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Putnam
1958 February Putnam
A5
A5
Part of
1958 February Putnam
Problems
(1)
Putnam 1958 February A5
Source: Putnam 1958 February
7/18/2022
Show that the integral equation
f
(
x
,
y
)
=
1
+
∫
0
x
∫
0
y
f
(
u
,
v
)
d
u
d
v
f(x,y) = 1 + \int_{0}^{x} \int_{0}^{y} f(u,v) \, du \, dv
f
(
x
,
y
)
=
1
+
∫
0
x
∫
0
y
f
(
u
,
v
)
d
u
d
v
has at most one solution continuous for
0
≤
x
≤
1
,
0
≤
y
≤
1.
0\leq x \leq 1, 0\leq y \leq 1.
0
≤
x
≤
1
,
0
≤
y
≤
1.
Putnam
calculus
integration
equation