
The foundations of calculus of variations 13
problem via its Euler-Lagrange equation form. Note that the form of the in-
tegrand dictates the use of the extended form.
∂f
∂y
=0,
∂
2
f
∂x∂y
=0,
∂
2
f
∂y∂y
=0,
and
∂
2
f
∂y
2
=
1
(1 + y
2
)
3/2
.
Substituting into the extended form gives
1
(1 + y
2
)
3/2
y
=0,
which simplifies into
y
=0.
Integrating twice, one obtains
y(x)=c
0
+ c
1
x,
clearly the equation of a line. Substituting into the boundary conditions we
obtain two equations,
y
0
= c
0
+ c
1
x
0
,
and
y
1
= c
0
+ c
1
x
1
.
The solution of the resulting linear system of equations is
c
0
= y
0
− c
1
x
0
,
and
c
1
=
y
1
− y
0
x
1
− x
0
.
It is easy to reconcile that
y(x)=y
0
−
y
1
− y
0
x
1
− x
0
x
0
+
y
1
− y
0
x
1
− x
0
x
is identical to
y(x