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Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
December 8, 1999
Chapter 2
Problem 2.
A point charge q is brought to a position a distance d away from an infinite plane conductor held at zero potential. Using the method of images, find:
(a) the surface-charge density induced on the plane, and plot it;
(b) the force between the plane and the charge by using Coulomb’s law for the force between the charge and its image;
(c) the total force acting on the plane by integrating σ^2 / 2 0 over the whole plane;
(d) the work necessary to remove the charge q from its position to infinity;
(e) the potential energy between the charge q and its image (com- pare the answer to part d and discuss).
(f ) Find the answer to part d in electron volts for an electron originally one angstrom from the surface.
(a) We’ll take d to be in the z direction, so the charge q is at (x, y, z) = (0, 0 , d). The image charge is −q at (0, 0 , −d). The potential at a point r is
Φ(r) =
q 4 π 0
[
|r − dk|
|r + dk|
]
The surface charge induced on the plane is found by differentiating this:
σ = − 0
dΦ dz
z=
= −
q 4 π
[
−(z − d) |r + dk|^3
(z + d) |r + dk|^3
]
z=
= − qd 2 π(x^2 + y^2 + d^2 )^3 /^2
We can check this by integrating this over the entire xy plane and verifying that the total charge is just the value −q of the image charge:
∫ (^) ∞
−∞
−∞
σ(x, y)dxdy = −
qd 2 π
0
∫ (^2) π
0
rdψdr (r^2 + d^2 )^3 /^2
= −qd
0
rdr (r^2 + d^2 )^3 /^2
= −
qd 2
d^2
u−^3 /^2 du
qd 2
∣−^2 u−^1 /^2
∞ d^2 = −q
(b) The point of this problem is that, for points above the z axis, it doesn’t matter whether there is a charge −q at (0, 0 , d) or an infinite grounded sheet at z = 0. Physics above the z axis is exactly the same whether we have the charge or the sheet. In particular, the force on the original charge is the same whether we have the charge or the sheet. That means that, if we assume the sheet is present instead of the charge, it will feel a reaction force equal to what the image charge would feel if it were present instead of the sheet. The force on the image charge would be just F = q^2 / 16 π 0 d^2 , so this must be what the sheet feels.
(c) Total force on sheet
0
∫ (^2) π
0
σ^2 dA
q^2 d^2 4 π 0
0
rdr (r^2 + d^2 )^3
= q^2 d^2 8 π 0
d^2
u−^3 du
q^2 d^2 8 π 0
∣−^
u−^2
∞
d^2
q^2 d^2 8 π 0
[
d−^4
]
Problem 2.
A straight-line charge with constant linear charge density λ is located perpendicular to the x − y plane in the first quadrant at (x 0 , y 0 ). The intersecting planes x = 0 , y ≥ 0 and y = 0, x ≥ 0 are conducting boundary surfaces held at zero potential. Consider the potential, fields, and surface charges in the first quadrant.
(a) The well-known potential for an isolated line charge at (x 0 , y 0 ) is Φ(x, y) = (λ/ 4 π 0 ) ln(R^2 /r^2 ), where r^2 = (x − x 0 )^2 + (y − y 0 )^2 and R is a constant. Determine the expression for the potential of the line charge in the presence of the intersecting planes. Verify explicitly that the potential and the tangential electric field vanish on the boundary surface.
(b) Determine the surface charge density σ on the plane y = 0, x ≥ 0. Plot σ/λ versus x for (x 0 = 2, y 0 = 1), (x 0 = 1, y 0 = 1), and (x 0 = 1, y 0 = 2).
(c) Show that the total charge (per unit length in z) on the plane y = 0, x ≥ 0 is
Qx = −
π λ tan−^1
x 0 y 0
What is the total charge on the plane x = 0?
(d) Show that far from the origin [ρ ρ 0 , where ρ =
√ x^2 +^ y^2 and^ ρ^0 = x^20 + y^20 ] the leading term in the potential is
Φ → Φasym =
4 λ π 0
(x 0 )(y 0 )(xy) ρ^4
Interpret.
(a) The potential can be made to vanish on the specified boundary surfaces by pretending that we have three image line charges. Two image charges have charge density −λ and exist at the locations obtained by reflecting the original image charge across the x and y axes, respectively. The third image charge has charge density +λ and exists at the location obtained by reflecting the original charge through the origin. The resulting potential in the first quadrant is
Φ(x, y) = λ 4 π 0
ln
R^2
r 12
− ln
R^2
r^22
− ln
R^2
r^23
R^2
r 42
λ 2 π 0 ln
r 2 r 3 r 1 r 4
where
r^21 = [(x − x 0 )^2 + (y − y 0 )^2 ] r^22 = [(x + x 0 )^2 + (y − y 0 )^2 ]
r 32 = [(x − x 0 )^2 + (y + y 0 )^2 ] r^24 = [(x + x 0 )^2 + (y + y 0 )^2 ]. From this you can see that
- when x = 0, r 1 = r 2 and r 3 = r 4
- when y = 0, r 1 = r 3 and r 2 = r 4
and in both cases the argument of the logarithm in (2) is unity.
(b)
σ = − 0
d dy
λ 2 π
r 2
dr 2 dy
r 3
dr 3 dy
r 1
dr 1 dy
r 4
dr 4 dy
y=
We have dr 1 /dy = (y − y 0 )/r 1 and similarly for the other derivatives, so
σ = − λ 2 π
y − y 0 r^22
y + y 0 r^23
y − y 0 r 12
y + y 0 r^24
y=
= − y 0 λ π
(x − x 0 )^2 + y^20
(x + x 0 )^2 + y^20 )
(c) Total charge per unit length in z
Qx =
0
σdx
y 0 λ π
[∫ ∞
0
dx (x − x 0 )^2 + y 02
0
dx (x + x 0 )^2 + y^20
]
For the first integral the appropriate substitution is (x − x 0 ) = y 0 tan u, dx = y 0 sec^2 udu. A similar substitution works in the second integral.
λ π
[∫
π/ 2
tan−^1 − x y^00
du −
∫ (^) π/ 2
tan−^1 x y^00
du
]
λ π
[
π 2
− tan−^1
−x 0 y 0
π 2
x 0 y 0
]
2 λ π
tan−^1
x 0 y 0
The calculations are obviously symmetric with respect to x 0 and y 0. The total charge on the plane x = 0 is (3) with x 0 and y 0 interchanged:
Qy = − 2 λ π
tan−^1 y 0 x 0
Since tan−^1 x − tan−^1 (1/x) = π/2 the total charge induced is
Q = −λ
λ 4 π 0
[
16 xyx 0 y 0 (x^2 + y^2 )^2
]
4 λ π 0
(xy)(x 0 y 0 ) (x^2 + y^2 )^2
Problem 2.
A point charge is placed a distance d > R from the center of an equally charged, isolated, conducting sphere of radius R.
(a) Inside of what distance from the surface of the sphere is the point charge attracted rather than repelled by the charged sphere?
(b) What is the limiting value of the force of attraction when the point charge is located a distance a(= d−R) from the surface of the sphere, if a R?
(c) What are the results for parts a and b if the charge on the sphere is twice (half) as large as the point charge, but still the same sign?
Let’s call the point charge q. The charged, isolated sphere may be replaced by two image charges. One image charge, of charge q 1 = −(R/d)q at radius r 1 = R^2 /d, is needed to make the potential equal at all points on the sphere. The second image charge, of charge q 2 = q − q 1 at the center of the sphere, is necessary to recreate the effect of the additional charge on the sphere (the “additional” charge is the extra charge on the sphere left over after you subtract the surface charge density induced by the point charge q). The force on the point charge is the sum of the forces from the two image charges:
F =
4 π 0
[
qq 1 [ d − R 2 d
] 2 +^
qq 2 d^2
]
q^2 4 π 0
[
−dR [d^2 − R^2 ]^2
d^2 + dR d^4
]
As d → R the denominator of the first term vanishes, so that term wins, and the overall force is attractive. As d → ∞, the denominator of both terms looks like d^4 , so the dR terms in the numerator cancel and the overall force is repulsive.
(a) The crossover distance is found by equating the two bracketed terms in (5):
dR [d^2 − R^2 ]^2
d^2 + dR d^4 d^4 R = (d + R)[d^2 − R^2 ]^2 0 = d^5 − 2 d^3 R^2 − 2 d^2 R^3 + dR^4 + R^5
I used GnuPlot to solve this one graphically. The root is d/R=1.6178.
(b) The idea here is to set d = R + a = R(1 + a/R) and find the limit of (4) as a → 0.
F =
q^2 4 π 0
[
−R^2 (1 + (^) Ra ) [ R^2 (1 + (^) Ra )^2 − R^2
] 2 +^
R^2
[
(1 + a R )^2 + (1 + a R )
]
R^4 (1 + (^) Ra )^4
]
q^2 4 π 0
[
−R^2 − aR 4 a^2 R^2
(2R + 3a)(R − 4 a) R^4
]
The second term in brackets approaches the constant 2/R^2 as a → 0. The first term becomes − 1 / 4 a^2. So we have
F → −
q^2 16 π 0 a^2
Note that only the first image charge (the one required to make the sphere an equipotential) contributes to the force as d → a. The second image charge, the one which represents the difference between the actual charge on the sphere and the charge induced by the first image, makes no contribution in this limit. That means that the limiting value of the force will be as above regardless of the charge on the sphere.
(c) If the charge on the sphere is twice the point charge, then q 2 = 2q − q 1 = q(2 + R/d). Then (5) becomes
F =
q^2 4 π 0
[
dR [d^2 − R^2 ]^2
2 d^2 + dR d^4
]
and the relevant equation becomes
0 = 2d^5 − 4 d^3 R^2 − 2 d^2 R^3 + 2dR^4 + R^5.
Again I solved graphically to find d/R = 1.43. If the charge on the sphere is half the point charge, then
F =
q^2 4 π 0
[
dR [d^2 − R^2 ]^2
d^2 + 2dR 2 d^4
]
and the equation is
0 = d^5 − 2 d^3 R^2 − 4 d^2 R^3 + dR^4 + 2R^5.
The root of this one is d/R=1.88.
its image is
P E =
4 π 0
qq′ |r − r′|
4 π 0
−q^2 a r(r − a^2 /r)
4 π 0
−q^2 a r^2 − a^2
Result (7) is only half of (8). This would seem to violate energy conservation. It would seem that we could start with the point charge at infinity and allow it to fall in to a distance r from the sphere, liberating a quantity of energy (8), which we could store in a battery or something. Then we could expend an energy equal to (7) to remove the charge back to infinity, at which point we would be back where we started, but we would still have half of the energy saved in the battery. It would seem that we could keep doing this over and over again, storing up as much energy in the battery as we pleased. I think the problem is with equation (8). The traditional expression q 1 q 2 / 4 π 0 r for the potential energy of two charges comes from calculating the work needed to bring one charge from infinity to a distance r from the other charge, and it is assumed that the other charge does not move and keeps a constant charge during the process. But in this case one of the charges is a fictitious image charge, and as the point charge q is brought in from infinity the image charge moves out from the center of the sphere, and its charge increases. So the simple expression doesn’t work to calculate the potential energy of the configuration, and we should take (7) to be the correct result.
(b) In this case there are two image charges: one of the same charge and location as in part a, and another of charge Q − q′^ at the origin. The work needed to remove the point charge q to infinity is the work needed to remove the point charge from its image charge, plus the work needed to remove the point charge from the extra charge at the origin. We calculated the first contribution above. The second contribution is
r
q(Q − q′)dy 4 π 0 y^2
4 π 0
r
[
qQ y^2
q^2 a y^3
]
dy
4 π 0
∣−^
qQ y
q^2 a 2 y^2
∞
r
= −
4 π 0
[
qQ r
q^2 a 2 r^2
]
so the total work done is
W =
4 π 0
[
q^2 a 2(r^2 − a^2 )
q^2 a 2 r^2
qQ r
]
Review of Green’s Functions
Some problems in this and other chapters use the Green’s function technique. It’s useful to review this technique, and also to establish my conventions since I define the Green’s function a little differently than Jackson. The whole technique is based on the divergence theorem. Suppose A(x) is a vector valued function defined at each point x within a volume V. Then ∫
V
(∇ · A(x′)) dV ′^ =
S
A(x′) · dA′^ (9)
where S is the (closed) surface bounding the volume V. If we take A(x) = φ(x)∇ψ(x) where φ and ψ are scalar functions, (9) becomes
∫
V
[
(∇φ(x′)) · (∇ψ(x′)) + φ(x′)∇^2 ψ(x′)
]
dV ′^ =
S
φ(x′)
∂ψ ∂n
x′
dA′
where ∂ψ/∂n is the dot product of ∇~ψ with the outward normal to the surface area element. If we write down this equation with φ and ψ switched and subtract the two, we come up with
∫
V
[
φ∇^2 ψ − ψ∇^2 φ
]
dV ′^ =
S
[
φ
∂ψ ∂n
− ψ
∂φ ∂n
]
dA′. (10)
This statement doesn’t appear to be very useful, since it seems to require that we know φ over the whole volume to compute the left side, and both φ and ∂φ/∂n on the boundary to compute the right side. However, suppose we could choose ψ(x) in a clever way such that ∇^2 ψ = δ(x − x 0 ) for some point x 0 within the volume. (Since this ψ is a function of x which also depends on x 0 as a parameter, we might write it as ψx 0 (x).) Then we could use the sifting property of the delta function to find
φ(x 0 ) =
V
[
ψx 0 (x′)∇^2 φ(x′^ )
]
dV ′^ +
S
[
φ(x′)
∂ψx 0 ∂n
x′
− ψx 0 (x′)
∂φ ∂n
x′
]
dA′.
If φ is the scalar potential of electrostatics, we know that ∇^2 ψ(x′) = −ρ(x′)/ 0 , so we have
φ(x 0 ) = −
V
ψx 0 (x′)ρ(x′^ )dV ′^ +
S
[
φ(x′)
∂ψx 0 ∂n
x′
− ψx 0 (x′)
∂φ ∂n
x′
]
dA′. (11) Equation (11) allows us to find the potential at an arbitrary point x 0 as long as we know ρ within the volume and both φ and ∂φ/∂n on the boundary. boundary. Usually we do know ρ within the volume, but we only know either φ or ∂φ/∂n on the boundary. This lack of knowledge can be accommodated by choosing ψ such that either its value or its normal derivative vanishes on the boundary surface, so that the term which we can’t evaluate drops out of the surface integral. More specifically,
Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
December 8, 1999
Chapter 2: Problems 11-
Problem 2.
A line charge with linear charge density τ is placed parallel to, and a distance R away from, the axis of a conducting cylinder of radius b held at fixed voltage such that the potential vanishes at infinity. Find
(a) the magnitude and position of the image charge(s);
(b) the potential at any point (expressed in polar coordinates with the origin at the axis of the cylinder and the direction from the origin to the line charge as the x axis), including the asymptotic form far from the cylinder;
(c) the induced surface-charge density, and plot it as a function of angle for R/b=2,4 in units of τ / 2 πb;
(d) the force on the charge.
(a) Drawing an analogy to the similar problem of the point charge outside the conducting sphere, we might expect that the potential on the cylinder can be made constant by placing an image charge within the cylinder on the line conducting the line charge with the center of the cylinder, i.e. on the x axis. Suppose we put the image charge a distance R′^ < b from the center of the cylinder and give it a charge density −τ. Using the expression quoted in Problem 2.3 for the potential of a line charge, the potential at a point x due to the line charge and its image is
Φ(x) =
τ 4 π 0
ln
R^2
|x − Rˆi|^2
τ 4 π 0
ln
R^2
|x − R′ˆi|^2
τ 4 π 0
ln
|x − R′ˆi|^2 |x − Rˆi|^2
We want to choose R′^ such that the potential is constant when x is on the cylinder surface. This requires that the argument of the logarithm be equal to some constant γ at those points:
|x − R′ˆi|^2 |x − Rˆi|^2
= γ
or b^2 + R′^2 − 2 R′b cos φ = γb^2 + γR^2 − 2 γRb cos φ.
For this to be true everywhere on the cylinder, the φ term must drop out, which requires R′^ = γR. We can then rearrange the remaining terms to find
R′^ =
b^2 R
This is also analogous to the point-charge-and-sphere problem, but there are differences: in this case the image charge has the same magnitude as the original line charge, and the potential on the cylinder is constant but not zero.
(b) At a point (ρ, φ), we have
τ 4 π 0
ln ρ^2 + R′^2 − 2 ρR′^ cos φ ρ^2 + R^2 − 2 ρR cos φ
For large ρ, this becomes
τ 4 π 0
ln
1 − 2 R
′ ρ cos^ φ 1 − 2 R ρ cos φ
Using ln(1 − x) = −(x + x^2 /2 + · · ·), we have
τ 4 π 0
2(R − R′) cos φ ρ
=
τ 2 π 0
R(1 − b^2 /R^2 ) cos φ ρ
(c)
σ = − 0
∂ρ
r=b = − τ 4 π
[
2 b − 2 R′^ cos φ b^2 + R′^2 − 2 bR′^ cos φ
2 b − 2 R cos φ b^2 + R^2 − 2 bR cos φ
]
τ 2 π
[
b − b
2 R cos^ φ b^2 + (^) Rb^42 − 2 b R^3 cos φ
b − R cos φ b^2 + R^2 − 2 bR cos φ
]
Referring to equation (2.71), we know the bn are all zero, because the ln term and the negative powers of ρ are singular at the origin. We are left with
Φ(ρ, φ) = a 0 +
∑^ ∞
n=
ρn^ {an sin(nφ) + bn cos(nφ)}. (1)
Multiplying both sides successively by 1, sin n′φ, and cos n′φ and integrating at ρ = b gives
a 0 =
2 π
∫ (^2) π
0
Φ(b, φ)dφ (2)
an =
πbn
∫ (^2) π
0
Φ(b, φ) sin(nφ)dφ (3)
bn =
πbn
∫ (^2) π
0
Φ(b, φ) cos(nφ)dφ. (4)
Plugging back into (1), we find
Φ(ρ, φ) =
π
∫ (^2) π
0
Φ(b, φ′)
∑^ ∞
n=
( (^) ρ b
)n [sin(nφ) sin(nφ′) + cos(nφ) cos(nφ′)]
dφ′
π
∫ (^2) π
0
Φ(b, φ′)
∑^ ∞
n=
( (^) ρ b
)n cos n(φ − φ′)
The bracketed term can be expressed in closed form. For simplicity define x = (ρ/b) and α = (φ − φ′). Then
1 2
∑^ ∞
n=
xn^ cos(nα) =
∑^ ∞
n=
[
xneinα^ + xne−inα
]
[
1 − xeiα^
1 − xe−iα^
]
[
1 − xe−iα^ − xeiα^ + 1 1 − xeiα^ − xe−iα^ + x^2
]
[
1 − x cos α 1 + x^2 − 2 x cos α
]
x cos α − x^2 1 + x^2 − 2 x cos α
=
[
1 − x^2 1 + x^2 − 2 x cos α
]
Plugging this back into (5) gives the advertised result.
Problem 2.
(a) Two halves of a long hollow conducting cylinder of inner radius b are separated by small lengthwise gaps on each side, and are kept at different potentials V 1 and V 2. Show that the potential inside is given by
Φ(ρ, φ) =
V 1 + V 2
V 1 − V 2
π
tan−^1
2 bρ b^2 − ρ^2
cos φ
where φ is measured from a plane perpendicular to the plane through the gap. (b) Calculate the surface-charge density on each half of the cylinder.
This problem is just like the previous one. Since we are looking for an expression for the potential within the cylinder, the correct expansion is (1) with expansion coefficients given by (2), (3) and (4):
a 0 =
2 π
∫ (^2) π
0
Φ(b, φ)dφ
2 π
[
V 1
∫ (^) π
0
dφ + V 2
∫ (^2) π
π
dφ
]
V 1 + V 2
an =
πbn
[
V 1
∫ (^) π
0
sin(nφ)dφ + V 2
∫ (^2) π
π
sin(nφ)dφ
]
nπbn
[
V 1 |cos nφ|π 0 + V 2 |cos nφ|^2 ππ
]
nπbn^
[V 1 (cos nπ − 1) + V 2 (1 − cos nπ)]
0 , n even 2(V 1 − V 2 )/(nπbn) , n odd
bn =
πbn
[
V 1
∫ (^) π
0
cos(nφ)dφ + V 2
∫ (^2) π
π
cos(nφ)dφ
]
nπbn
[
V 1 |sin nφ|π 0 + V 2 |sin nφ|^2 ππ
]
With these coefficients, the potential expansion becomes
Φ(ρ, φ) =
V 1 + V 2
2(V 1 − V 2 )
π
n odd
n
( (^) ρ b
)n sin nφ. (6)
Problem 2.
(a) Show that the Green function G(x, y; x′^ , y′) appropriate for Dirichlet boundary conditions for a square two-dimensional region, 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1, has an expansion
G(x, y; x′, y′) = 2
∑^ ∞
n=
gn(y, y′) sin(nπx) sin(nπx′)
where gn(y, y′) satisfies ( ∂^2 ∂y^2
− n^2 π^2
gn(y, y′) = δ(y′^ − y) and gn(y, 0) = gn(y, 1) = 0.
(b) Taking for gn(y, y′) appropriate linear combinations of sinh(nπy′) and cosh(nπy′) in the two regions y′^ < y and y′^ > y, in accord with the boundary conditions and the discontinuity in slope required by the source delta function, show that the explicit form of G is
G(x, y; x′, y′) =
− 2
∑^ ∞
n=
nπ sinh(nπ)
sin(nπx) sin(nπx′) sinh(nπy<) sinh[nπ(1 − y>)]
where y< (y>) is the smaller (larger) of y and y′.
(I have taken out a factor − 4 π from the expressions for gn and G, in accordance with my convention for Green’s functions; see the Green’s functions review above.)
(a) To use as a Green’s function in a Dirichlet boundary value problem G must satisfy two conditions. The first is that G vanish on the boundary of the region of interest. The suggested expansion of G clearly satisfies this. First, sin(nπx′) is 0 when x′^ is 0 or 1. Second, g(y, y′) vanishes when y′^ is 0 or 1. So G(x, y; x′, y′) vanishes for points (x′, y′) on the boundary. The second condition on G is
∇^2 G =
∂^2
∂x′^2
∂^2
∂y′^2
G = δ(x − x′) δ(y − y′). (8)
With the suggested expansion, we have
∂^2 ∂x′^2
G = 2
∑^ ∞
n=
gn(y, y′) sin(nπx)
[
−n^2 π^2 sin(nπx′^ )
]
∂^2
∂y′^2
G = 2
∑^ ∞
n=
∂′^2
∂y^2
gn(y, y′) sin(nπx) sin(nπx′)
We can add these together and use the differential equation satisfied by gn to find
∇^2 G = δ(y − y′) · 2
∑^ ∞
n=
sin(nπx) sin(nπx′)
= δ(y − y′) · δ(x − x′)
since the infinite sum is just a well-known representation of the δ function.
(b) The suggestion is to take
gn(y, y′) =
An 1 sinh(nπy′) + Bn 1 cosh(nπy′), y′^ < y; An 2 sinh(nπy′) + Bn 2 cosh(nπy′), y′^ > y. (9)
The idea to use hyperbolic sines and cosines comes from the fact that sinh(nπy) and cosh(nπy) satisfy a homogeneous version of the differential equation for gn (i.e. satisfy that differential equation with the δ function replaced by zero). Thus gn as defined in (9) satisfies its differential equation (at all points except y = y′) for any choice of the As and Bs. This leaves us free to choose these coefficients as required to satisfy the boundary conditions and the differential equation at y = y′. First let’s consider the boundary conditions. Since y is somewhere between 0 and 1, the condition that gn vanish for y′^ = 0 is only relevant to the top line of (9), where it requires taking Bn 1 = 0 but leaves An 1 undetermined for now. The condition that gn vanish for y′^ = 1 only affects the lower line of (9), where it requires that
0 = An 2 sinh(nπ) + Bn 2 cosh(nπ) = (An 2 + Bn 2 )enπ^ + (−An 2 + Bn 2 )e−nπ^ (10)
One way to make this work is to take
An 2 + Bn 2 = −e−nπ^ and − An 2 + Bn 2 = enπ^.
Then
Bn 2 = enπ^ + An 2 → 2 An 2 = −enπ^ − e−nπ so An 2 = − cosh(nπ) and Bn 2 = sinh(nπ).
With this choice of coefficients, the lower line in (9) becomes
gn(y, y′) = − cosh(nπ) sinh(nπy′)+sinh(nπ) cosh(nπy′) = sinh[nπ(1−y′)] (11)
for (y′^ > y). Actually, we haven’t completely determined An 2 and Bn 2 ; we could multiply (11) by an arbitrary constant γn and (10) would still be satisfied. Next we need to make sure that the two halves of (9) match up at y′^ = y:
An 1 sinh(nπy) = γn sinh[nπ(1 − y)]. (12)
