Special case of the Euler-Lagrange equations
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} and b {\displaystyle b} are constants and u ′ ( x ) = d u d x {\displaystyle u'(x)={\frac {du}{dx}}} .[1]
If ∂ L ∂ x = 0 {\displaystyle {\frac {\partial L}{\partial x}}=0} , then the Euler–Lagrange equation reduces to the Beltrami identity,
L − u ′ ∂ L ∂ u ′ = C , {\displaystyle L-u'{\frac {\partial L}{\partial u'}}=C\,,}
where C is a constant.[2] [note 1]
Derivation [ edit ] By the chain rule , the derivative of L is
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} , we write
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''\,.} We have an expression for ∂ L ∂ u {\displaystyle {\frac {\partial L}{\partial u}}} from 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'}}\,} that we can substitute in the above expression for d L d x {\displaystyle {\frac {dL}{dx}}} to obtain
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 [ edit ] Solution to the brachistochrone problem [ edit ] 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)} that minimizes the integral
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} , so the Beltrami identity applies,
L − y ′ ∂ L ∂ y ′ = C . {\displaystyle L-y'{\frac {\partial L}{\partial y'}}=C\,.} Substituting for L {\displaystyle L} and simplifying,
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} being half the above constant, 1 2 C 2 {\displaystyle {\frac {1}{2C^{2}}}} , and ϕ {\displaystyle \phi } being a variable. These are the parametric equations for a cycloid .[3]
Solution to the catenary problem [ edit ] A chain hanging from points forms a catenary . Consider a string with uniform density μ {\displaystyle \mu } of length l {\displaystyle l} suspended from two points of equal height and at distance D {\displaystyle D} . By the formula for arc length ,
l = ∫ S d S = ∫ s 1 s 2 1 + y ′ 2 d x , {\displaystyle l=\int _{S}dS=\int _{s_{1}}^{s_{2}}{\sqrt {1+y'^{2}}}dx,} where
S {\displaystyle S} is the path of the string, and
s 1 {\displaystyle s_{1}} and
s 2 {\displaystyle s_{2}} are the boundary conditions.
The curve has to minimize its potential energy
U = ∫ S g μ y ⋅ d S = ∫ s 1 s 2 g μ y 1 + y ′ 2 d x , {\displaystyle U=\int _{S}g\mu y\cdot dS=\int _{s_{1}}^{s_{2}}g\mu y{\sqrt {1+y'^{2}}}dx,} and is subject to the constraint
∫ s 1 s 2 1 + y ′ 2 d x = l , {\displaystyle \int _{s_{1}}^{s_{2}}{\sqrt {1+y'^{2}}}dx=l,} where
g {\displaystyle g} is the force of gravity.
Because the independent variable x {\displaystyle x} does not appear in the integrand, the Beltrami identity may be used to express the path of the string as a separable first order differential equation
L − y ′ ∂ L ∂ y ′ = μ g y 1 + y ′ 2 + λ 1 + y ′ 2 − [ μ g y y ′ 2 1 + y ′ 2 + λ y ′ 2 1 + y ′ 2 ] = C , {\displaystyle L-y\prime {\frac {\partial L}{\partial y\prime }}=\mu gy{\sqrt {1+y\prime ^{2}}}+\lambda {\sqrt {1+y\prime ^{2}}}-\left[\mu gy{\frac {y\prime ^{2}}{\sqrt {1+y\prime ^{2}}}}+\lambda {\frac {y\prime ^{2}}{\sqrt {1+y\prime ^{2}}}}\right]=C,} where
λ {\displaystyle \lambda } is the
Lagrange multiplier .
It is possible to simplify the differential equation as such:
g ρ y − λ 1 + y ′ 2 = C . {\displaystyle {\frac {g\rho y-\lambda }{\sqrt {1+y'^{2}}}}=C.} Solving this equation gives the hyperbolic cosine , where C 0 {\displaystyle C_{0}} is a second constant obtained from integration
y = C μ g cosh [ μ g C ( x + C 0 ) ] − λ μ g . {\displaystyle y={\frac {C}{\mu g}}\cosh \left[{\frac {\mu g}{C}}(x+C_{0})\right]-{\frac {\lambda }{\mu g}}.} The three unknowns C {\displaystyle C} , C 0 {\displaystyle C_{0}} , and λ {\displaystyle \lambda } can be solved for using the constraints for the string's endpoints and arc length l {\displaystyle l} , though a closed-form solution is often very difficult to obtain.
References [ edit ] ^ 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.