Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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14.6 Variational Methods and the Euler–Lagrange Equations 641

−h

dθ

− h

= −F





dθ

+ u =





. (14.6.44)

This is the diﬀerential equation of the central orbit and can be solved by

standard methods.

In particular, if the law of force is the attractive inverse square, F (r)=

µ/r

so that the potential V (r)=−µ/r, the diﬀerential equation (14.6.44)

becomes

dθ

+ u =

, (14.6.45)

if the particle is projected initially from distance a with velocity V at an

angle β that the direction of motion makes with the outward radiu s vector.

Thus, the constant h in (14.6.43) is h = Vasin β. The angle φ between the

tangent and radius vector of the orbit at any point is given by

cot φ =

dθ

= u

dθ





= −

dθ

. (14.6.46)

At t = 0, the initial conditions are

u =

dθ

= −

cot β when θ =0. (14.6.47)

The general solution of equation (14.6.45) is

u =

[1 + e cos (θ + α)] , (14.6.48)

where e and α are constants to be determined from the initial data.

Finally, the solution can be written as

=1+e cos (θ + α) , (14.6.49)

where

l =

=(Vasin β)

/µ. (14.6.50)

This represents a conic section of semi-latus rectum l and eccentricity

e with its axis inclined at an angle α to the radius vector at the point of

projection.

The initial conditions (14.6.47) lead to

642 14 Numerical and Approximation Methods

=1+e cos α, −

cot β = −e sin α, (14.6.51)

which give

tan α =

l cot β

l − a



− 1



cot

β =

cosec

β −

+1,

=1−

2aV

sin

. (14.6.52)

Thus, the conic is an ellipse, parabola, or hyperbola accordingly as e<=> 1

that is, V

< = > 2µ/a.

To derive the Hamilton equations, we introduce the concept of general-

ized momentum, p

and generalized force, F

∂L

∂ ˙q

∂L

∂q

. (14.6.53ab)

Consequently, the Lagr ange equations (14.6.34) become

∂L

∂q

=˙p

. (14.6.54)

The Hamiltonian function H i s deﬁned by

H =



i=1

˙q

− L. (14.6.55)

In general, L = L (q

, ˙q

,t) is a function of q

,˙q

and t,where ˙q

enters

through the kinetic energy as a quadratic term. Hence, equation (14.6.53a)

will give p

as a linear function of ˙q

. This system of linear equations in-

volving p

and ˙q

can be solved to determine ˙q

in terms of p

,andthen,the

˙q

can, in principle, be eliminated from (14.6.55). This means that H can

always be expressed as a function of p

, q

and t so that H = H (p

,t).

Thus,

dH =



∂H

∂p



∂H

∂q

∂H

∂t

dt. (14.6.56)

On the other hand, diﬀerentiating H in (14.6.55) with respect to t gives



˙q



˙q

−



∂L

∂q

−



∂L

∂ ˙q

˙q

−

∂L

∂t

(14.6.57)

14.6 Variational Methods and the Euler–Lagrange Equations 643

dH =



d ˙q



˙q

−



∂L

∂q

−



∂L

∂ ˙q

d ˙q

−

∂L

∂t

dt, (14.6.58)

which becomes, in view of (14.6.53a),

dH =



˙q

−



∂L

∂q

−

∂L

∂t

dt. (14.6.59)

Evidently, two expressions of dH in (14.6.56) and (14.6.59) must be

equal so that the coeﬃcients of the corresponding diﬀerentials can b e

equated to obtain

˙q

∂H

∂p

, −

∂L

∂q

∂H

∂q

, −

∂L

∂t

∂H

∂t

. (14.6.60abc)

Using the Lagrange equation (14.6.54), the ﬁrst two of the above equations

become

˙q

∂H

∂p

, ˙p

= −

∂H

∂q

. (14.6.61ab)

These are commonly known as the Hamilton canonical equations of motion.

They play a fundamental role in advanced analytical dynamics.

Finally, the Lagrange–Hamilton theory can be used to derive the law

of conservation of energy. In general, the Lagran gian L is independent of

time t and hence, (14.6.60c) implies that H = constant. Again, T involved

in L = T − V is given by

T =



i=1



j=1

˙q

, (14.6.62)

where the coeﬃcients a

are symmetric functions of the generalized coor-

dinates q

,thatis,a

= a

On the other hand, V is, in general, ind ependent of q

and hence,

∂L

∂ ˙q

∂T

∂ ˙q



j=1

˙q

. (14.6.63)

Thus, the Hamiltonian H becomes

H =



i=1

˙q

− L =



i=1

⎛

⎝



j=1

˙q

⎞

⎠

˙q

− L =2T − L = T + V. (14.6.64)

Thus, H is equal to the total energy. It has already been observed that,

if L does not contain t explicitly, H is a constant. This means that the

sum of the potential and kinetic energies is constant. This is the law of the

conservation of energy.

644 14 Numerical and Approximation Methods

Example 14.6.12. Use the Hamiltonian equations to derive the equations of

motion for the problem stated in Example 14.6.11.

The Lagrangian L for this problem is given by (14.6.40) with q

= r

and q

= θ. It follows from the deﬁnition (14.6.53a) of the generalized

momentum that

= m ˙q

= m ˙r, p

= mq

˙q

= mr

θ. (14.6.65)

Expressing the results of the kinetic energy (14.6.38) and the potential

energy (14.6.39) in terms of p

and p

the Hamiltonian H = T + V can be

written as

H =





+ m



F (q

) dq

. (14.6.66)

Then, equations (14.6.65) and the Hamilton equation (14.6.61b) give

= m ˙r, p

= mr

θ, (14.6.67)

˙p

+ mF (q

) , ˙p

=0. (14.6.68)

Clearly, these equations are identi cal with the equations of motion

(14.6.42ab).

Example 14.6.13. Derive the equation of a simple p end ulu m by using (i) the

Lagrange equations and (ii) the Hamilton equations.

We consider the motion of simple pendulum of mass m attached at the

end of a rigid massless string of length l that pivots about a ﬁxed point.

We suppose that the pendulum makes an angle θ with its vertical position.

The force F acting on the mass m is F = −mg sin θ, so that the potential

V is obtained f rom F = −∇V as V = mgl (1 − cos θ). The kinetic energy

T =

Thus the Lagrangian L is

L = T − V =

− mgl (1 − cos θ)=L



θ,



. (14.6.69)

The Lagrange equation is

∂L

∂θ

−



∂L

∂



=0, (14.6.70)

−mgl sin θ −





=0,

θ + ω

sin θ =0,ω

= g/l. (14.6.71ab)

14.6 Variational Methods and the Euler–Lagrange Equations 645

This is the equation of the simple pendulum.

To derive the same equation from the Hamilton equations, we choose

= l (˙q

=0) and q

= θ as the generalized (polar) coordinates. The

kinetic and potential energies are

T =

˙q

,V= mgl (1 − cos q

) . (14.6.72ab)

Thus, H = T + V and L = T − V are given by

(H, L)=

˙q

mgl (1 − cos q

) . (14.6.73ab )

From the deﬁnition of the generalized momentum, we ﬁnd that

∂L

∂ ˙q

= ml

˙q

so that the Hamiltonian H in terms of p

and q

H =

+ mgl (1 − cos q

) .

Thus, the Hamilton equation (14.6.61ab) gives

θ + ω

sin θ =0,ω

. (14.6.74)

The variational methods can b e further extended for function als de-

pending on functions or more independent variables in the form

I [u (x, y)] =



F (x, y, u, u

) dx dy (14.6.75)

where the values of the f unction u (x, y) are prescribed on the boundary

∂D of a ﬁnit e domain D in the (x, y)-plane. We assume that F is diﬀeren-

tiable and the surface u = u (x, y) giving an extremum is also continuously

diﬀerentiable twice.

The ﬁrst variation δI of I is deﬁned by

δI [u, ε]=I (u + ε) − I (u) (14.6.76)

which is, by Taylor’s expansion theorem



[εF

+ ε

] dx dy (14.6.77)

where ε ≡ ε (x, y) is small and p = u

and q = u

.Accordingtothe

variational pri nciple, δI = 0 for all admissible values of ε. The partial

integration of (14.6.77) combi ned with ε =0on∂D gives

646 14 Numerical and Approximation Methods

0=δI =





−

∂

∂x

−

∂

∂y



ε (x , y) dx dy. (14.6.78)

This is true for all arbitrary ε, and hence, the integrand must vanish, that

∂

∂x

∂

∂y

− F

=0. (14.6.79)

This is the Euler–Lagrange equation which is the second-order partial di f-

ferential equation to be satisﬁed by the extremizing function u (x, y).

Example 14.6.14. Derive the equation of motion for the free vibration of an

elastic string of length l.

The potential energy V of the string is

V =

∗



dx (14.6.80)

where u = u (x, y) is the displacement of the string from its equilibrium

position and T

∗

is the constant tension of the string.

The kinetic energy T is

T =



ρu

dx (14.6.81)

where ρ is the constant line-density of the string.

According to the Hamilton principle

δI = δ



(T − V ) dt = δ





ρu

− T

∗



dx dt = 0 (14.6.82)

which has the for m



L (u

)=0, (14.6.83)

where

L =



ρu

− T

∗



Then the Euler–Lagrange equation is given by

∂

∂t

(ρu

) −

∂

∂x

∗

)=0, (14.6.84)

− c

=0,c

= T

∗

/ρ. (14.6.85)

This is the wave equation of motion of the string.

14.7 The Rayleigh–Ritz Approximation Method 647

Example 14.6.15. Derive th e Laplace equation from the functional

I (u)=





+ u



dx dy

with a boundary condition u = f (x, y)on∂D.

The variational principle gives

δI = δ





+ u



dx dy =0.

This leads to the Euler–Lagrange equation

+ u

=0 in D.

Similarly, the functional

I [u (x, y, z)] =





+ u



dx dy dz

will lead to the three-dimensional Laplace equation

∇

u = u

+ u

=0.

14.7 The Rayleigh–Ritz Approximation Method

We consider the boundary-value problem governed by the diﬀerential equa-

tion

Au = f in D (14.7.1)

with the boundary condition

B (u)=0 on ∂D (14.7.2)

where A is a self-adjoint diﬀerential operator in a Hilbert space H and

f ∈ H.

In general, the determination of the exact solu tion of the problem is of-

ten a diﬃcult task. However, it can be shown that the solution of (14.7.1)–

(14.7.2) is equivalent to ﬁnding the minimum of a functional I (u) associated

with the diﬀerential system. In other words, the solution can be ch aracter-

ized as the function which minimizes (or maximizes) the functional I (u). A

simple and eﬃcient method for an approximate soluti on of the extremum

problem was independently formulated by Lord Rayleigh and W. Ritz.

We next prove a fundamental result which states that the solution of

the equation (14. 7.1) is equivalent to ﬁnding the minimum of the quadratic

functional

648 14 Numerical and Approximation Methods

I (u) ≡ A u, u−2 f,u. (14.7.3)

Suppose that u = u

is the solution of (14.7.1) so that Au

= f. Conse-

quently,

I (u) ≡ A u, u−2 Au

,u = A (u − u

) ,u−Au

,u.

Since the inner product is symmetrical and Au

,u = u

,Au =

Au, u

, I (u) can be written as

I (u)=A (u − u

) ,u−Au, u

 + Au

,u−Au

,

= A (u − u

) ,u− u

−Au

,

= A (u − u

) ,u− u

 + I (u

) . (14.7.4)

Since A is a positive operator, A (u − u

) ,u− u

≥0 where equality

holds if and only if u − u

= 0. It follows that

I (u) ≥ I (u

) , (14.7.5)

where equality holds if and only if u = u

. We conclude from this inequality

that I (u) assumes its mini mum at the solution u = u

of equation (14.7.1).

Conversely, the function u = u

that minimizes I (u) is a solution of

equation (14.7.1). Clearly, I (u) ≥ I (u

), that is, in particular, I (u

+ αv) ≥

I (u

) for any real α and any function v. Explicitly,

I (u

+ αv)=A (u

+ αv) ,u

+ αv−2 f,u

+ αv,

= Au

 +2α Au

,v+ α

Av, v−2 f, u

−2α f,v.

This means that I (u

+ αv) is a quadratic expression in α.SinceI (u)is

minimum at u = u

,thenδI (u

,v) = 0, that is,



dα

I (u

+ αv)



α=0

=2Au

,v−2 f,v

=2Au

− f, v.

This is true for any arbitrary but ﬁxed v. Hence, Au

−f =0.Thisproves

the assertion.

In the Rayleigh–Ritz method an approximate solution of (14.7.1)–

(14.7.2) is sought in the form

(x)=



i=1

(x) , (14.7.6)

where a

, a

, ..., a

are n unk nown coeﬃcients to be determined so that

I (u

) is minimum, and φ

, φ

, ..., φ

represent a linearly independent and

14.7 The Rayleigh–Ritz Approximation Method 649

complete set of arbitrarily chosen functions that satisfy (14.7.2). This set

of functions is often called a trial set. We substitute (14.7.6) into (14.7.3)

to obtain

I (u

;



i=1

A (φ

) ,



j=1

− 2

;



i=1

Then the necessary condition for I to obtain a minimum (or maximum) is

that

∂I

∂a

,...,a

)=0,j=1, 2,...,n, (14.7.7)

∂I

∂a

⎡

⎣

;



i=1

A (φ

) ,



j=1

− 2

;



i=1

⎤

⎦

=0,



i=1

A (φ

) ,φ

a



i=1

A (φ

) ,φ

a

− 2 f, φ

 =0,



i=1

A (φ

) ,φ

a

=2f,φ

.

Therefore,



i=1

A (φ

) ,φ

a

= f,φ

,j=1, 2,...,n. (14.7.8)

This is a linear system of n equations for the n unknown coeﬃcients a

Once a

, a

, ..., a

are determined, the approximate solution is given by

(14.7.6).

In particular, when

A (φ

) ,φ

 =

⎧

⎨

⎩

0,i= j

1,i= j,

(14.7.9)

equation (14.7.8) gives a

= φ

,f, (14.7.10)

so that the Rayleigh–Ritz approximate series (14.7.6) becomes

650 14 Numerical and Approximation Methods

(x)=



i=1

φ

,fφ

(x) . (14.7.11)

This is similar to the Fourier series solution with known Fourier coeﬃcients

In the limit n →∞, a limit function can be obtained from (14.7.6) as

u (x) = lim

n→∞



i=1

(x)=

∞



i=1

(x) , (14.7.12)

provided that the series converges. Under certain assumptions imposed on

the functional I (u) and the tri al functions φ

, φ

, ..., φ

, the limit function

u (x) represents an exact solution of the problem. In any event, (14.7.6) or

(14.7.11) gives a reasonable approximate solution.

In the simplest case corresponding to n = 1, the Rayleigh–Ritz method

gives a simple form of the functional

I (u

)=I (a

)=a

Aφ

,φ

−2a

f,φ

,

where a

is readily determined from the n ecessary condition for extremum

∂

∂a

I (a

)=2a

Aφ

,φ

−2 f, φ

,

f,φ



Aφ

,φ



. (14.7.13)

The corresponding minimum value of the functional is given by

I (a

)=−

f,φ



Aφ

,φ



. (14.7.14)

Thus, the essence of the Rayleigh–Ritz method is as follows. For a given

boundary-value problem, an approximate series solution is sought so that

the trial functions φ

satisfy the boundary conditions. We solve the system

of algebraic equations (14.7.7) to determine the coeﬃcients a

We now illustrate the method by several examples.

Example 14.7.1. Find an approximate solution of t he Dirichlet problem

∇

u ≡ u

+ u

=0 in D

u = f on ∂D,

where D ⊂ R

,andf is a given function.

This problem is equivalent to ﬁnding the minimum of the associated

functional