Contents
7 Boundary Value Problems
7.1 One dimensional wave equation . . . . . . . . . . .
7.2 One dimensional heat equation . . . . . . . . . . .
7.3 The Laplace equation . . . . . . . . . . . . . . . . .
7.4 Strum Liouville Boundary Value Problem (SLBVP)
7.4.1 Orthogonality of eigenfunctions . . . . . . .
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Note: Comments/suggestions on these lecture notes are welcome on the e-mail: sukuyd@gmail.com
to Dr. Suresh Kumar.
1
Chapter 7
Boundary Value Problems
In this chapter, we shall discuss the solution of some boundary value problems.
7.1
One dimensional wave equation
Consider an elastic string of negligible mass and length π tied at the two ends along x-axis at the
points (0, 0) and (π, 0). Suppose the string is pulled in the shape y = f (x) in the xy-plane and
released. Then it can be shown that the vibrations of the string in the xy-plane are governed by
the one dimensional wave equation
1 ∂ 2y
∂ 2y
= 2 2,
∂x2
a ∂t
(7.1)
where a is some positive constant, and y(x, t) is the displacement/vibration of the string along
y-axis direction. The wave equation is subjected to the following four conditions.
The first condition is
y(0, t) = 0,
(7.2)
since the left end of the string is tied at (0, 0) for all the time, and hence it can not have displacement
along the y-axis.
The second condition is
y(π, t) = 0
(7.3)
since the right end of the string is tied at (π, t) for all the time, and hence it can not have
displacement along the y-axis.
The third condition is
∂y
= 0,
∂t
at t = 0,
(7.4)
since the string is in rest at t = 0.
The fourth condition is
y(x, 0) = f (x),
(7.5)
since the string is in the shape y = f (x) at t = 0.
2
Once the string is released from the initial shape y(x, 0) = f (x), we are interested to find the
distance or displacement of the string from the x-axis at any time t. It is equivalent to saying that
we are interested to solve (7.1) for y(x, t) subject to the four conditions (7.2)-(7.5).
Assume that (7.1) possesses a solution of the form
y(x, t) = u(x)v(t),
(7.6)
where u(x) and v(t) are to be determined. Plugging (7.6) into (7.1), we get
1 v ′′ (t)
u′′ (x)
= 2
= λ,
u(x)
a v(t)
(7.7)
where λ is some constant. This yields following two equations
u′′ (x) − λu(x) = 0,
(7.8)
v ′′ (t) − λa2 v(t) = 0.
(7.9)
Now, let us first solve (7.8). Later, we shall look for the solution of (7.9). Considering (7.6), the
condition y(0, t) = 0 in (7.2) gives u(0)v(t) = 0 or u(0) = 0. Similarly, y(π, t) = 0 in (7.3) gives
u(π) = 0. Further, we see that the nature of solution of (7.8) depends on the values of λ.
√
√
(i) When λ > 0, the solution reads as u(x) = c1 e λx + c2 e− λx . Using the conditions u(0) = 0
and u(π) = 0, we get c1 = 0 = c2 , and hence u(x) = 0. This leads to the trivial solution
y(x, t) = u(x)v(t) = 0, which is not of our interest.
(ii) When λ = 0, the solution reads as u(x) = c1 x + c2 . Again, using the conditions u(0) = 0
and u(π) = 0, we get c1 = 0 = c2 , which leads to the trivial solution y(x, t) = u(x)v(t) = 0.
(iii) When λ < 0, say, λ = −n2 , the solution reads as u(x) = c1 sin nx + c2 cos nx. Applying
the condition u(0) = 0, we get c2 = 0. The condition u(π) = 0 then implies that c1 sin nπ = 0.
Obviously, for a non-trivial solution we must have c1 ̸= 0. Then the condition c1 sin nπ = 0 forces
n to be a positive integer. Thus,
un (x) = sin nx,
(7.10)
is non-trivial solution of (7.8) for each positive integer n.
Now, the solution of (7.9) with λ = −n2 reads as v(t) = c1 sin nat + c2 cos nat. The condition
in (7.4) leads to u(x)v ′ (0) = 0 or v ′ (0) = 0, which in turn gives c1 = 0. So
vn (t) = cos nat,
(7.11)
is non-trivial solution of (7.8).
In view of (7.6), (7.10) and (7.11), we can say that
yn (x, t) = un (x)vn (t) = sin nx cos nat,
(7.12)
3
is a solution of (7.1) for each positive integer n. It follows that
y(x, t) =
∞
X
bn yn (x, t) =
n=1
∞
X
bn sin nx cos nat,
(7.13)
n=1
is also a solution of (7.1). To determine bn , we use the fourth condition y(x, 0) = f (x) given in
(7.5). Then (7.13) gives
f (x) =
∞
X
bn sin nx.
(7.14)
n=1
Notice that the series on right hand side in (7.14) is the Fourier sine series of f (x) in the interval
[0, π]. So we have
Z
2 π
bn =
f (x) sin nxdx.
(7.15)
π 0
Hence,
y(x, t) =
∞
X
bn sin nx cos nat,
(7.16)
n=1
with bn from (7.15) is the solution of (7.1) subject to the four conditions (7.2)-(7.5).
7.2
One dimensional heat equation
Consider a uniform rod of length π aligned along x-axis from (0, 0) to (π, 0). Suppose that the two
ends of the rod are kept at zero temperature all the time, and f (x) represents the temperature
function at time t = 0. Then it can be shown that the temperature w(x, t) of the road is governed
by the one dimensional heat equation
∂ 2w
1 ∂w
= 2
,
2
∂x
a ∂t
(7.17)
where a is some positive constant. The heat equation is subjected to the following three conditions.
The first condition is
w(0, t) = 0,
(7.18)
since the left end of the rod is kept at zero temperature for all t.
The second condition is
w(π, t) = 0
(7.19)
since the right end of the rod is kept at zero temperature for all t.
The third condition is
w(x, 0) = f (x),
(7.20)
4
since the temperature of the rod is given by f (x) at t = 0.
Having known the temperature of the rod at t = 0, we are interested to find the temperature
of the rod at any time t. It is equivalent to saying that we are interested to solve (7.17) for w(x, t)
subject to the three conditions (7.18)-(7.20).
Assume that (7.17) possesses a solution of the form
w(x, t) = u(x)v(t),
(7.21)
where u(x) and v(t) are to be determined. Plugging (7.21) into (7.17), we get
1 v ′ (t)
u′′ (x)
= 2
= λ,
u(x)
a v(t)
(7.22)
where λ is some constant. This yields following two equations
u′′ (x) − λu(x) = 0,
(7.23)
v ′ (t) − λa2 v(t) = 0,
(7.24)
Following the strategy discussed in the previous section, the non-trivial solution of (7.23) subject
to the conditions (7.18) and (7.19), reads as
un (x) = sin nx,
(7.25)
where n a positive integer with λ = −n2 .
2 a2 t
Now, the solution of (7.24) with λ = −n2 reads as v(t) = c1 e−n
2 a2 t
vn (t) = e−n
,
. So
(7.26)
is non-trivial solution of (7.24).
In view of (7.21), (7.25) and (7.26),
2 a2 t
wn (x, t) = un (x)vn (t) = sin nxe−n
,
(7.27)
is a solution of (7.17) for each positive integer n. It follows that
w(x, t) =
∞
X
bn wn (x, t) =
n=1
∞
X
2 a2 t
bn sin nxe−n
,
(7.28)
n=1
is also a solution of (7.17). To determine bn , we use the third condition w(x, 0) = f (x) given in
(7.20). Then (7.28) gives
f (x) =
∞
X
bn sin nx.
(7.29)
n=1
5
Notice that the series on right hand side in (7.29) is the Fourier sine series of f (x) in the interval
[0, π]. So we have
Z
2 π
f (x) sin nxdx.
(7.30)
bn =
π 0
Hence,
w(x, t) =
∞
X
bn sin nxe−n
2 a2 t
,
(7.31)
n=1
with bn from (7.30) is the solution of (7.17) subject to the three conditions (7.18)-(7.20).
7.3
The Laplace equation
The steady state temperature w(x, y) (independent of time) in the two dimensional xy-plane is
governed by
∂ 2w ∂ 2w
+
= 0,
∂x2
∂y 2
(7.32)
known as the Laplace equation. With the transformations x = r cos θ and y = r sin θ, the polar
form of (7.32) reads as
1 ∂ 2w
∂ 2 w 1 ∂w
+
+
= 0.
∂r2
r ∂r
r2 ∂θ2
(7.33)
For,
∂w
∂w ∂x ∂w ∂y
∂w
∂w
=
+
= cos θ
+ sin θ
.
∂r
∂x ∂r
∂y ∂r
∂x
∂y
2
2
∂ 2w
∂ 2w
∂ 2w
2 ∂ w
2 ∂ w
=
cos
θ
+
cos
θ
sin
θ
+
sin
θ
+
sin
θ
cos
θ
∂r2
∂x2
∂x∂y
∂y 2
∂x∂y
∂w
∂w ∂x ∂w ∂y
∂w
∂w
=
+
= −r sin θ
+ r cos θ
.
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
2
2
∂ 2w
∂ 2w
∂w 2
∂ 2w
∂w
2
2 ∂ w
2
2 ∂ w
2
=
r
sin
θ
−r
sin
θ
cos
θ
−r
cos
θ
sin
θ
−r
cos
θ
+r
cos
θ
−r
sin
θ
.
∂θ2
∂x2
∂x∂y
∂x
∂y 2
∂x∂y
∂y
2
2
Substituting the values of ∂∂rw2 , ∂w
and ∂∂θw2 into (7.33), we get (7.32).
∂r
Suppose the steady state temperature is given on the boundary of a unit circle r = 1, say
w(1, θ) = f (θ). Then the problem of finding the temperature at any point (r, θ) inside the circle
is a Dirichlet’s problem for a circle. Now we shall solve (7.33) subject to the condition
w(1, θ) = f (θ).
(7.34)
Assume that (7.33) possesses a solution of the form
w(r, θ) = u(r)v(θ),
(7.35)
6
where u(r) and v(θ) are to be determined. Plugging (7.35) into (7.33), we get
v ′′ (θ)
r2 u′′ (r) + ru′ (r)
=−
= λ,
u(r)
v(θ)
(7.36)
where λ is some constant. This yields following two equations
v ′′ (θ) + λv(θ) = 0,
(7.37)
r2 u′′ (r) + ru′ (r) − λu(r) = 0.
(7.38)
The non-trivial solution of (7.37) reads as
vn (θ) = an cos nθ + bn sin nθ,
(7.39)
where λ = n2 ; an , bn are constants such that both the terms on right hand side of (7.39) do not
vanish together for n = 1, 2, 3, ......
Notice that (7.38) is a Cauchy-Euler DE with λ = n2 . So it transforms to
d2 u
− n2 u = 0,
ds2
(7.40)
where r = es . Solution of this equation is
u(z) = c1 ens + c2 e−ns
for n = 1, 2, 3, .....
where c1 and c2 are constants. In terms of r, the solution read as
u(r) = c1 rn + c2 r−n
for n = 1, 2, 3, .....
Since we are interested in solutions which are well defined inside the circle r = 1, we discard the
second term r−n as it is not finite at r = 0. Thus, the solution of our interest is
un (r) = rn , n = 1, 2, 3, .....
(7.41)
In view of (7.35), (7.39) and (7.41),
wn (r, θ) = un (r)vn (θ) = rn (an cos nθ + bn sin nθ),
(7.42)
is a solution of (7.33) for n = 1, 2, ....... It follows that
∞
X
wn (x, t) =
n=0
∞
X
rn (an cos nθ + bn sin nθ),
(7.43)
n=1
is also a solution of (7.33). Since
a0
2
is also a solution of (7.33), so
∞
w(r, θ) =
a0 X n
+
r (an cos nθ + bn sin nθ),
2
n=1
is also a solution of (7.33).
7
(7.44)
To determine a0 , an and bn , we use the third condition w(1, θ) = f (θ) given in (7.34). Then
(7.44) gives
∞
a0 X
+
(an cos nθ + bn sin nθ),
f (θ) =
2
n=1
(7.45)
Notice that the series on right hand side in (7.45) is the Fourier series of f (θ) in the interval [−π, π].
So we have
Z
1 π
an =
f (ϕ) cos nϕdϕ, (n = 0, 1, 2, ....)
(7.46)
π −π
1
bn =
π
Z
π
f (ϕ) sin nϕdϕ, (n = 1, 2, 3, ......).
(7.47)
−π
Thus, (7.44) with an from (7.46) and bn from (7.47) is the solution of (7.33) subject to the
condition (7.34). Thus, the Dirichlet problem for the unit circle is solved.
Now substituting an from (7.46) and bn from (7.47) into (7.44), we get
#
"
Z
∞
1 π
1 X n
r cos n(θ − ϕ) dϕ.
(7.48)
w(r, θ) =
f (ϕ)
+
π −π
2 n=1
Let z = rei(θ−ϕ) = r[cos(θ − ϕ) + i sin(θ − ϕ)] so that z n = rn [cos n(θ − ϕ) + i sin n(θ − ϕ)]. Then
we have
"
#
∞
∞
1 X n
1 X n
+
r cos n(θ − ϕ) = Re
+
z
2 n=1
2 n=1
z
1
+
= Re
2 1−z
1+z
= Re
2(1 − z)
(1 + z)(1 − z̄)
= Re
2(1 − z)(1 − z̄)
(1 + z − z̄ − |z|2 )
= Re
2(1 − z − z̄ + |z|2 )
1 − r2
=
2[1 − 2r cos(θ − ϕ) + r2 ]
So (7.48) becomes
1
w(r, θ) =
2π
Z
π
−π
1 − r2
f (ϕ)dϕ,
1 − 2r cos(θ − ϕ) + r2
(7.49)
known as the Poission integral. It expresses the value of the harmonic function w(r, θ) at all points
inside the circle r = 1 in terms of its values on the circumference of the circle. In particular, at
r = 0, we have
Z π
1
w(0, θ) =
f (ϕ)dϕ,
(7.50)
2π −π
which shows that the value of the harmonic function w at the center of the circle is the average of
its values on the circumference.
8
7.4
Strum Liouville Boundary Value Problem (SLBVP)
Let p(x) ̸= 0, p′ (x), q(x) and r(x) be continuous functions on [a, b]. Then the DE
d
[p(x)y ′ ] + [λq(x) + r(x)]y = 0,
dx
with the boundary conditions
(7.51)
c1 y(a) + c2 y ′ (a) = 0,
(7.52)
d1 y(b) + d2 y ′ (b) = 0,
(7.53)
and
where neither both c1 and c2 nor both d1 and d2 are zero, is called a SLBVP. We see that y = 0 is
trivial solution of (7.51). The values of λ for which (7.51) has non-trivial solutions, are known as
its eigenvalues while the corresponding non-trivial solutions are known as eigenfunctions.
Ex. 7.4.1. Find eigenvalues and eigenfunctions of the SLBVP
y ′′ + λy = 0,
y(0) = 0, y(π) = 0.
Solution 7.4.1. Eigenvalues are λ = n2 , where n is a positive integer.
Eigenfunctions are yn = sin nx.
7.4.1
Orthogonality of eigenfunctions
Consider the SLBVP given by (7.51), (7.52) and (7.52). If ym and yn are any two distinct eigenfunctions corresponding to the eigenvalues λm and λn , then
Z b
q(x)ym (x)yn (x)dx = 0.
a
In other words, any two distinct eigen functions ym and yn of the SLBVP are orthogonal with
respect to the weight function q(x). Let us prove this result.
Since ym and yn are eigenfunctions corresponding to the eigenvalues λm and λn , we have
′ ′
(pym
) + (λm q + r)ym = 0
(7.54)
(pyn′ )′ + (λn q + r)yn = 0.
(7.55)
and
Multiplying (7.54) by yn and (7.55) by ym , and subtracting we get
′ ′
yn (pym
) − ym (pyn′ )′ + (λm − λn )qym yn = 0
(7.56)
Moving the first two terms on right hand side, and then integrating from a to b, we have
Z b
Z b
Z b
′ ′
′ ′
yn (pym
) dx
(λm − λn )
qym yn dx =
ym (pyn ) dx −
a
a
a
Z b
Z b
′
′
b
′
′ b
′
= [ym (pyn )]a −
ym (pyn )dx − [yn (pym )]a +
yn′ (pym
)dx
a
a
′
′
(a)]
= p(b)[ym (b)yn′ (b) − yn (b)ym
(b)] − p(a)[ym (a)yn′ (a) − yn (a)ym
= p(b)W (b) − p(a)W (a)
9
′
where W (x) = ym (x)yn′ (x) − yn (x)ym
(x) is Wronskian of ym and yn .
Z
∴ (λm − λn )
b
qym yn dx = p(b)W (b) − p(a)W (a).
(7.57)
a
Notice that the eigen functions ym and yn are particular solutions of the SLBVP given by (7.51),
(7.52) and (7.52). So we have
′
c1 ym (a) + c2 ym
(a) = 0,
(7.58)
c1 yn (a) + c2 yn′ (a) = 0,
(7.59)
′
(b) = 0,
d1 ym (b) + d2 ym
(7.60)
d1 yn (b) + d2 yn′ (b) = 0.
(7.61)
Now by the given, c1 and c2 are not zero together. So the homogeneous system given by (7.58)
′
(a) = W (a) must be
and (7.59) has a non-trivial solution. It follows that ym (a)yn′ (a) − yn (a)ym
′
zero. Likewise, (7.60) and (7.61) lead to ym (b)yn′ (b) − yn (b)ym
(b) = W (b) = 0. So (7.57) becomes
Z
(λm − λn )
b
qym yn dx = 0.
(7.62)
a
Also, λm ̸= λn . So we get
Z b
qym yn dx = 0,
(7.63)
a
the desired result.
Remark 7.4.1. The orthogonality property of eigenfunctions can be used to write a given function
as the series expansion of eigenfunctions.
Remark 7.4.2. A DE in the form
d
[p(x)y ′ ] + [λq(x) + r(x)]y = 0
dx
is called in self adjoint form.
10