5Second-order differential equations
IA Differential Equations
5.6 Impulses and point forces
5.6.1 Dirac delta function
Consider a ball bouncing on the ground. When the ball hits the ground at some
time
T
, it experiences a force from the ground for some short period of time.
The force on the ball exerted by the ground
F
(
t
) is 0 for most of the time, except
during the short period (T − ε, T + ε).
Often we don’t know (or we don’t wish to know) the details of
F
(
t
) but we
can note that it only acts for a short time of
O
(
ε
) that is much shorter than the
overall time
O
(
t
2
−t
1
) of the system. It is convenient mathematically to imagine
the force acting instantaneously at time t = T, i.e. consider the limit ε → 0.
Newton’s second law gives
m¨x
=
F
(
t
)
− mg
. While we cannot solve it, we
can integrate the equation from T −ε to T + ε. So
Z
T +ε
T −ε
m
d
2
x
dt
2
dt =
Z
T +ε
T −ε
F (t) dt −
Z
T +ε
T −ε
mg dt
m
dx
dt
T +ε
T −ε
= I − 2εmg
∆p = I − O(ε)
Where ∆
p
is the change in momentum and the impulse
I
=
R
T +ε
T −ε
F
(
t
) d
t
is the
area under the force curve. Note that the impulse
I
is the only property of
F
that influences the macroscopic behaviour of the system. If the contact time 2
ε
is small, we’ll neglect it and write
∆p = I
Assuming that
F
only acts on a negligible amount of time
ε
, all that matters to
us is its integral I, i.e. the area under the force curve.
wlog, assume
T
= 0 for easier mathematical treatment. We can consider a
family of functions D(t; ε) such that
lim
ε→0
D(t; ε) = 0 for all t 6= 0;
lim
ε→0
Z
∞
−∞
D(t; ε) dt = 1.
So we can replace the force in our example by
ID
(
t
;
ε
), and then take the limit
as ε → 0.
For example, we can choose
D(t; ε) =
1
ε
√
π
e
−t
2
/ε
2
t
D
ε = 1
ε = 0.5
This has height O(1/ε) and width O(ε).
It can be checked that this satisfies the properties listed above. Note that as
ε → 0, D(0; ε) → ∞. Therefore lim
ε→0
D(0; ε) does not exist.
Definition (Dirac delta function). The Dirac delta function is defined by
δ(x) = lim
ε→0
D(x; ε)
on the understanding that we can only use its integral properties. For example,
when we write
Z
∞
−∞
g(x)δ(x) dx,
we actually mean
lim
ε→0
Z
∞
−∞
g(x)D(x; ε) dx.
In fact, this is equal to g(0).
More generally,
R
b
a
g
(
x
)
δ
(
x − c
) d
x
=
g
(
c
) if
c ∈
(
a, b
) and 0 otherwise,
provided g is continuous at x = c.
This gives a convenient way of representing and making calculations involving
impulsive or point forces. For example, in the previous example, we can write
m¨x = −mg + Iδ(t − T ).
Example.
y
00
−y
= 3
δ
(
x −
π
2
) with
y
= 0 at
x
= 0
, π
. Note that our function
y
is split into two parts by x =
π
2
.
First consider the region 0
≤ x <
π
2
. Here the delta function is 0, and we have
y
00
−y
= 0 and
y
= 0 at
x
= 0. Then
y
=
Ce
x
+
De
−x
=
A sinh x
+
B cosh x
and
obtain
B
= 0 from the boundary condition. In the region
π
2
< x ≤ π
, we again
obtain
y
=
C sinh
(
π − x
) +
D cosh
(
π − x
) and (from the boundary condition),
D = 0.
When
x
=
π
2
, first insist that
y
is continuous at
x
=
π
2
. So
A
=
C
. Then
note that we have to solve
y
00
− y = 3δ
x −
π
2
But remember that the delta function makes sense only in an integral. So we
integrate both sides from
π
2
−
to
π
2
+
. Then we obtain
[y
0
]
π
2
+
π
2
−
−
Z
π
2
+
π
2
−
y dx = 3
Since we assume that y is well behaved, the second integral is 0. So we are left
with
[y
0
]
π
2
+
π
2
−
= 3
So we have
−C cosh
π
2
− A cosh
π
2
= 3
A = C =
−3
2 cosh
π
2
Then we have
y =
(
−3 sinh x
2 cosh
π
2
0 ≤ x <
π
2
−3 sinh(π−x)
2 cosh
π
2
π
2
< x ≤ π
x
y
Note that at
x
=
π
2
, our final function has continuous
y
, discontinuous
y
0
and
infinite
y
00
. In general, differentiating a function makes it less continuous. This
is why we insisted at first that
y
has to be continuous. Otherwise,
y
0
would look
like a delta function, and y
00
would be something completely unrecognizable.
Hence the discontinuity is always addressed by the highest order derivative
since differentiation increases the discontinuity.