We would like to study the extremum or the local extremum of the function S : Xα,β → R, q 7→ Z b a L(q(t), ˙q(t), t)dt where ˙q(t

Let L : U ⊆ R³ → R be a smooth function defined on an open subset of R³. For each α, β in R, we set

X_α,β= {q(x) ∈ C¹[a, b] : q(a) = α, q(b) = β}.

We would like to study the extremum or the local extremum of the function S : Xα,β → R, q 7→

Z b a

L(q(t), ˙q(t), t)dt

where ˙q(t) = dq/dt. Because Xα,β is not a normed space, we can not define/compute the derivative of S. Hence we are not able to use theory of differentiation to find the (local) extremum of S. To remedy this situation, we introduce the following vector subspace of C¹[a, b] :

X = {f ∈ C¹[a, b] : f (a) = f (b) = 0}.

Proposition 1.1. For each f ∈ C¹[a, b], we set

kf k_C1 = kf k∞+ kf⁰k∞.

Then (C¹[a, b], k · k_C1) is a real Banach space and X is a closed vector subspace of C¹[a, b];

hence X is also a real Banach space.

Proof. Exercise. 

Although X_α,β is not a normed space, it is a metric subspace of C¹[a, b]. Recall that the induced metric of a normed space (V, k · k_V) is defined to be

d(v, w) = kv − wk_V.

Hence the metric d_C1 on C¹[a, b] associated to k · k_C1 is d_C1(f, g) = kf − gk_C1. On the other hand, any nonempty subset N of a metric space (M, d) has a natural metric dN defined to be d_N(x, y) = d(x, y) for x, y ∈ N. The metric space (N, d_N) is called the metric subspace of (M, d). Hence we equip Xα,β with the subspace metric dXαβ induced from (C¹[a, b], d_C¹) so that X_αβ becomes a metric subspace of C¹[a, b]. Therefore we can talk about open sets and closed sets of X_αβ. Notice that C¹[a, b] is complete and X_αβ is a closed subspace of C¹[a, b], X_αβ is also a complete metric space.

Definition 1.1. Let (M, d) be a metric space. A function F : M → R has a local maximum (minimum) at x₀ if there exists δ > 0 so that

F (x) ≤ F (x₀), (F (x) ≥ F (x₀)) for any x ∈ B(x0, δ).

Suppose that S has a local extremum at q0, i.e. there exists δ > 0 so that either S(q) ≥ S(q0) or S(q) ≤ S(q0)

whenever d_Xαβ(q, q0) = kq − q0k_C1 < δ with q ∈ X_α,β. We define a function S : X → R, S(f ) = S(q0+ f ).

Then S has a local extremum at f = 0.

Proposition 1.2. Let V be a normed vector space over R and S : V → R be a differentiable function. Suppose that S has a local extremum at v₀ ∈ V. Then

(DS)(v₀)(h) = 0 for any h ∈ V .

1

Proof. Let h ∈ V and consider the function F_h(t) = S(v0+ th) for |t| < δ. If S has a local extremum at v₀, F_h has a local extremum at t = 0. By Calculus, F_h⁰(0) = 0. By chain rule,

F_h⁰(0) = DS(v0)(h). We prove our assertion. 

Let us compute DS(0). To do this, S(h) − S(0) =

Z b a



L(q₀+ h, ˙q₀+ ˙h, t) − L(q₀, ˙q₀, t) dt.

We use Taylor’s Theorem to obtain that L(q₀+ h, ˙q₀+ ˙h, t) − L(q₀, ˙q₀, t) = ∂L

∂x(q₀, ˙q₀, t)h(t) + ∂L

∂y(q₀, ˙q₀, t) ˙h(t)

+ A(h, ˙h, t)h(t)²+ 2B(h, ˙h, t)h(t) ˙h(t) + C(h, ˙h, t) ˙h(t)². Choose δ > 0 so that

|A(h, ˙h, t)|, |B(h, ˙h, t)|, |C(h, ˙h, t)| ≤ M for some M > 0 whenever khk_C1 < δ. Then

A(h, ˙h, t)h(t)²+ 2B(h, ˙h, t)h(t) ˙h(t) + C(h, ˙h, t) ˙h(t)²  

≤ M (|h(t)|²+ 2|h(t)|| ˙h(t)| + | ˙h(t)|²)

≤ M (|h(t)| + |h⁰(t)|)²

≤ M (khk_∞+ kh⁰k_∞)² = M khk²_C1. Let T : X → R be the function

T (h) = Z b

a

 ∂L

∂x(q0, ˙q0, t)h(t) +∂L

∂y(q0, ˙q0, t) ˙h(t)

 dt.

Lemma 1.1. The function T : X → R is a bounded linear map.

Proof. The linearity of T is left to the readers. By continuity of Lx and Ly, the functions t 7→ Lx(q0(t), ˙q0(t), t) and t 7→ Ly(q0(t), ˙q0(t), t) are continuous on [0, 1]. By compactness of [0, 1], we can find M > 0 such that

|L_x(q0(t), ˙q0(t), t)| , |Ly(q0(t), ˙q0(t), t)| ≤ M for any t ∈ [0, 1]. By triangle inequality

|T (h)| ≤ M Z 1

0

(|h(t)| + |h⁰(t)|)dt ≤ M khk_C¹. This shows that T is a bounded linear map.

 We see that |S(h) − S(0) − T (h)| ≤ M khk²_C1 when 0 < khk_C¹ < δ. This implies that

khklim_C1→0

|S(h) − S(0) − T (h)|

khk_C1

= 0.

In other words, DS(0) = T. If h ∈ X, by integration by parts, DS(0)(h) =

Z b a

 ∂L

∂x(q₀, ˙q₀, t)h(t) +∂L

∂y(q₀, ˙q₀, t) ˙h(t)

 dt

= h(t)∂L

∂y    

t=b t=a

+ Z b

a

 ∂L

∂x(q0, ˙q0, t) − d dt

∂L

∂y(q0, ˙q0, t)

 h(t)dt

= Z _b

a

 ∂L

∂x(q₀, ˙q₀, t) − d dt

∂L

∂y(q₀, ˙q₀, t)

 h(t)dt.

Lemma 1.2. Let f : [a, b] → R be a continuous function. Suppose that Z b

a

f (x)h(x)dx = 0

for any C¹ function h : [a, b] → R with h(a) = h(b) = 0. Then f is the zero function.

Proof. Exercise. 

If S has a local extremum at f = 0. we have DS(0)(h) = 0 for any h ∈ X. In other words, Z _b

a

 ∂L

∂x(q0, ˙q0, t) − d dt

∂L

∂y(q0, ˙q0, t)



h(t)dt = 0 for any h ∈ X. By Lemma 1.2, we find

∂L

∂x(q0, ˙q0, t) − d dt

∂L

∂y(q0, ˙q0, t) = 0.

This equation is called the Euler-Lagrange equation for S. In physics literature, the Euler- Lagrange equation is denoted by

∂L

∂q(q₀, ˙q₀, t) − d dt

∂L

∂ ˙q(q₀, ˙q₀, t) = 0.

Example 1.1. Let L : R³ → R be the function L(x, y, z) = 1

2my²− V (x)

where V ∈ C^∞(R). Compute the Euler Lagrange-equation for the function S : Xα,β → R defined by

S(q) = Z b

a

L(q, ˙q, t)dt.

Solution. By the formula,

∂L

∂x = −V⁰(x), ∂L

∂y = my.

Hence

∂L

∂x(q₀, ˙q₀, t) − d dt

∂L

∂y(q₀, ˙q₀, t) = −V⁰(q₀) − m¨q₀= 0.

In other words, m¨q₀ = −V⁰(q₀).

Example 1.2. Let V = {f ∈ C¹[1, 2] : f (1) = 0, f (2) = 1} and define S : V → R by S(q) =

Z 2 1

p1 + ( ˙q(t))²

t dt.

Compute the Euler-Lagrange equation for S.

The Lagrangian of S is

L(x, y, z) =

p1 + y²

z .

The Euler-Lagrange equation for S is

∂L

∂x(q, ˙q, t) − d dt

∂L

∂y(q, ˙q, t) = −d dt

˙ q

tp1 + ( ˙q)² = 0.

By integrating this equation, we find

˙ q

tp1 + ( ˙q)² = C.

Therefore

˙

q² = (Ct)²

1 − (Ct)² ⇒ ˙q = Ct p1 − (Ct)². Integrating this new equation, we find

q = 1 C

p1 − (Ct)²+ C₁⇒ t²+ (q − C₁)² = 1 C².

Now let us consider a more general setting. Let U be an open subset of Rⁿ× Rⁿ× R and L : U → R be a smooth function. We consider L as a function of (x1, · · · , x_n, y₁, · · · , y_n, z).

Let PA,B be the set of all C¹-functions q : [a, b] → Rⁿ such that q(a) = A and q(b) = B, where A, B ∈ Rⁿ. We would like to study the (local) extremum of the function

S : P_A,B → R, q 7→ S(q) = Z b

a

L(q(t), ˙q(t), t)dt.

On C¹([a, b], Rⁿ), we define a norm by kf k_C1 = sup

t∈[a,b]

kf (t)k_Rⁿ+ sup

t∈[a,b]

kf⁰(t)k_Rⁿ.

Proposition 1.3. C¹([a, b], Rⁿ) with k · k_C1 defined above is a Banach space over R. If X is the subset of C¹([a, b], Rⁿ) consisting of functions satisfying f (0) = f (1) = 0, then X is a closed vector subspace of C¹([a, b], Rⁿ).

Proof. The proof is similar as before. 

Assume that q₀ is a local extremum of S. We define a function S : X → R by S(f ) = S(f + q₀).

We can show that DS(0)(h) =

Z b a

n

X

i=1

∂L

∂x_i(q₀, ˙q₀, t)h_i+

n

X

i=1

∂L

∂y_i(q₀, ˙q₀, t) ˙h_i(t)

! dt.

Using integration by parts and the fact that h ∈ X, we have DS(0)(h) =

Z b a

n

X

i=1

 ∂L

∂xi

(q₀, ˙q₀, t) − d dt

∂L

∂yi

(q₀, ˙q₀, t)



h_i(t)dt.

If q₀ is a local extremum for S, 0 is a local extremum for S. In this case, we can show that

(1.1) ∂L

∂x_i(q₀, ˙q₀, t) − d dt

∂L

∂y_i(q₀, ˙q₀, t) = 0 for 1 ≤ i ≤ n.

We call (1.2) the Euler-Lagrange equation for S. In physics literature, they use

(1.2) ∂L

∂qi

(q₀, ˙q₀, t) − d dt

∂L

∂ ˙qi

(q₀, ˙q₀, t) = 0 for 1 ≤ i ≤ n, where q = (q1, · · · , qn).

Example 1.3. Let V : R³ → R be a function and consider the function S : PA,B → R defined by

S(q) = Z b

a

L(q(t), ˙q(t), t)dt where L is given by

L(q(t), ˙q(t), t) = 1

2mk ˙q(t)k²

R³− V (q).

The Euler Lagrange equation for S is given by

∂L

∂q_i − d dt

∂L

∂ ˙q_i = −∂V

∂q_i − d

dtm ˙q_i = 0, 1 ≤ i ≤ 3.

The system of equation is equivalent to

−∇V (q) = m¨q.

Example 1.4. Consider the function S : P_A,B → R defined by S(γ) =

Z ₁

0

p( ˙x(t))²+ ( ˙y(t))²dt

where γ(t) = (x(t), y(t)) for t ∈ [0, 1]. Find the curve γ minimizing the function S.

The Lagrangian of S is given by

L(x, y, ˙x, ˙y, t) =p

( ˙x(t))²+ ( ˙y(t))².

The Euler-Lagrange equation for S is the following system of differential equations











∂L

∂x − d dt

∂L

∂ ˙x = 0

∂L

∂y − d dt

∂L

∂ ˙y = 0 By direct calculation, we have

∂L

∂x = ∂L

∂y = 0, ∂L

∂ ˙x = x˙

p( ˙x)²+ ( ˙y)², ∂L

∂ ˙y = y˙ p( ˙x)²+ ( ˙y)². Therefore the Euler-Lagrange equation for S is given by













 d dt

˙ x

p( ˙x)²+ ( ˙y)² = 0 d

dt

˙ y

p( ˙x)²+ ( ˙y)² = 0

In other words, we find that there exist c1, c2 ∈ R so that















˙ x

p( ˙x)²+ ( ˙y)² = c₁

˙ y

p( ˙x)²+ ( ˙y)² = c₂ This implies that

c²₁+ c²₂ = x˙ p( ˙x)²+ ( ˙y)²

!2

+ y˙

p( ˙x)²+ ( ˙y)²

!2

= 1.

Therefore c1, c2can not be zero at the same time. Without loss of generality, we may assume that c₁ 6= 0. This implies that ˙y = a ˙x where a = c₂/c₁. Integrating this equation, we find y = ax + b which is the straight line on R². Using the conditions q(0) = A and q(1) = B, we would be able to determine a, b.

Example 1.5. Let E, F, G be real valued smooth functions on R² such that E > 0 and EG − F²> 0 on R². Define L : R²× R²× R → R by

L(x₁, x₂, y₁, y₂, z) = E(x₁, x₂)y₁²+ 2F (x₁, x₂)y₁y₂+ G(x₁, x₂)y²₂. Define S : P_A,B→ R by

S(q) = Z 1

0

L(u(t), v(t), ˙u(t), ˙v(t), t)dt where q(t) = (u(t), v(t)).

(1) Compute the Euler-Lagrange equation for S.

(2) Solve for the Euler-Lagrange equation when E(x1, x2) = 1 and F (x1, x2) = 0 and G(x₁, x₂) = sin²x₂.

The Euler Lagrange equation for S is given by











∂L

∂u − d dt

∂L

∂ ˙u = 0

∂L

∂v − d dt

∂L

∂ ˙v = 0 By direct calculation, we have

Lu= Eu(u(t), v(t))(u⁰(t))²+ 2Fu(u(t), v(t))u⁰(t)v⁰(t) + Gu(u(t), v(t))(v⁰(t))² L_v = E_v(u(t), v(t))(u⁰(t))²+ 2F_v(u(t), v(t))u⁰(t)v⁰(t) + G_v(u(t), v(t))(v⁰(t))² Lu˙ = 2E(u(t), v(t))u⁰(t) + 2F (u(t), v(t))v⁰(t)

Lv˙ = 2F (u(t), v(t))u⁰(t) + 2G(u(t), v(t))v⁰(t).

We obtain the following system of differential equations

(1.3)









 d

dt 2Eu⁰(t) + 2F v⁰(t)

= E_u(u⁰(t))²+ 2F_uu⁰(t)v⁰(t) + G_u(v⁰(t))² d

dt 2F u⁰(t) + 2Gv⁰(t)

= E_v(u⁰(t))²+ 2F_vu⁰(t)v⁰(t) + G_v(v⁰(t))².

The system of differential equations (1.3) is called the geodesic equation for (E, F, G). Let us find the solution for (1.3) when E = 1 and F = 0 and G = sin²x₂. We obtain that







u⁰⁰(t) = 0

d

dt 2v⁰(t) sin²(v(t))

= 2(v⁰(t))²sin v(t) cos v(t).

The first equation implies that u(t) = at + b. The second equation implies that v⁰⁰(t) sin²(v(t)) + sin(v(t)) cos(v(t))(v⁰(t))²= 0.

When sin²v 6= 0, we obtain that

v⁰⁰(t) sin(v(t)) + (v⁰(t))²cos(v(t)) = 0.

This implies that (v⁰(t) sin(v(t)))⁰ = 0 and hence v⁰sin v = c for some constant c. This also implies that

d

dtcos(v(t)) = c⁰

for some constant c⁰. Therefore cos(v(t)) = c⁰t + c⁰⁰ for some constant c⁰ and c⁰⁰. You can also discuss the case when sin v(t) = 0. We leave it to the reader. Therefore the solution to the Euler-Lagrange equation for S is given by

u(t) = at + b, v(t) = cos⁻¹(c⁰t + c⁰⁰).