Euler-Lagrange Equation (1)

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Theorem of the Euler-Lagrange equation for the case of one dependent variable

Let \ J:C^{2}\left[x_{0},x_{1}\right]\rightarrow\mathbb{R} be a functional of the form

\ J\left(y\right)=\int_{x_{0}}^{x_{1}}f\left(x,y,y^{\prime}\right)dx ,

where \ f has continuous partial derivatives of second order and \ x_{0}<x_{1} .

Let \ S=\left\{y\in C^{2}\left[x_{0},x_{1}\right]:y\left(x_{0}\right)=y_{0}\;and\;y\left(x_{1}\right)=y_{1}\right\} ,

where \ y_{0} and \ y_{1} are given real numbers. If \ y\in S is a local extremum for \ J , then

\ \frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)-\frac{\partial f}{\partial y}=0

for all \ x\in\left[x_{0},x_{1}\right] .

The Proof for the case of one dependent variable

Here we consider a particular class of problem called the fixed endpoint variational problem and work in \ C^{2}\left[x_{0},x_{1}\right] .

Let \ J:C^{2}\left[x_{0},x_{1}\right]\rightarrow\mathbb{R} be a functional of the form

\ J\left(y\right)=\int_{x_{0}}^{x_{1}}f\left(x,y,y^{\prime}\right)dx ,

where \ f is a function assumed to have at least second-order continuous partial derivatives.

Given two values \ y_{0},y_{1}\in\mathbb{R} , the fixed endpoint variational problem consists of determining the functions \ y\in C^{2}\left[x_{0},x_{1}\right] such that \ y\left(x_{0}\right)=y_{0},y\left(x_{1}\right)=y_{1} , and \ J has a local extremum in \ S at y\in S , where

\ S=\left\{ y\in C^{2}\left[x_{0},x_{1}\right]:y\left(x_{0}\right)=y_{0}\;\textrm{and}\; y\left(x_{1}\right)=y_{1}\right\} ,


\ H=\left\{ \eta\in C^{2}\left[x_{0},x_{1}\right]:\eta\left(x_{0}\right)=\eta\left(x_{1}\right)=0\right\} .

We want to derive necessary conditions for \ J to have a local extremum in \ S at \ y .

Suppose that \ J has a local extremum in \ S at \ y .

Then there is an \ \epsilon>0 such that \ J\left(\hat{y}\right)-J\left(y\right) does not change sign for all \ \hat{y}\in S such that \ \left\Vert \hat{y}-y\right\Vert <\epsilon .

For any \ \hat{y}\in S there is an \ \eta\in H such that \ \hat{y}=y+\epsilon\eta , and for \ \epsilon small Taylor's theorem implies that

\ f\left(x,\hat{y},\hat{y}^{\prime}\right)=f\left(x,y+\epsilon\eta,y^{\prime}+\epsilon\eta^{\prime}\right)=f\left(x,y,y^{\prime}\right)+\epsilon\left(\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}\right)+O\left(\epsilon^{2}\right) .

Please note that the partial derivatives in the above expression are all evaluated at the point \ \left(x,y\left(x\right),y^{\prime}\left(x\right)\right) .


\ J\left(\hat{y}\right)-J\left(y\right)=\int_{x_{0}}^{x_{1}}f\left(x,\hat{y},\hat{y}^{\prime}\right)dx-\int_{x_{0}}^{x_{1}}f\left(x,y,y^{\prime}\right)dx

\ =\int_{x_{0}}^{x_{1}}\left(\left(f\left(x,y,y^{\prime}\right)+\epsilon\left(\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}\right)+O\left(\epsilon^{2}\right)\right)-f\left(x,y,y^{\prime}\right)\right)dx

\ =\epsilon\int_{x_{0}}^{x_{1}}\left(\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}\right)dx+O\left(\epsilon^{2}\right)

\ =\epsilon\delta J\left(\eta,y\right)+O\left(\epsilon^{2}\right) .

The quantity

\ \delta J\left(\eta,y\right)=\int_{x_{0}}^{x_{1}}\left(\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}\right)dx

is called the first variation of \ J .

It is clear that if \ \eta\in H then \ -\eta\in H , and \ \delta J\left(-\eta,y\right)=-\delta J\left(\eta,y\right) . For \ \epsilon small, the sign of \ J\left(\hat{y}\right)-J\left(y\right) is determined by the sign of the first variation, unless \ \delta J\left(\eta,y\right)=0 for all \ \eta\in H .

Since \ J has a local extremum in \ S at \ y , \ J\left(\hat{y}\right)-J\left(y\right) does not change sign for all \ \hat{y}\in S such that \ \left\Vert \hat{y}-y\right\Vert <\epsilon . Hence, \ \delta J\left(\eta,y\right)=\int_{x_{0}}^{x_{1}}\left(\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}\right)dx=0 for all \ \eta\in H .

Using integration by parts,

\ \int_{x_{0}}^{x_{1}}\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}dx=\left[\eta\frac{\partial f}{\partial y^{\prime}}\right]_{x_{0}}^{x_{1}}-\int_{x_{0}}^{x_{1}}\eta\frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)dx

\ =-\int_{x_{0}}^{x_{1}}\eta\frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)dx ,

since \ \eta\left(x_{0}\right)=\eta\left(x_{1}\right)=0 .

Hence, \ \delta J\left(\eta,y\right)=\int_{x_{0}}^{x_{1}}\eta\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)\right)dx=0 for all \ \eta\in H.

Note that \ \frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)=\frac{\partial f}{\partial y}-\frac{\partial^{2}f}{\partial x\partial y^{\prime}}-\frac{\partial^{2}f}{\partial y\partial y^{\prime}}y^{\prime}-\frac{\partial^{2}f}{\partial y^{\prime}\partial y^{\prime}}y^{\prime\prime},

and given that \ f has at least two continuous derivatives, we see that for any \ y\in C^{2}\left[x_{0},x_{1}\right] the function \ E:\left[x_{0},x_{1}\right]\rightarrow\mathbb{R} defined by

\ E\left(x\right)=\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)

is continuous on the interval \ \left[x_{0},x_{1}\right] . Note again that for a given function \ y the partial derivatives defining \ E are evaluated at the point \ \left(x,y\left(x\right),y^{\prime}\left(x\right)\right) .

Hence, \ \int_{x_{0}}^{x_{1}}\eta\left(x\right)E\left(x\right)dx=0 for all \ \eta\in H .

By the Fundamental Lemma of Calculus of Variations, \ E=0 .

Hence, \ \frac{d}{dx}\left(\frac{\partial f}{\partial y^{\prime}}\right)-\frac{\partial f}{\partial y}=0\qquad\left(1\right)

and the equation \left(1\right) is called the Euler-Lagrange equation.

Theorem of the Euler-Lagrange equation for the case of several dependent variables

Let \ J:\mathbf{C}^{2}\left[t_{0},t_{1}\right]\rightarrow\mathbb{R} be a functional of the form

\ J\left(\mathbf{q}\right)=\int_{t_{0}}^{t_{1}}L\left(t,\mathbf{q},\dot{\mathbf{q}}\right)dt ,

where \ \mathbf{q}=\left(q_{1},q_{2},\ldots,q_{n}\right), and \ L has continuous second order partial derivatives.

Let \ S=\left\{\mathbf{q}\in\mathbf{C}^{2}\left[t_{0},t_{1}\right]:\mathbf{q}\left(t_{0}\right)=\mathbf{q}_{0}\;and\;\mathbf{q}\left(t_{1}\right)=\mathbf{q}_{1}\right\} ,

where \ \mathbf{q}_{0},\mathbf{q}_{1}\in\mathbb{R}^{n} are given vectors.

If \ \mathbf{q} is a local extremum for \ J in \ S then

\ \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{k}}-\frac{\partial L}{\partial q_{k}}=0

for \ k=1,2,\ldots,n .

The proof for case of several dependent variables

Let \ \mathbf{C}^{2}\left[t_{0},t_{1}\right] denote the set of functions \ \mathbf{q}:\left[t_{0},t_{1}\right]\rightarrow\mathbb{R}^{n} such that for every \ \mathbf{q}=\left(q_{1},q_{2},\ldots,q_{n}\right) we have \ q_{k}\in C^{2}\left[t_{0},t_{1}\right] for \ k=1,2,\ldots,n .

Consider a functional of the form

\ J\left(\mathbf{q}\right)=\int_{t_{0}}^{t_{1}}L\left(t,\mathbf{q},\dot{\mathbf{q}}\right)dt ,

where \ \dot{} denotes differentiation with respect to \ t , and \ L is a function having continuous partial derivatives of second order.

Given two vectors \ \mathbf{q}_{0},\mathbf{q}_{1}\in\mathbb{R}^{n} , the fixed endpoint problem consists of determining the local extrema for \ J subject to the conditions \ \mathbf{q}\left(t_{0}\right)=\mathbf{q}_{0} and \ \mathbf{q}\left(t_{1}\right)=\mathbf{q}_{1} .

Again, we have

\ S=\left\{\mathbf{q}\in\mathbf{C}^{2}\left[t_{0},t_{1}\right]:\mathbf{q}\left(t_{0}\right)=\mathbf{q}_{0}\;and\;\mathbf{q}\left(t_{1}\right)=\mathbf{q}_{1}\right\} and

\ H=\left\{\boldsymbol{\eta}\in\mathbf{C}^{2}\left[t_{0},t_{1}\right]:\boldsymbol{\eta}\left(t_{0}\right)=\boldsymbol{\eta}\left(t_{1}\right)=0\right\} .

For \ \epsilon small Taylor's theorem implies that

\ L\left(t,\hat{\mathbf{q}},\dot{\hat{\mathbf{q}}}\right)=L\left(t,\mathbf{q}+\epsilon\boldsymbol{\eta},\dot{\mathbf{q}}+\epsilon\dot{\boldsymbol{\eta}}\right)

\ =L\left(t,\mathbf{q},\dot{\mathbf{q}}\right)+\epsilon\sum_{k=1}^{n}\left(\eta_{k}\frac{\partial L}{\partial q_{k}}+\dot{\eta}_{k}\frac{\partial L}{\partial\dot{q}_{k}}\right)+O\left(\epsilon^{2}\right) ,

where \ \hat{\mathbf{q}}=\mathbf{q}+\epsilon\boldsymbol{\eta} .


\ J\left(\hat{\mathbf{q}}\right)-J\left(\mathbf{q}\right)=\int_{t_{0}}^{t_{1}}L\left(t,\hat{\mathbf{q}},\dot{\hat{\mathbf{q}}}\right)dt-\int_{t_{0}}^{t_{1}}L\left(t,\mathbf{q},\dot{\mathbf{q}}\right)dt

\ =\epsilon\int_{t_{0}}^{t_{1}}\sum_{k=1}^{n}\left(\eta_{k}\frac{\partial L}{\partial q_{k}}+\dot{\eta}_{k}\frac{\partial L}{\partial\dot{q}_{k}}\right)dt+O\left(\epsilon^{2}\right) .

Therefore the first variation for this functional is

\ \delta J\left(\boldsymbol{\eta},\mathbf{q}\right)=\int_{t_{0}}^{t_{1}}\sum_{k=1}^{n}\left(\eta_{k}\frac{\partial L}{\partial q_{k}}+\dot{\eta}_{k}\frac{\partial L}{\partial\dot{q}_{k}}\right)dt .

If \ J has a local extremum at \ \mathbf{q} then similar arguments imply that \ \delta J\left(\boldsymbol{\eta},\mathbf{q}\right)=0 for all \ \boldsymbol{\eta}\in H.

Define \ H_{1}=\left\{\left(\eta_{1},0,\ldots,0\right)\in H\right\} . For any \ \boldsymbol{\eta}\in H_{1} the above condition reduces to

\ \int_{t_{0}}^{t_{1}}\left(\eta_{1}\frac{\partial L}{\partial q_{1}}+\dot{\eta}_{1}\frac{\partial L}{\partial\dot{q}_{1}}\right)dt=0 .

From the one variable case, we know this condition leads to the Euler-Lagrange equation

\ \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{1}}-\frac{\partial L}{\partial q_{1}}=0 .

Similarly, we have

\ \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{k}}-\frac{\partial L}{\partial q_{k}}=0 for \ k=1,2,\ldots,n .

Special cases

Case 1

Suppose \ f does not depend on \ y.

Then, the Euler-Lagrange equation simplifies to

{d\over dx}\left({\partial f\over\partial {y'}}\right) = 0

and hence

{\partial f\over\partial {y'}} = C

where \ C is some constant.

Solving for \ y', we get

\ y' = g(x, C)

for some function \ g. This can be solved by a quadrature.

Case 2

We will use a more compact notation to improve readability.

Suppose the integrand does not depend on \ x, in which case \ f = f(y, y').

First, we notice the following

{d\over dx}(f-y'f_{y'})=(f_yy'+f_{y'}y'')-(y''f_{y'}+y'(f_{yy}y'+f_{yy'}y''))

again by the chain rule. The second and third terms cancel, and we obtain

{d\over dx}(f-y'f_{y'})=f_yy'+f_{yy}{y'}^2-f_{yy'}y''=(f_y-f_{y'y}y'-f_{y'y'}y'')y' (*)

Now, from the Euler-Lagrange equation, we get

0=f_y-{d\over dx}f_{y'} = f_y-f_{y'y}y'-f_{y'y'}y''

by the chain rule.

However, the right side of this equation (which is zero) can be directly substituted into the right side of (*). Hence (*) is also zero. That is,

{d\over dx}(f-y'f_{y'})=0


\ f-y'f_{y'} = C

where \ C is a constant.

Case 3

Suppose \ f does not depend on \ y'.

Then, the Euler-Lagrange equation takes the form

\ {\partial f\over y}(x,y)=0


Straight Lines

Result: The shortest line between two points \ P and \ Q in \ \mathbb R^n is straight.

Proof: Consider a continuously differentiable function \ \vec x:[t_1, t_2] \to \mathbb R^n that parametrizes a curve with the required endpoints.

We will let \ x_i denote the ith component of the function for \ i = 1, 2, ..., n.

We have the arc length function

\ S = \int_{t_1}^{t_2} {\left(\sum_{i = 1}^n \dot{x_i}^2\right)}^{1/2} dt

where \ f is the integrand.

Now, for every \ i = 1, 2, ..., n

{\partial f \over \partial x_i} = 0


{\partial f \over \partial \dot{x_i}} = \dot{x_i} {\left(\sum_{i = 1}^n \dot{x_i}^2\right)}^{-1/2}

Substituting into the Euler-Lagrange equation, we get \dot{x_i} = 0 for \ i = 1, 2, ..., n, which describes a straight line.

We have only shown that the straight line with the required endpoints an extremum. Nonetheless, in 3 (or less) dimensions, it is intuitively clear that such a line has the minimum length.


See Geodesic.


See Catenary.


For a discussion on the brachistochrone property of the cycloid, see Cycloid.


Result: The frictionless tunnel connecting two points \ A and \ B on the surface of a (spherical, homogeneous) planet taking the shape of a hypocycloid is the one for which a ball dropped (initially from rest) at one point moves to the other the fastest.

Proof: We will study the system partly in polar co-ordinates \ (r, \theta) with the planet centered at the origin and partly in Cartesian co-ordinates \ (x,y), whereby we assume that the problem is 2-dimensional.

Out problem is to minimize the total time

\ t_{AB} = \int_A^B \frac1vds


\ v denotes the speed of the ball at time \ t
\ s denotes the total distance traveled by time \ t

We now find \ v by first finding the potential energy \ U(r), where we set \ U(0) = 0.

Consider a spherical surface of radius \ r concentric to the planet such that \ r<R, where \ R is the radius of the planet. Then, applying Gauss' law on this surface, we obtain

4\pi r^2 F = - 4\pi G {\left(\frac43 \pi r^3 \rho\right)}m


\ F is the gravitational force (positive outward) at radius \ r
\ \rho is the planet's density
\ m is the mass of the ball
\ G is the gravitational constant

The left side is the surface flux of the gravitational force and the right side is the total mass contained in the surface.


\ \rho = {M\over\frac43\pi r^3}

we obtain

\ F = -\frac{GMm}{R^3}r


U(r) = -\int_0^r F(u)du = -\int_0^r -\frac{GMm}{R^3}udu = \frac12\frac{GMm}{R^3}r^2

To find \ v, we apply the conservation of mechanical energy, so that

\frac12 mv^2 = U(0) - U(t) = \frac12\frac{GMm}{R^3}R^2-\frac12\frac{GMm}{R^3}r^2

where \ m is the mass of the ball, and we remind that \ v(0) = 0 and \ r(0) = R.

Letting \ g = \frac{GM}{R^2}, we get

\ v = \sqrt{\frac{g(R^2 - r^2)}R} = \sqrt{\frac{g(R^2 - (x^2 + y^2))}R}


\ ds = \sqrt{}dt

Hence, we conclude that

\ t_{AB} = \int_A^B \frac1vds = \int_0^{t} \sqrt{\frac{\left({\dot x}^2+{\dot y}^2\right)R}{g(R^2-x^2-y^2)}} du


\cdot denotes the time derivative

and we let \ f denote the integrand and minimized the integral using the Euler-Lagrange equation.

Since \ f is independent of \ t, therefore our equation reduces to (case 2 from above)

\ f - {x'}{\partial f\over\partial x'} = C_1
\ f - {y'}{\partial f\over\partial y'} = C_2

where \ C_1 and \ C_2 are constants.

Differentiating then substituting, we obtain

\ A = -\frac{{\dot y}^2 z}{{\dot x}^2 - {\dot y}^2}
\ B = \frac{{\dot x}^2 z}{{\dot x}^2 - {\dot y}^2}


\ z = \sqrt{\frac{\left({\dot x}^2+{\dot y}^2\right)R}{g(R^2-x^2-y^2)}}

Adding both equations then squaring both sides, we get

\ R({\dot x}^2+{\dot y}^2)=gC^2(R^2-x^2-y^2)

where \ C = A + B

The following cycloid with outer radius \ R and inner radius \ b satisfies the required equation

\vec r = (R-b)\left(\cos t, \sin t\right) + b\left(\cos {\frac{R - b}b t}, -\sin{\frac{R - b}b t}\right)

This completes the proof.


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