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2.1 Poisson’s Equation in One Dimension
and observe that
x
y

F(y) dy
f(z) dz dy =
0
0
x

x
yF (y) dy
=[yF(y)] −
0
0
x

yf(y) dy
= xF(x) −
0

=
0 0 x x (x − y)f(y) dy, 41
where we have used integration by parts. Now (2.5) can be rewritten in the
following form:
x

u(x)= c 1 + c 2 x − (x − y)f(y) dy. (2.6)
0
Note that c 1 and c 2 are arbitrary constants, and that so far we have only
used the differential equation of (2.1) and not the boundary conditions
given by
u(0) = u(1)=0.
These conditions are taken into account by choosing c 1 and c 2 properly.
The condition u(0) = 0 implies that c 1 = 0, and then u(1) = 0 implies that

1
c 2 = (1 − y)f(y) dy.
0
Hence, the constants c 1 and c 2 are uniquely determined from the boundary
conditions. This observation is an important one; since any solution of the
differential equation

−u (x)= f(x)
can be written on the form (2.6) and the constants involved in (2.6) are
uniquely determined by the boundary conditions of (2.1), it follows that
the problem (2.1) has a unique solution.
We observe that if we use the derived expressions for c 1 and c 2 in (2.6),
we are allowed to write the solution u in the following form:
x
1
u(x)= x (1 − y)f(y) dy − (x − y)f(y) dy. (2.7)
0 0
Example 2.1 Consider the problem (2.1) with f(x) = 1. From (2.7) we
easily obtain
x
1 1
u(x)= x (1 − y) dy − (x − y) dy = x(1 − x).
2
0 0