APMA 2101 Supplementary Handout
Introduction to Applied Mathematics
Inhomogeneous ODE BVPs:
Green’s Functions and Eigenfunction Expansions
1
Green’s Function for the Laplacian
The Laplacian operator (often written in coordinateinvariant and dimensioninvariant form as
∇
2
) appears
as a term in the partial differential equations describing most of the conservation laws of continuum me
chanics, electrodynamics, and in many other PDE systems. The potential equation with sources is
∇
2
u
=
f
.
This inhomogeneous problem is also known as “Poisson’s equation,” whereas the homogeneous problem is
also known as “Laplace’s equation.” Here
f
can be gravitational mass density, electrical charge density, or
other types of sources in other areas of physics and engineering.
N.B.: As it turns out, the Laplacian with a “
+
” sign has all negative eigenvalues (think of taking two
derivatives of the sine function to land back at the negative of the original). In order to be able to deal with
an operator with all positive eigenvalues, we will include a ‘
‘

” sign in our definition, referring to
∇
2
as
the Laplacian throughout the balance of these notes. Observe that the Green’s function for
∇
2
is just the
negative of the Green’s function for
∇
2
, by linearity.
In one dimension, the Poisson problem reduces simply to

d
2
dx
2
u
(
x
) =
f
(
x
)
.
(1)
We examine this equation as a boundary value problem on the interval 0
≤
x
≤
1, with homogeneous
boundary conditions on the value of
u
, namely
u
(0) = 0 and
u
(1) = 0.
We note that the fundamental solutions of the Laplacian operator in one dimension are
u
1
(
x
) = 1 and
u
2
(
x
) =
x
, and we construct the general solution to the inhomogeneous problem by the method of variation
of parameters (a simple exercise, since the Wronskian of the fundamental set is equal to 1) as
u
(
x
) =
c
1
+
c
2
x
+
Z
x
0
f
(
ξ
) (
ξ

x
)
dξ.
If we evaluate the boundary conditions at
x
= 0 and
x
= 1, we see that
u
(0) =
c
1
and
u
(1) =
c
1
+
c
2
·
1 +
Z
1
0
f
(
ξ
)(
ξ

1)
dξ,
which means that
c
1
= 0 and
c
2
=

R
1
0
f
(
ξ
)(
ξ

1)
dξ.
(Note that
c
2
is independent of
x
, but
c
2
balances the
forcing
f
across the entire interval.) Upon plugging
c
1
and
c
2
into the general solution for
u
(
x
), splitting
the range of integration into the two subranges (0
, x
) and (
x,
1) and regrouping terms, we derive
u
(
x
) =
Z
x
0
f
(
ξ
)
ξ
(1

x
)
dξ
+
Z
1
x
f
(
ξ
)
x
(1

ξ
)
dξ.
This can be rewritten in the compact and insightproducing form:
u
(
x
) =
Z
1
0
f
(
ξ
)
G
(
x, ξ
)
dξ,
(2)
where we have introduced the “Green’s function” for the onedimensional Laplacian operator,

d
2
dx
2
:
G
(
x, ξ
)
≡
x
(1

ξ
)
,
x
≤
ξ
ξ
(1

x
)
,
x
≥
ξ
.
(3)
1
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Observe that we have, in effect, inverted the differential operator

d
2
dx
2
! We have written the solution
u
(
x
)
as a linear functional of the forcing term
f
(
x
).
The inverse operator to the differential operator in (1) is
the integral operator in (2). The range of integration spans the domain of definition of the boundary value
problem. The kernel of the integral is a function of two variables, the dummy variable of integration that
ranges over all source points, and the free variable at which the function is evaluated.
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 Spring '08
 Keyes
 Boundary value problem, Laplacian, Dirac delta function, Boundary conditions, Green's function

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