92
Lecture 18
Parametric Problems (optional)
z w w
,
,
,
x y z
etc. These are differentiations with all except one variable held constant. Naturally, we can also perform integrations to multivariable functions but with all except
R
one variable held constant, for example, f (x, y) dy means that x is held constant
Rb
and the integration is for the variable y only; similarly, a f (x, y, z)dx means y and
z are held constant, and the definite integral is performed for the variable x only.
For z = f (x, y) or w = f (x, y, z), we have defined partial derivatives like
Thus, if f (x, y) = x2 + y 2 + xy, then
Z b
Z b
f (x, y)dy =
(x2 + y 2 + xy)dy
a
a
b
3
y
x
= x2 y +
+ y2
3
2
a
b3 a3 x 2
= x2 (b a) +
+ (b a2 )
3
2
Rb
We see that a f (x, y)dy is a function of x, which we can denote as F (x), where a
and b are regarded as given constants.
Let us note that F (x) = 2x(ba)+ 21 (b2 a2 ) by differentiation of the expression
3
3
of F (x), i.e. F (x) = x2 (b a) + b a
+ x2 (b2 a2 ).
3
On the other hand, we have
fx (x, y) = 2x + y
and
b
Z
b
b
y 2
fx (x, y)dy =
(2x + y)dy = 2xy +
2
a
a
b2 a2
= 2x(b a) +
2
Therefore, we have
F (x) =
fx (x, y)dy, i.e.
d
dx
Z
f (x, y)dy =
f (x, y)dy
x
(1)
This is to say, whether we perform integration first and differentiation second, or
differentiation first and integration second, we may arrive at the same result.
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Let us note, however, that our above discussion is not a rigorous proof that
identity (1) holds for any function f (x, y) as we merely checked that this identity
happens to be true for the particular function f (x, y) = x2 + y 2 + xy.
Nevertheless, identity (1) might be true in general, and it is an interesting problem to find an answer for this. The following theorem is a result of some research
along this line.
Theorem 1. Suppose that for every x (c, d), where c < d , the
following hold:
(i)
Rb
a
f (x, y)dy and
Rb
f (x, y)dy
a x
exist,
(ii) fxx (x, y) exists and satisfies
|fxx (x, y)| g(y) for a y b, c < x < d,
where g(y) is some function with the property that
Z
b
a
g(y)dy = K <
Then for each x (c, d),
d
dx
f (x, y)dy =
f (x, y)dy
x
Remark. Let us note that condition (i) in Theorem 1 is very natural. In fact,
it is necessary for identity (1) to make any sense. However, condition (ii) makes a
2
restriction to the function f (x, y). For example, if f (x, y) = x2 g(y), then fxx (x, y) =
Rb
g(y) and such an f (x, y) satisfies (ii) only if a g(y)dy < . Nevertheless, many
functions satisfy (ii).
Proof of Theorem 1. Let us denote F (x) =
F (x) =
i.e.
or
b
a
f (x, y)dy. We need to show
f (x, y)dy,
x
F (x + h) F (x)
lim
=
h0
h
Rb
F (x + h) F (x)
lim
h0
h
b
a
f (x, y)dy,
x
f (x, y)dy = 0
x
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Our strategy of achieving this is to use a squeezing argument: if we can find a
function (h) satisfying
Z b
F (x + h) F (x)
f (x, y)dy (h)
h
a x
and (h) 0 as h 0, then we must have
Z b
F (x + h) F (x)
f (x, y)dy = 0
lim
h0
h
a x
which implies that the limit of the quantity inside the absolute value sign is 0. A
key step in this strategy is to rewrite and change the quantity inside the absolute
value sign through using inequalities, to arrive at a simpler expression which can be
used as (h). The criterion is that (h) 0 should be evident.
Let us now start this process. We have
Z b
F (x + h) F (x)
f (x, y)dy
h
a x
Z b
Z b
Z b
1
f (x + h, y)dy
f (x, y)dy
f (x, y)dy
=
h
a
a x
Z b a
f (x + h, y) f (x, y)
f (x, y) dy (by properties of integrals)
h
x
Za "
#
b f (x, y)h + f (x + h, y) h2
x
xx
2
fx (x, y) dy (by using Taylors formula)
=
a
h
Z b
h
=
fxx (x + h, y)dy
a 2
Z
|h| b
|fxx (x + h, y)|dy
2 a
Z
|h| b
g(y)dy by condition (ii)
2 a
|h|
= K
2
Clearly we can take (h) = K |h|
. This finishes the squeezing process and hence
2
proved what we want.
2
d
Example 1 Find
dx
Solution f (x, y) =
exy
dy (x > 0).
y
exy
, fx (x, y) = exy , fxx (x, y) = yexy
y
For any d > c > 0, fxx (x, y) = yexy yecy when c < x < d and g(y) = yecy
is clearly integrable on [1, ). Hence condition (ii) of Theorem 1 is satisfied. It is
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easily seen that condition (i) is also satisfied (please check this). Therefore we can
use theorem 1 to conclude
Z xy
Z
d
e
exy
dy
dy =
dx 1
y
x
y
1
Z
xy
ex
e
.
=
exy dy =
=
x
x
1
1
From the last part of the above calculation, we observe that if we change the
integration limits to from 0 to , then
Z
xy
1
e
exy dy =
= .
x
x
0
0
Therefore, by using Theorem 1 (please check that the conditions are satisfied),
Z
Z
Z
d
xy
1
xy
=
e dy =
e dy =
yexy dy
x
dx 0
x
0
0
Z
Z
Z
d
1
xy
xy
=
=
ye dy =
(ye )dy =
y 2exy dy
2
x
dx 0
x
0
0
1
=
x2
2
x3
Z
(1) n!
=
(1)n y n exy dy
xn+1
0
n
This last identity can be written as
Z
n!
y n exy dy = n+1 , n = 1, 2 . . .
x
0
This turns out to be a useful formula. For example, if we let x = m, we obtain
Z
n!
y n emy dy = n+1 , n = 1, 2, . . . , m = 1, 2, . . .
m
0
This formula may not be as easily proved by other methods (can you?)
