is said to be bounded if there exist a constant C such that kAvk ≤ Ckvk for all v ∈ V .
Proposition 2. A linear operator A : V → W is bounded if and only if it is continuous.
Examples
1. Integral operator: Let k(x, y) be a continuous function defined on [a, b] × [a, b]. the operator defined by
Af (x) = Z b
a
k(x, y)f (y) dy maps f ∈ C[a, b] into Af ∈ C[a, b]. This operator is bounded.
2. Consider the interval [0, 1]. Let k(x, y) be defined as k(x, y) =
(1 − y)x for x < y (1 − x)y for x > y You can show that Af ∈ C1[0, 1] if f ∈ C[a, b].
For a bounded operator A : V → W , we can define its operator norm by kAk := supkAf kW
kf kV .
One can show that the set B(V, W ) := {A : V → W is a bounded operator} endow with this operator norm is a normed linear space. If W is complete, so is B(V, W ).
Unbounded Operators The unbounded operators are those differential operators. They usually define on a dense subspace of a Banach space. Here is the definition:
Examples
1. u 7→ Du is defined on C1(a, b) whose range is in C0(a, b). We may think D : C1(a, b) ⊂ C0(a, b) → C0(a, b).
2. D can also be thought as a mapping defined on H1(a, b) ⊂ L2(a, b) to L2(a, b).
3.4 Variation of Functionals
As we have seen in Section 1 of this chapter that variational problems involve a functional J defined on an admissible set A in a Banach space V . Our goal is to look for extrema of J in A. A vector y0 is said to be a local minimum of J in A if J (y0) ≤ J (y) for all y in a small neighborhood of y0in A.
48 CHAPTER 3. CALCULUS OF VARIATIONS Tangent space Suppose y0is a local extremum in A ⊂ V . A function h ∈ V such that y0+ h ∈ A for all sufficiently small , such a function h is called a variation of A at y0. In other words, h lies on the tangent space of A at y0. For example, let us consider the set
A = {y : [a, b] 7→ R3 | y ∈ C1[a, b] and y(a) = A, y(b) = B}
where A and B are two points in R3. The admissible variations are
T (y0) = {h : [a, b] 7→ R3| h ∈ C1[a, b] and h(a) = 0, h(b) = 0}
Sometimes, we denote a variation h by δy.
Directional Derivatives The directional derivative of J at y0in a direction h ∈ T (y0) is defined to be dJ (y0+ h)/d at = 0, if it exists. We denote it by δJ (y0, h). If δJ (y0, h) exists for all the admissible variations h ∈ T , one can show that δJ (y0, αh) = αδJ (y0, h). However, it may not be linear in h. Even if it is linear in h, it may not be continuous in h. In most applications, it is indeed a bounded linear in h. In this case, we can express δJ (y0, h) by δJ (y0) · h. The functional δJ (y0) is called the first variation (or the Gˆateaux derivative) of J at y0. Notice that the linear functional δJ (y0) may not be bounded.
Another relevant definition is the Fr´echet’s derivative. A functional J uis said to be Fr´echet differentiable at y0 if there exists a bounded linear functional ` such that for any sufficiently small h, we have
J (y0+ h) = J (y0) + `(h) + o(khk).
The functional ` is also denoted by δJ , or DJ . It is clear that Fr´echet differentiability implies Gˆateaux differentiability. Conversely, if J is Gˆateaux differentiable at y0, Further, the corresponding G´ateaux derivative δJ (y0, h) is continuous in both y0 and h, then J is also Fr´echet differentiable.
Necessary Conditions for Extremals
Theorem 3.4. Let J : A → R be a functional. If y be its local minimum in A, then δJ(y, h) = 0 for all admissible variationh.
Let now study one dimensional variational problem. Let L : R × Rn× Rn → R called the Lagragian. We look for minimum of the functional
J (y) = Z b
a
L(x, y(x), y0(x)) dx in the admissible class
A = {y : [a, b] → Rn|y ∈ C1, y(a) = y0, y(b) = y1}.
The argument of L is (x, y, v) ∈ R × Rn× Rn. Later, we shall take partial derivative of L with respect to v. However, since y0 is used for the argument v, we commonly use Ly0 to represent Lv.
3.4. VARIATION OF FUNCTIONALS 49 Now, let us compute the first variation of J .
d
h)(b) = y1for all small . Thus, admissible variation h should satisfies h(a) = h(b) = 0.
Thus, a local minimum y should satisfy δJ (y) · h =
A fundamental lemma states that
Lemma 3.1. If f is continuous on [a, b] and ifRb boundary constraint if is small enough. Using this h, we get
Z b
a
f (x)h(x) dx ≥ C
2 > 0.
This contradicts to our assumption.
Thus, we get a necessary condition for y being a local minimum. That is y should satisfies
− d
dxLy0(x, y, y0) + Ly(x, y, y0) = 0
50 CHAPTER 3. CALCULUS OF VARIATIONS with the boundary condition
y(a) = y0, y(b) = y1.
The above equation is called the Euler-Lagrange equation for the functional J . The variational problem is transformed to solving a differential equation with certain boundary conditions.
Example 1. For the problem of minimizing arc length, the functional is J (y) =
Z b a
q
1 + y02dx,
where y(a) = y0, y(b) = y1. The corresponding Euler-Lagrange equation is
− d
dxLy0 = d dx
y0 p1 + y02
!
= 0.
This yields
y0
p1 + y02 = Const.
Solving y0, we further get
y0 = C (a constant).
Hence y = Cx + D. Applying boundary condition, we get C = y1− y0
b − a , D = by0− ay1 b − a . Thus, the minimal arc length curve is a straight line.
Example 2 In classical mechanics, we define the Lagrangian L to be L(t, y, ˙y) := 1
2m ˙y2− V (y),
where V is the potential. The corresponding Euler-Lagrange equation is
−d
dtm ˙y − Vy = 0.
This is exactly the Newton’s law of motion, where ∂L/∂ ˙y = m ˙y = p is the momentum and
∂L/∂y = −Vyis the force. This equation admits an time invariant quantity, namely the total energy is unchanged along physical trajectory. To see this, we multiply the above Newton’s equation by ˙y . We get
m¨y ˙y = −Vyy˙ This can be rewritten as
d dt
1
2m ˙y2+ V (y)
= 0.
3.4. VARIATION OF FUNCTIONALS 51 Thus, we have
1
2m ˙y2+ V (y) = E(aConst).
The quantity12m ˙y2+ V (y) is the total energy. The advantage of the existence of first integral is that the Euler-Lagrange equation is integrated once and the first integral is a first order equation, which is much easier to solve. To be precise, we can solve y0in terms of y from the above energy equality:
y0 = ±p
2(E − V (y))/m Uing separation of variable, we get
dy
p2(E − V (y))/m = ±dt This can be integrated to get y(t).
Such a property is in general true for a Lagrangian which is independent of time. The invariant is called the first integral. We have the following theorem.
Theorem 3.5 (The first integral). If L is independent of x, then the quantity I(y, y0) := −y0Ly0(y, y0) + L(y, y0)
is independent ofx along the solution of the Euler-Largange equation.
Example: Barchistochrone The Barchistochrone problem is to minimize the travel time from (0, 0) to (a, −h). The functional of the travel time for a path y(·) connecting the above two points is given by
T (y) = Z S
0
ds v =
Z a
0
p1 + y02
√−2gy dx.
The Lagrangian
L(y, y0) =
p1 + y02
√−2gy
is independent of x. Thus, the solution admits the first integral which is independent of x:
p1 + y02
√−2gy − y0 y0
p1 + y02√
−2gy
!
= Const.
which can be simplified to
y02= 1 + ky
−ky where k is a constant. Using separation of variable
√−ky
√1 + kydy = ±dx.
52 CHAPTER 3. CALCULUS OF VARIATIONS We make a substitution:
y = −1 ksin2 φ
2, then the above differential equation becomes
sinφ2 This is the parametric form of a cycloid. The portion we want is to take
x = 1
2k(φ − sin φ).
Example In electrostatics, given a charge density function f in Ω, the boundary ∂Ω is a conductor.
We are interested in the induced electric potential φ. The energy induced by f is −R f φ dx and the energy induced by φ isR |∇φ|2/2 dx. The total energy is
E[φ] = Z
(1
2|∇φ|2− f φ) dx.
The variation of E with respect to φ is δE · h = There are two kinds of natural boundary conditions:
1. Dirichlet boundary condition: We impose φ(x) = g(x) for x ∈ ∂Ω. In this case, the admissible φ should satisfy this boundary. Its variation h should satisfies h(x) = 0 for x ∈ ∂Ω. In this case, the variational of the energy is
δE · h = Z
Ω
(−∇2φ · h − f · h) dx
3.4. VARIATION OF FUNCTIONALS 53 The Euler-Lagrange is
−∇2φ = f.
This together with the Dirichlet boundary condition φ(x) = g(x), x ∈ ∂Ω
constitute the well-known Dirichlet problem. Indeed, solving this Dirichlet is equivalent to solving
minφ∈AE[φ], A = {φ ∈ C1(Ω), φ(x) = g(x), x ∈ ∂Ω}
We have seen one side of the equivalent. Indeed, if φ solves the Dirichlet problem, then for any ψ ∈ A, we express ψ = φ + h with h vanishing on ∂Ω. Then
This shows that if φ solves the Dirichlet problem, φ is the minimum of E in A.
2. The Neumann boundary condition. The second natural boundary condition is to prescribe
∇φ(x) · n = σ(x) on the boundary. In this case, σ is the surface charge. So, we should modify our energy functional to be
E1[φ] =
Now, we choose the admissible class to be
A = {φ ∈ C1(Ω)}.
Its variation class is still the same A, no restriction on the boundary. The variation of E with respect to a variation h now read
δE·h =
We can first choose those h which are vanish on ∂Ω. This leads to the Euler-Lagrange equa-tion
−∇2φ(x) = f (x), in Ω.
54 CHAPTER 3. CALCULUS OF VARIATIONS Next, we have
Z
∂Ω
(φn− σ)h dx = 0 for arbitrary function h. This leads to
φn= σ on ∂Ω.
This PDE with the boundary condition above is called the Neumann problem.
Remark. The condition y ∈ C2 in Lemma ?? is only for the classical solutions of the Euler-Lagrange equations. Indeed, the condition for having the functional J well-defined is weaker, say we only need y ∈ C1. In this case, the Euler-Lagrange equation is still meaningful, but in weak sense. This means that the equation is valid when it is tested by smooth functions. More precisely, the Euler-Lagrange
− d
dxLy0(x, y, y0) + Ly(x, y, y0) = 0 is valid if
Z b a
Ly0(x, y, y0) · h0+ Ly(x, y, y0) · h dx = 0,
for any smooth compact supported function h defined on (a, b). In the latter definition, we only need y ∈ C1. The logic of the approach to this problem is that we find solution in weak class. Then if it is indeed a smooth function, we prove so-called the regularity result.
Homework You can show that if φ solves the Neumann problem, then φ is the minimum of the energy functional E1.
Homework
1. pp. 167: 7, 9, 13.
2. pp. 176: 8, 12, 14.