Chapter 4
1D Linear Second-Order Equations
In this chapter, our primary focus is on introducing the series solution approach for 1D heat
and wave diﬀerential equations. Building upon the foundations laid in the preceding chapters,
where we explored the derivation of these equations in higher dimensions and analyzed
their overarching behavior, we now delve into the speciﬁc context of solving heat and wave
equations within b o unded media extending along the x-axis, deﬁned as the interval (x
0
; x
1
).
As we navigate this chapter, we encounter a crucial factor that heavily inﬂuences the
nature of the solution: the conditions at the boundary points x=x
0
and x=x
1
. This prompts
the eme rgence of boundary value problems a s a pivo tal concept in th is context. To enhance
our understanding, we introduce a renowne d theorem, the Sturm-Liouville theorem, w hich
assumes a critical role in our exploration of the e igenfunction expan sion method. This
method becomes instrumental in representing the solutions of 1D partial diﬀerential equa-
tions.
4.1 From PDE to e igenvalue problem
This section serves as a bridge between partial diﬀerential equations (PDEs) and eigenvalue
problems.
4.1.1 Outline of the method
We'll focus on the context of an open interval =(x
0
; x
1
), where we encounter a PDE that
governs the function u=u(x; t):
u
t
= L[u]
where L takes the f o rm of a 1D diﬀerential o perator:
L := a(x)
@
2
@x
2
+ b(x)
@
@x
+ c(x): (4.1)
In our scenarios, a; b, and c are continuou s functions deﬁned within . This equation ﬁnds
practical applications, like modeling temperature distribution along an extended conductive
rod within . However, to establish a complete formulation, we need to determine conditions
at both endp o ints, x=x
0
and x=x
1
. This is because the diﬀerential equation is conﬁned to
the open interval .
1
We proceed by assuming that the function u adhere s to a particular condition at th ese
endpoints:
αu + βu
x
= 0:
Additionally, the initial temperature distribution, u(x; 0), along the rod must be s peciﬁed.
The interplay between the thermal energy distribution along the rod and heat diﬀusivity
generates the dynamic heat behavior of the system. By combining these elements, we present
a comprehensive initial-boundary value problem:
8
<
:
u
t
= L[u] on
αu + βu
x
= 0 on bnd(Ω)
u(x; 0) = f(x) initial condition
:
This synthesis of the diﬀerential operator L, boundary conditions, and initial condition lays
the foundation fo r solving various heat and wave equations, gradually unraveling the intricate
dynamics of these systems.
As L is a linear operator that operates solely on the spatial variable x, we can treat the
solution of the equation u
t
= L[u] as an ordinary diﬀerential equation for u and express it as:
u
t
= L[u], as an ordinary diﬀerential equation fo r u and write it as: u(x; t) = φ(x) e
¡λt
where
φ(x) is an unknown function and λ is a constant. To verify its validity, substituting this into
the equation gives:
¡λφ(x) e
¡λt
= e
¡λt
L[φ];
which leads to: L[φ] = ¡λφ. To satisfy the prescribed boundary conditions for u, we arrive
at the subsequent boundary value problem, often termed the eigenvalue problem:
L[φ] = ¡λφ on
αφ + βφ
0
= 0 ob bnd(Ω)
Remark 4.1. The use of terminology ca n be justiﬁed by drawing parallels with the e igen-
value problem in linear algebra. Remember that a vector v in R
n
is an eigenvector of a linear
transformation T :R
n
!R
n
if T [v]+λv=0. In the context of our problem, the operator L acts
linearly on the space of smooth functions deﬁned on . Consequently, a function φ: !R
is termed an eigenfunction if L[φ](x) + λφ(x)=0 holds for all x in . Following the analogy
with the eigenvalue problem in linear algebra, the va lue λ is referred to as the associated
eigenvalue of the eigenfunction φ. It's important to note that if φ is an eigenfunction of
the problem, then c φ(x) for any non-zero constant c is also an eigenfunction. Just like
eigenvectors in linear algebra, eigenfunctions are typically considered as non-trivial functions.
Solving an eigenvalue problem involves determining the eigenfunctions φ(x) and eigen-
values λ. When the problem can be solved, the solution of the equation u
t
= L[u] while
adhering to the boundary condition αu + β u
x
= 0 can be expressed as follows: u(x; t) =
φ(x) e
¡λt
.
Example 4.1. Consider the interval (0; 1) and let represent this interval. Let's explore
the eigenvalue problem φ
00
= ¡λφ, where φ(x) is deﬁned on . The boundary conditions
are set as φ(0) = 0 and φ(1) = 0. We want to ﬁnd values of λ and corresponding non-trivial
functions φ(x) that satisfy this equation.
2 1D Linear Second-Order Equations
First, let's show that the problem h as a solution when λ is greater th a n zero. To do this,
we multiply the equation by φ(x) and integrate it over the interval (0; 1). This gives us an
equation involving integrals and the boundary term:
¡
Z
0
1
jφ
0
(x)j
2
dx + φ
0
(x) φ(x)
0
1
= ¡λ
Z
0
1
jφ(x)j
2
dx;
where we used the integration by parts formula for the integral at the le f t-hand side. The
boundary term φ
0
(x) φ(x)j
0
1
is equal zero due to the given boundary cond itions. This gives
the following formula for λ:
λ =
Z
0
1
jφ
0
(x)j
2
dx
Z
0
1
jφ(x)j
2
dx
0:
If we were to assume λ = 0, we'd ﬁnd that φ(x) turns out to be a constant function. But
considering the boundary conditions and continuity, we'd be left with the trivial function
φ(x) = 0. Therefore, λ must b e greater than zero.
With this understanding, we proceed to solve the equation φ
00
= ¡λφ as an ordinary
diﬀerential equation. The general soluti on for φ(x) can be expressed a s:
φ(x) = A cos( λ
p
x) + B sin( λ
p
x):
Plugging in the boundary condition φ(0) = 0, we ﬁnd that A = 0. Applying the second
boundary condition, we determine that the eigenvalues λ are given by λ = n
2
π
2
for n = 1; 2; ···.
For each eigenvalue λ
n
, we obtain an associated eigenfunction φ
n
(x) = B
n
sin(nπx),
where B
n
is a non-zero constant. This set of eigenvalues and eigenfunctions forms an inﬁnite
collection: φ
1
(x) =sin(πx), λ
1
= π
2
, φ
2
(x) =sin(2πx), λ
2
= 4π
2
, φ
3
(x) =sin(3πx), λ
3
= 9π
2
and
so on. These eigenfunctions provide the building blocks for s o lv ing more complex diﬀerential
equations and understanding the behav ior of solutions on the interval (0; 1).
Exercise 4.1. Let's explore another approach to demonstrate that λ > 0 in the previously discussed
eigenvalue p roblem. Thi s will involve direct calculations. Begin by considering the algebraic character-
istic equation for the diﬀerential equation φ
00
= ¡λφ, which results in r = ± ¡λ
p
.
Task 1: Suppose λ < 0. Proceed by establishing the g ener al solution for this ordinary diﬀerential
equation (ODE). Afterward, apply the given boundary conditions and deduce that the only possible
outcome is φ = 0.
Task 2 : Assume that λ = 0. Derive the general solution for this ODE and arrive at the conclusion
that φ = 0 is again the only feasible solution.
Task 3: Now f ocus on the case where λ> 0. Determin e the non-trivial eigenfunctions that correspond
to these positive eigenvalues.
Exercise 4.2. Building upon the insights gained from the previous example, let's apply that knowledge
to draw conclusions about the heat problem given as:
u
t
= u
xx
x 2(0; 1)
u(0; t) = u(1; t) = 0
:
Task: Utilize the outcomes from the previous example and deduce that this heat problem possesses an
inﬁnite number of solutions, denoted as u
n
(x; t) = e
¡n
2
π
2
t
sin(nπx) for any positive integer n = 1; 2; ···.
These solutions are often referred to as “s eparated solutions” due to their form.
4.1 From PDE to eigenvalue problem 3
Additionally, employing the sup erposition principle, you can express the general solution as:
u(x; t) =
X
n=1
1
c
n
u
n
(x; t);
Here, the constants c
n
can take any arbitrary values, provided that the inﬁnite series converges to a valid
function.
4.1.2 Separation of variables
The derivation of the eigenva lue problem discussed a bove can also be expressed in terms of
the separation o f variables technique. By taking u(x; t) as the separated function:
u(x; t) = φ(x) U(t);
and substituting this into the heat equation, we arrive at:
U
0
U
=
L[φ]
φ
;
where L[φ] represents the operator applied to φ(x). This equation is possible only when the
ratios are equal to the same dimensionless constant. Let's denote this constant by ¡λ. The
minus sign is used for historical reasons and does not carry a physical meaning, but it will
become clear why this choice is more appropriate for subsequent calculations.
By introducing ¡λ as the eigenvalue, we obtain the following ordinary diﬀerential equa-
tion for U (t):
U
0
= ¡λU ;
which is solved as U(t)=Ce
¡λt
, where C is a c o nstant. The eigenvalue λ and eigenfunction
φ(x) are determined by solving the second ordinary diﬀerential equation:
L[φ] = ¡λφ;
subject to the boundary conditions: αφ + βφ
0
= 0 at the boundary points.
The derivation holds true for gene ral 1 D pa rtial diﬀerential equations. Consider the
following second-order PD E for u(x; t) on the interval x 2[x
0
; x
1
]:
8
<
:
a
1
(t)u
tt
+ a
2
(t) u
t
= L[u]
α
1
u(x
0
; t) + β
1
u
x
(x
0
; t) = 0
α
2
u(x
1
; t) + β
2
u
x
(x
1
; t) = 0
;
where a
1
(t) a nd a
2
(t) a re continuous functions. The equation is general enough to encompass
all types of 1D heat and wave equations. To nd a solution to this PDE, we consider the
solution in the separated fo rm as u(x; t)=U(t)φ(x). By substituting this separated function
into the PDE, we obtain the equality
a
1
(t)U
00
+ a
2
(t)U
0
U
=
L[φ]
φ
:
4 1D Linear Second-Order Equations
Obviously, the equality holds only if the ratio is a dimensionless constant since x and t are
indep endent variables. Therefore, we can write
a
1
(t)U
00
+ a
2
(t)U
0
U
=
L[φ]
φ
= ¡λ;
where λ is a constant. Here, the negative sign does not carry any signiﬁcant physical meaning;
it is merely a historical convention. Therefore, the separated form of the solution leads to
the following ordinary diﬀerential equations:
1. For U(t), we have the equation:
a
1
(t) U
00
+ a
2
(t) U
0
= ¡λU ;
2. For φ(x), we have the eigenvalue problem:
8
<
:
L[φ] = ¡λφ on
α
1
φ(x
0
) + β
1
φ
0
(x
0
) = 0
α
2
φ(x
0
) + β
2
φ
0
(x
1
) = 0
:
4.1.3 Series solution
Superposition solutions hold value when the coeﬃcients c
n
are carefully chosen to ensure
that the inﬁnite series converges into a smooth function u(x; t). The convergence of inﬁnite
function s eries is a complex topic beyond this book's scope. Nevertheless, we provide illus-
trative examples to convey the concept of f unction series convergence.
Example 4.2. Consider the interval (0; 1), w here we delve into the followin g inﬁnite series:
X
n=1
1
1
n
sin(x):
Upon observation, it's evident that this series doesn't converge to a continuous function
within the interval (0; 1). A clear instance a rises when x =
1
2
, res ulting in the series becoming
a numerical series:
X
n=1
1
(¡1)
n¡1
2n ¡1
:
This numerical series exhibits conditional convergence, lacking absolute convergence. Now
explore the s eries:
X
n=1
1
1
n
2
sin(x):
We ascertain that this series demonstrates a bso lute convergence. This is evident from the
inequality:
X
n=1
1
1
n
2
sin(nπx)
X
n=1
1
1
n
2
< 1:
The ﬁgure below depicts the truncated series of the above inﬁnite series up to 100 terms :
4.1 From PDE to eigenvalue problem 5
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
1.2
These examples underscore the intricate convergence behaviors of inﬁnite series and their
implications on the continuity of functions within a speciﬁed interval.
In the follow ing sections, we will delve into other ty pes of convergence. However, the
discussed form of convergence remains fundamentally crucial for comprehending initial value
problems. Consider th e subsequent heat problem as an example:
u
t
= u
xx
x 2(0; 1)
u(0; t) = u(1; t) = 0
;
alongside the initial condition u(x; 0) = f(x). Given the general series solution for this heat
problem as:
u(x; t) =
X
n=1
1
c
n
e
¡λ
n
t
sin(nπx):
The signiﬁcance of the convergenc e of this inﬁnite series as t approaches 0 becomes evident.
This convergence property implies:
lim
t!0
u(x; t) = lim
t!0
X
n=1
1
c
n
e
¡λ
n
t
sin(nπx) =
X
n=1
1
c
n
sin(nπx) = f (x):
Consequently, the coeﬃcients c_n must be chosen carefully, ensuring that:
lim
N !1
X
n=1
N
c
n
sin(nπx) = f (x):
As an illustration, consider the case where f (x) = x(1 ¡ x). For this scenario, appropriate
values for c
n
are determined as follows:
c
n
=
(
8
n
3
π
3
n: odd
0 n: even
:
With this speciﬁc selection of c
n
, the series:
X
n=1
1
c
n
e
¡λ
n
t
sin(nπx);
6 1D Linear Second-Order Equations
successfully converges to the initial condition as t approaches 0, aligning with the prescribed
values of c
n
.
Exercise 4.3. Let be the interval Ω: (0; 1):
a) Solve the following eigenvalue problem
φ
00
= ¡λφ on
φ
0
(0) = φ
0
(1) = 0 on bnd(Ω)
:
b) Use this eigenvalue problem and solve the following heat problem
u
t
= ku
xx
on
u
x
(0; t) = u
x
(1; t) = 0
;
for k > 0.
Exercise 4.4. Let be the interval Ω: (0; 1):
a) Solve the following eigenvalue problem
φ
00
= ¡λφ on
φ(0) = φ
0
(1) = 0 on bnd(Ω)
:
b) Use this eigenvalue problem and solve the following heat problem
u
t
= ku
xx
on
u(0; t) = u
x
(1; t) = 0
;
for k > 0.
c) Find the steady state solution
lim
t!1
u(x; t):
Exercise 4.5. Let be the interval Ω: (0; 1):
a) Solve the following eigenvalue problem
φ
00
= ¡λφ on
φ
0
(0) = φ(1) = 0 on bnd(Ω)
:
b) Use this eigenvalue problem and solve the following heat problem
u
t
= ku
xx
on
u
x
(0; t) = u(1; t) = 0
;
for k > 0.
c) Find the steady state solution
lim
t!1
u(x; t):
Exercise 4.6. Let be the interval Ω: (0; 1):
a) Fin d eigenvalues and eigenfunctions of the following eigenvalue problem
8
>
>
>
>
<
>
>
>
>
:
φ
00
= ¡λφ
φ
¡
¡
π
2
= 0
φ
¡
π
2
= 0
:
b) Use these eigenfunct ions and solve the following wave equation
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
u
tt
= c
2
u
xx
x 2
¡
¡
π
2
;
π
2
u
¡
¡
π
2
; t
= u
¡
π
2
; t
= 0
u(x; 0) = cos(x) + sin(2x)
u
t
(x; 0) = 0
:
4.1 From PDE to eigenvalue problem 7
4.1.4 Problems
Problem 4.1. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem
(
φ
00
= ¡λφ
φ
¡
1
2
= φ
¡
3
4
= 0
:
Problem 4.2. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem
8
<
:
φ
00
= ¡λφ
φ(¡1) = φ(1)
φ
0
(¡1) = φ
0
(1)
:
Problem 4.3. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem
8
>
>
>
>
<
>
>
>
>
:
φ
00
= ¡λφ
φ
¡
¡1
2
= 0
φ
0
¡
1
2
= 0
:
Problem 4.4. Find the general ser ies solution of the following wave equation
(
u
tt
= c
2
u
xx
x 2(0; 1)
u(0; t) = u(1; t) = 0
:
Problem 4.5. Consider the following damped wave equation
(
u
tt
+ 2ξu
t
= c
2
u
xx
x 2(0; 1)
u(0; t) = u(1; t) = 0
:
a) For what value of damping factor ξ, the system is in the underdamped state? The underdamped
state is the state where the t-compo nent of the solution is in the form
U(t) = e
σt
[A cos(!t) + sin(!t)];
for some real parameters σ and !.
b) Find the general series solution of the problem when the system is under damped.
Problem 4.6. Consider the following eigenvalue problem on x 2(1; e)
8
<
:
L[φ] = ¡λφ
φ(1) = 0
φ(e) = 0
;
where L stands for the diﬀerential operator
L := x
2
d
2
dx
2
+ x
d
dx
:
a) Use the transformation x = e
s
and convert the given eigenvalue problem to the the following one
8
<
:
'
00
(s) = ¡λ'(s)
'(0) = 0
'(1) = 0
:
Solve t he transformed problem and ﬁnd eigenvalues and eigenfunctions of the oper ator L.
b) Use this results and solve the following problem
8
>
>
<
>
>
:
u
t
= x
2
u
xx
+ xu
x
u(1; t) = 0
u(e; t) = 0
:
8 1D Linear Second-Order Equations