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3 Sufficient Conditions of Extrema of Functionals
Example 3.5.2 Determine whether the functional J [y] =
1
0 e
x
y
2
+
1
2
y
2
dx has
an extremum, the boundary conditions are y(0) = 1, y(1) = e.
Solution The Euler equation of the functional is
2e
x y − e
x y
− e
x y
= 0
or
y
+ y
− 2y = 0
The general solution of the equation is y = c 1 e
x
+ c 2 e
−2x , from the boundary
conditions y(0) = 1, y(1) = e, we get c 1 = 1, c 2 = 0, thus, the extremal curve of
the functional is y = e
x . It is included in the extremal curve field of y = ce
x . The
Weierstrass function is
E = e
x
y
2
+
1
2
y
2
− e
x
y
2
+
1
2
p
2
− (y
− p)e
x p =
1
2
e
x
(y
− p)
2
It is observed that for any y
, there is E ≥ 0. According to Theorem 3.5.2, the
functional gets a strong minimum on the extremal curve y = e
x .
Example 3.5.3 Determine whether the functional J [y] =
1
0 e
y y
2 dx has extremum,
the boundary conditions are y(0) = 0, y(1) = ln 4.
Solution Since the integrand does not contain x, so the first integral of the Euler
equation of the functional is
e
y y
2
− 2e
y y
2
= −c
2
or
e
y
2 dy = cdx
Integrate it, we get
e
y
2 = c 1 x + c 2
Taking logarithm, we have
y = 2 ln(c 1 x + c 2 )
From the boundary conditions y(0) = 0, y(1) = ln 4, we get c 1 = c 2 = 1, so the
extremal curve is
y = 2 ln(x + 1)
3 Sufficient Conditions of Extrema of Functionals
Example 3.5.2 Determine whether the functional J [y] =
1
0 e
x
y
2
+
1
2
y
2
dx has
an extremum, the boundary conditions are y(0) = 1, y(1) = e.
Solution The Euler equation of the functional is
2e
x y − e
x y
− e
x y
= 0
or
y
+ y
− 2y = 0
The general solution of the equation is y = c 1 e
x
+ c 2 e
−2x , from the boundary
conditions y(0) = 1, y(1) = e, we get c 1 = 1, c 2 = 0, thus, the extremal curve of
the functional is y = e
x . It is included in the extremal curve field of y = ce
x . The
Weierstrass function is
E = e
x
y
2
+
1
2
y
2
− e
x
y
2
+
1
2
p
2
− (y
− p)e
x p =
1
2
e
x
(y
− p)
2
It is observed that for any y
, there is E ≥ 0. According to Theorem 3.5.2, the
functional gets a strong minimum on the extremal curve y = e
x .
Example 3.5.3 Determine whether the functional J [y] =
1
0 e
y y
2 dx has extremum,
the boundary conditions are y(0) = 0, y(1) = ln 4.
Solution Since the integrand does not contain x, so the first integral of the Euler
equation of the functional is
e
y y
2
− 2e
y y
2
= −c
2
or
e
y
2 dy = cdx
Integrate it, we get
e
y
2 = c 1 x + c 2
Taking logarithm, we have
y = 2 ln(c 1 x + c 2 )
From the boundary conditions y(0) = 0, y(1) = ln 4, we get c 1 = c 2 = 1, so the
extremal curve is
y = 2 ln(x + 1)