sin(c µm + κk t) sin( µm µm x) cos( κk y)] .
We determine the constants using the orthogonality relations of those eigenfunctions. We get from the first initial conditions:
f (x) = u(x, y, 0) =
=
∞
X
√
Am,0 sin( µm x)+
m=1
∞
X
m=1
∞
X
√
√
Am,k sin( µm x) cos( κk y)
k=1
so
Am,k
4
=
LM
Z
2
=
LM
Z
and
Am,0
L
0
L
0
Z
Z
M
√
√
f (x, y)) sin( µm x) cos( κk y)dydx ,
M
√
f (x, y) sin( µm )dydx .
0
0
The second initial condition gives
g(x) = ut (x, y, 0) =
21
=
∞
X
√
√
c µm Bm,0 sin( µm x)
m=1
+
∞
X
m=1
hence
∞
X
√
√
√
c µm + κk Bm,k sin( µm x) cos( κk y) ,
k=1
Bm,k =
4
√
c µm + κk LM
for m, k = 1, 2, 3, . . . and
Bm,0 =
2
√
Z
LM µm
for m = 1, 2, 3, . . . .
L
0
Z
Z
L
0
M
√
√
g(x, y) sin( µm x) cos( κk y)dydx ,
0
Z
M
√
g(x, y) sin( µm x)dydx ,
0
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