X
Y
Z
Figure 7.50
Exercise 7.7.10
By calculating partial derivatives, verify that
(
X, Y
) = (0
,
0)
is a critical
point of the function
Z
=
f
(
X, Y
) = 0
.
5(
X
2

Y
2
)
.
So how do we classify critical points as maxima, minima or saddle points?
We will use
the deep relationship between critical points of functions and equilibrium points of di
ff
erential
equations.
Given any function
Z
=
f
(
X, Y
)
we can define a new vector field on
(
X, Y
)
space by
X
0
=
dX
dt
=
@
f
@
X
and
Y
0
=
d Y
dt
=
@
f
@
Y
(Recall that
X
0
is the change of
X
with respect to time,
dX
dt
.) This new vector field, derived
from the function
Z
=
f
(
X, Y
)
, is called the
gradient vector field
of
f
, called “
grad
f
” and often
written as “
r
f
”.
Exercise 7.7.11
Compute
r
f
for the functions in
Exercise 7.7.9
.
What are the equilibrium points of this vector field?
By definition, they are points where
X
0
= 0
and
Y
0
= 0
, that is,
@
f
@
X
= 0
and
@
f
@
Y
= 0
.
But we just said that a critical point of the function
f
is a point where
@
f
@
X
= 0
and
@
f
@
Y
= 0
.
Therefore, the critical points of
f
are exactly the equilibrium points of the vector field
r
f
.
Exercise 7.7.12
Verify that at the critical points you found in
Exercise 7.7.9
,
r
f
= 0
.
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7.7.
Optimization
379
If
Z
=
f
(
X, Y
)
is a height function, we can define the gradient vector field
r
f
as
X
0
=
@
f
@
X
Y
0
=
@
f
@
Y
Critical points of
f
(maxima, minima, saddles) exactly correspond to equilibrium points
(stable, purely unstable, saddle) of the gradient vector field
r
f
.
We will now make the key connection that will enable us to identify critical points as maxima,
minima or saddles.
First, let’s consider three simple height functions. We will plot the function
f
and project it
down onto the
X
and
Y
axes, where we have calculated and plotted the vector field
r
f
. The
first example is a hill (Figure 7.51 left). The function is
Z
=
f
(
X, Y
) = 5

X
2
2

Y
2
4
The vector field
r
f
is then
X
0
=
@
f
@
X
Y
0
=
@
f
@
Y
So
X
0
=

X
Y
0
=

0
.
5
Y
This is obviously a linear vector field that has a stable equilibrium point at
(0
,
0)
.
X
Y
Z
f
f
Figure 7.51
The second example is a bowl (Figure 7.51 middle):
Z
=
f
(
X, Y
) =
X
2
2
+
Y
2
4
The vector field
r
f
is then
X
0
=
@
f
@
X
=
X
Y
0
=
@
f
@
Y
= 0
.
5
Y
380
Multivariable Systems
which again is a linear di
ff
erential equation, with an unstable equilibrium point at
(0
,
0)
.
The third example is a saddle (Figure 7.51 right):
Z
=
f
(
X, Y
) = 0
.
5(
X
2

Y
2
)
The vector field
r
f
is then
X
0
=
@
f
@
X
=
X
Y
0
=
@
f
@
Y
=

Y
which again is a linear di
ff
erential equation, this time with a saddle point at
(0
,
0)
.
So in this example,
•
maxima of
f
correspond to stable equilibrium points (stable nodes) of
r
f
.
•
minima of
f
correspond to purely unstable equilibrium points (unstable nodes) of
r
f
.
•
saddle points of
f
correspond to saddle points of
r
f
.
This is true in general, due to the definition of the gradient vector field. Since
Z
=
f
(
X, Y
)
,
we know that the change in
Z
Δ
Z
=
@
f
@
X
·
Δ
X
+
@
f
@
Y
·
Δ
Y
But, from the definition of
r
f
we know that
8
>
>
<
>
>
:
X
0
=
@
f
@
X
=
)
Δ
X
Δ
t
=
@
f
@
X
=
)
Δ
X
=
@
f
@
X
·
Δ
t
Y
0
=
@
f
@
Y
=
)
Δ
Y
Δ
t
=
@
f
@
Y
=
)
Δ
Y
=
@
f
@
Y
·
Δ
t
If we substitute these expressions for
Δ
X
and
Δ
Y
in the
Δ
Z
equation, we get
Δ
Z
=
@
f
@
X
·
Δ
X
+
@
f
@
Y
·
Δ
Y
=
@
f
@
X
·
@
f
@
X
·
Δ
t
+
@
f
@
Y
·
@
f
@
Y
·
Δ
t
=
✓
@
f
@
X
◆
2
·
Δ
t
+
✓
@
f
@
Y
◆
2
·
Δ
t
Since
Δ
t >
0
,
✓
df
dX
◆
2
>
0
,
and
✓
df
d Y
◆
2
>
0
the whole
Δ
Z
expression is positive.
Therefore,
Z
will always increase following the gradient
function
r
f
.