Special case of the Euler-Lagrange equations
Eugenio Beltrami The Beltrami identity , named after Eugenio Beltrami , is a special case of the Euler–Lagrange equation in the calculus of variations .
The Euler–Lagrange equation serves to extremize action functionals of the form
I
[
u
]
=
∫
a
b
L
[
x
,
u
(
x
)
,
u
′
(
x
)
]
d
x
,
{\displaystyle I[u]=\int _{a}^{b}L[x,u(x),u'(x)]\,dx\,,}
where
a
{\displaystyle a}
b
{\displaystyle b}
u
′
(
x
)
=
d
u
d
x
{\displaystyle u'(x)={\frac {du}{dx}}}
[1]
If
∂
L
∂
x
=
0
{\displaystyle {\frac {\partial L}{\partial x}}=0}
L
−
u
′
∂
L
∂
u
′
=
C
,
{\displaystyle L-u'{\frac {\partial L}{\partial u'}}=C\,,}
where C [2] [note 1]
Derivation By the chain rule , the derivative of L
d
L
d
x
=
∂
L
∂
x
d
x
d
x
+
∂
L
∂
u
d
u
d
x
+
∂
L
∂
u
′
d
u
′
d
x
.
{\displaystyle {\frac {dL}{dx}}={\frac {\partial L}{\partial x}}{\frac {dx}{dx}}+{\frac {\partial L}{\partial u}}{\frac {du}{dx}}+{\frac {\partial L}{\partial u'}}{\frac {du'}{dx}}\,.}
Because
∂
L
∂
x
=
0
{\displaystyle {\frac {\partial L}{\partial x}}=0}
d
L
d
x
=
∂
L
∂
u
u
′
+
∂
L
∂
u
′
u
″
.
{\displaystyle {\frac {dL}{dx}}={\frac {\partial L}{\partial u}}u'+{\frac {\partial L}{\partial u'}}u''\,.}
Then by combining this with the Euler–Lagrange equation,
∂
L
∂
u
=
d
d
x
∂
L
∂
u
′
.
{\displaystyle {\frac {\partial L}{\partial u}}={\frac {d}{dx}}{\frac {\partial L}{\partial u'}}\,.}
We get the following expression,
d
L
d
x
=
u
′
d
d
x
∂
L
∂
u
′
+
u
″
∂
L
∂
u
′
.
{\displaystyle {\frac {dL}{dx}}=u'{\frac {d}{dx}}{\frac {\partial L}{\partial u'}}+u''{\frac {\partial L}{\partial u'}}\,.}
By the product rule , the right side is equivalent to
d
L
d
x
=
d
d
x
(
u
′
∂
L
∂
u
′
)
.
{\displaystyle {\frac {dL}{dx}}={\frac {d}{dx}}\left(u'{\frac {\partial L}{\partial u'}}\right)\,.}
By integrating both sides and putting both terms on one side, we get the Beltrami identity,
L
−
u
′
∂
L
∂
u
′
=
C
.
{\displaystyle L-u'{\frac {\partial L}{\partial u'}}=C\,.}
Applications Solution to the brachistochrone problem The solution to the brachistochrone problem is the cycloid. An example of an application of the Beltrami identity is the brachistochrone problem , which involves finding the curve
y
=
y
(
x
)
{\displaystyle y=y(x)}
I
[
y
]
=
∫
0
a
1
+
y
′
2
y
d
x
.
{\displaystyle I[y]=\int _{0}^{a}{\sqrt {{1+y'^{\,2}} \over y}}dx\,.}
The integrand
L
(
y
,
y
′
)
=
1
+
y
′
2
y
{\displaystyle L(y,y')={\sqrt {{1+y'^{\,2}} \over y}}}
does not depend explicitly on the variable of integration
x
{\displaystyle x}
L
−
y
′
∂
L
∂
y
′
=
C
.
{\displaystyle L-y'{\frac {\partial L}{\partial y'}}=C\,.}
Substituting for
L
{\displaystyle L}
y
(
1
+
y
′
2
)
=
1
/
C
2
(constant)
,
{\displaystyle y(1+y'^{\,2})=1/C^{2}~~{\text{(constant)}}\,,}
which can be solved with the result put in the form of parametric equations
x
=
A
(
ϕ
−
sin
ϕ
)
{\displaystyle x=A(\phi -\sin \phi )}
y
=
A
(
1
−
cos
ϕ
)
{\displaystyle y=A(1-\cos \phi )}
with
A
{\displaystyle A}
1
2
C
2
{\displaystyle {\frac {1}{2C^{2}}}}
ϕ
{\displaystyle \phi }
cycloid .[3]
Notes References
^ Courant R , Hilbert D (1953). Methods of Mathematical Physics . Vol. I (First English ed.). New York: Interscience Publishers, Inc. p. 184. ISBN 978-0471504474 ^ Weisstein, Eric W. "Euler-Lagrange Differential Equation." From MathWorld --A Wolfram Web Resource. See Eq. (5).
^ This solution of the Brachistochrone problem corresponds to the one in — Mathews, Jon; Walker, RL (1965). Mathematical Methods of Physics . New York: W. A. Benjamin, Inc. pp. 307–9.
![{\displaystyle I[u]=\int _{a}^{b}L[x,u(x),u'(x)]\,dx\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90e683a73b0dc65988f0967cf70d6f3a9d6be522)
![{\displaystyle I[y]=\int _{0}^{a}{\sqrt {{1+y'^{\,2}} \over y}}dx\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6033bb9e89143c834ff902dd4ba14acae9eee035)
