5.5 Constrained variational problems

It is quite common that we have subsidiary conditions on a minimisation problem, i.e., we want to know the minimum provided that certain other conditions hold. Let us first analyse this problem for ordinary functions first

5.5.1 Lagrange’s undetermined multipliers

To find the stationary points of a function f(x) subject to constraints {g}_{k}(x) = 0, (k = 1,2,\mathop{\mathop{…}}), we can solve an extended problem and find the unconstrained stationary points of the extended functional

w.r.t to x and {λ}_{i}.

Let’s look at a somewhat explicit example. We wish to minimise a function f(x,y) subject to a single constraint g(x,y) = 0. We thus need to minimise the extended function

and find

The last line clearly states that the constraint must be implemented. The new terms (proportional to λ) in the first two lines say that a constrained minimum is reached when the gradient of the function is parallel to the gradient of the constraint. This says that there is no change in the function, unless we violate our constraint condition–clearly a sensible definition of a stationary point.

Example 5.10: 

Find stationary points of f(x,y) under the subsidiary condition {x}^{2} + {y}^{2} = 1.

Solution: 

Look for stationary points of

F = xy − λ({x}^{2} + {y}^{2} − 1),

which are given by the solution(s) of

\eqalignno{ {∂F\over ∂x} & = y − 2λx = 0, & & \cr {∂F\over ∂y} & = x − 2λy = 0, & & \cr {∂F\over ∂λ} & = {x}^{2} + {y}^{2} − 1 = 0.\ & & }

The first two conditions give {x}^{2} = {y}^{2}, and from the constraint we find (x,y) = (±1∕\sqrt{2},±1∕\sqrt{2}). The values of λ associated with this are λ = ±1∕2.

5.5.2 Generalisation to functionals

We look for stationary points of I[y] subject to a constraint J[y] = C, (can also be multiple constraints) where I, J are given functionals of y(x) and C is a given constant.

To do this: we solve for the stationary points of an extended funtional,

with respect to variations in the function y(x) and λ. We then have

which can be dealt with as an unconstrained variational problem. Its solution can be slightly tricky; one way is to solve the problem

for fixed λ to find y(x) as a function of λ, and then use the constraint J[y] = C to find the allowed value(s) of λ, and thus the solution.

Figure 5.7: The isoperimetric problem.

Example 5.11: 

Find a closed curve of fixed length L = 2πl which encloses the maximum area A. (The isoperimetric problem, see Fig. 5.7.)

Solution: 

Describe the curve in polar coordinates by (θ,r(θ)), assuming the origin lies within the curve. We then find that

\eqalignno{ A[r] & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2π}{1\over 2}{r}^{2}dθ, & & \cr L[r] & =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dl ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2π}({r}^{2}{dθ}^{2} + d{r}^{2}) & & \cr & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2π}\sqrt{{r}^{2 } +{ r' }^{2}}dθ = 2πl\quad . & & }

We now need to find stationary points of

I[r,λ] = A[r] − λ(L[r] − 2πl) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2π}({r}^{2}∕2 − λ\sqrt{{r}^{2 } +{ r' }^{2}})dθ + 2πl.

Minimising with respect to r(θ), we find a problem where there is no explicit dependence of the function we usually call “F” on θ, and thus

F − r'{∂F\over ∂r'} = K.

Explicitly,

\eqalignno{ K & = {1\over 2}{r}^{2} − λ\sqrt{{r}^{2 } +{ r' }^{2}} − r'\left (−λ(−1∕2)2r'∕\sqrt{{r}^{2 } +{ r' }^{2}}\right ) & & \cr & = {1\over 2}{r}^{2} − λ {{r}^{2}\over \sqrt{{r}^{2 } +{ r' }^{2}}}\quad , &\text{(5.20)} }

together with the constraint

L ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2π}\sqrt{{r}^{2 } +{ r' }^{2}}dθ = 2πl.

Unfortunately, this equation is not easy to solve in general, but we can guess one solution: Look at λ = 0 when we find r = l, K = {l}^{2}∕2. This describes a circle through the origin. By translational invariance we see any other circle also satisfies this condition.

Figure 5.8: The catenary

Example 5.12: 

What is the equilibrium curve for a flexible “chain” of length l and density ρ per unit length, when we hang it between two points A and B. (The catenary, see Fig. 5.8.)

Solution: 

We need to minimise the gravitational potential energy

E[y] ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{B}ρdl\kern 1.66702pt y ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}ρgy\sqrt{1 +{ y' }^{2}}dx,

subject to the constraint of constant length L[y] = l, with

L[y] ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{A}^{B}dl ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}\sqrt{1 +{ y' }^{2}}dx.

Thus we need to find stationary points of G[y] = E[y] − λ(L[y] − l).

As usual the variation with respect to y is simple, since the integrand

F(y,y',λ) = (ρgy − λ)\sqrt{1 +{ y' }^{2}}

has no explicit dependence on x we can use the first integral,

F − y'{∂F\over ∂y'} = C,

or explicitly,

(ρgy − λ)\sqrt{1 +{ y' }^{2}} − y'(ρgy − λ) {y'\over \sqrt{1 +{ y' }^{2}}} = {ρgy − λ\over \sqrt{1 +{ y' }^{2}}} = C.

This can be solved by making a shift on y,

u = y − λ∕(gρ),\qquad α = C∕(gρ).

The new function u satisfies

\eqalignno{ u & = α\sqrt{1 +{ u' }^{2}} &\Rightarrow & & & & \cr {u}^{2}∕{α}^{2} & = 1 +{ u'}^{2} &\Rightarrow & & & & \cr {u'}^{2} & = ({u}^{2} − {α}^{2})∕{α}^{2} &\Rightarrow & & & & \cr {αu'\over \sqrt{{u}^{2 } − {α}^{2}}} & = 1 &\Rightarrow & & & & \cr \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {αdu\over \sqrt{{u}^{2 } − {α}^{2}}} & =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx\quad . & & & & }

Now use u = α\mathop{cosh}\nolimits w, du = α\mathop{sinh}\nolimits w\kern 1.66702pt dw to find

\eqalignno{ \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {{α}^{2}\mathop{ sinh}\nolimits w\kern 1.66702pt dw\over α\mathop{sinh}\nolimits w} & = x &\Rightarrow & & & & \cr αw & = x − {x}_{0} &\Rightarrow & & & & \cr u & = α\mathop{cosh}\nolimits ((x − {x}_{0})∕α) &\Rightarrow & & & & \cr y & = α\mathop{cosh}\nolimits ((x − {x}_{0})∕α) + λ∕(ρg)\quad . & & & & }

The three constants α, {x}_{0} and λ are determined by the condition that line goes through A and B and has length

l ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}{\left (1 + {α}^{2}{\mathop{ sinh}\nolimits }^{2}(x − {x}_{ 0})∕α\right )}^{1∕2}dx.

The point {x}_{0} is where the curve gets closest to the ground. If y(a) = y(b), {x}_{0} = (a + b)∕2 by symmetry.

A good demonstration can be found on ”Catenary: The Hanging Chain” on The Wolfram Demonstrations Project.

5.5.3 Eigenvalue problems

Consider the eigenvalue equation for the function u(x)

where L is an Hermitian operator, ρ(x) is a positive, real weight function. We now look for the stationary points of

subject to the normalisation constraint

We first look for an unconstrained stationary point of

and vary λ to obtain a solution that satisfies the constraint. We get

where we have used hermiticity of the operator L.

A key difference with the previous examples is that we have a complex function u, and any variation thus falls into two parts,

The real and imaginary parts are independent small functions, i.e., we can vary those independently. In the same way we can conclude that the alternative orthogonal combination of these two variations provided by δu and δ{u}^{∗} vary independently, so we can select either of the two terms above, since they must both be zero independently. The function multiplying δ{u}^{∗} must thus satisfy

which shows u is an eigenfunction of L. We conclude that the stationary points are the eigenfunctions u = {u}_{0},{u}_{1},\mathop{\mathop{…}} and the corresponding values of λ are the eigenvalues {λ}_{0},{λ}_{1},\mathop{\mathop{…}}.

Now suppose that there is a minimum eigenvalue {λ}_{0}. This implies that for a normalised u, I[u] ≥ {λ}_{0}. We show below how we can use that to our benefit.

5.5.4 The Rayleigh-Ritz method

Suppose the function {u}_{0} gives the minimum of

subject to the constraint

Now suppose {v}_{0} gives the unconstrained minimum of

This technique is very commonly used in quantum mechanics, where we then replace the functional dependence with a parametrised dependence by choosing a set of wave functions depending on set of parameters. In this case L = H, ρ = 1 and u = ψ.

Example 5.13: 

Find an approximation to the ground state of the quartic anharmonic oscillator

H = −{1\over 2} {{d}^{2}\over d{x}^{2}} + {1\over 2}{x}^{2} + λ∕2{x}^{4},

of the form ψ(x) =\mathop{ exp}\nolimits (−α{x}^{2}∕2).

Solution: 

The normalisation integral is

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}\mathop{exp}\nolimits (−α{x}^{2})dx = \sqrt{π∕α}.

By differentiating both sides w.r.t. α, we get two more useful integrals,

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{x}^{2}\mathop{ exp}\nolimits (−α{x}^{2})dx = {1\over 2α}\sqrt{π∕α}.

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{x}^{4}\mathop{ exp}\nolimits (−α{x}^{2})dx = {3\over 4{α}^{2}}\sqrt{π∕α}.

Thus the expectation value of the Hamiltonian requires the derivative

{{d}^{2}\over d{x}^{2}}\mathop{ exp}\nolimits (−α{x}^{2}) = ({x}^{2}{α}^{2} − α)\mathop{exp}\nolimits (−α{x}^{2}).

Thus the denominator of the variational functional becomes

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}(−{x}^{2}{α}^{2} + α + {x}^{2} + λ{x}^{4})∕2\mathop{exp}\nolimits (−α{x}^{2})dx = \sqrt{π∕α}\left (−α∕2 + α + 1∕(2α) + 3λ∕(4{α}^{2})\right )∕2.

And thus

K(α) = {1\over 4}\left (α + 1∕α + 3λ∕(2{α}^{2})\right ).

Minimising w.r.t. α, we find

1 − 1∕{α}^{2} − 3λ∕{α}^{3} = 0.

This equation can be solved in closed form, but is rather complicated. We find that α increases with λ, see Fig. 5.9.

Figure 5.9: The parameter α and the energy E as a function of anharmonicity λ.