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ODE, PDE & Applied Mathematics

Calculus of Variations

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1. Euler–Lagrange equation and standard functionals

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Lec-01 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Course opener: the variational problem and the Euler–Lagrange equation. (This NPTEL course titles every lecture identically — follow the numbering.)

Lec-02 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Derivation of the Euler–Lagrange equation continued.

Lec-03 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

First integrals: the Beltrami identity and cyclic cases.

Euler–Lagrange, first integrals and null Lagrangians free

In plain words

Ask for the best curve rather than the best point. The Euler–Lagrange equation plays exactly the role that setting a derivative to zero plays in ordinary calculus.

The equation

For J[y]=∫aJ[y] = \int_{a}ᵇ F(x, y, y′) dx with fixed endpoints: ∂F/∂y−d/\partial{}F/\partial{}y - d/dx(∂F/∂y′)=0(\partial{}F/\partial{}y') = 0.

Two first integrals

  • F independent of y⇒∂F/∂y′=y \Rightarrow \partial{}F/\partial{}y' = const (momentum conservation).
  • F independent of x ⇒ Beltrami: F−y′∂F/∂y′=F - y' \partial{}F/\partial{}y' = const (energy conservation).

Standard functionals

J[y] Euler–Lagrange Extremals
∫(y′)2\int(y')^{2} y″ = 0 straight lines
∫1+(y′)2(\int\sqrt{1 + (y')^{2}} (arc length) y″ = 0 straight lines
∫1+(y′)2/y(\int\sqrt{1 + (y')^{2}}/\sqrt{y} (brachistochrone) — cycloid
∫y1+(y′)2(\int y\sqrt{1 + (y')^{2}} (surface of revolution) — catenary
∫[(y′)2−y2]\int[(y')^{2} - y^{2}] y″ + y = 0 sines
∫(1+x3y′)y′\int(1 + x^{3}y')y' d/dx(1+2x3y′)=0(1 + 2x^{3}y') = 0 y=d−c/(2x2)y = d - c/(2x^{2})

Null Lagrangians — the trap

A term that is a total derivative d/dx G(x, y) contributes only boundary terms and does not affect the Euler–Lagrange equation. In ∫[(y′)2−y\int[(y')^{2} - y|y|y′ + xy] dx the middle term is −d/dx(|y|3/3)^{3}/3), so the equation reduces to 2y″ = x.

Several unknowns and constraints

  • Vector case: one Euler–Lagrange equation per component. For ∫[(y′)2+(z′)2+y′z′]:2y′′+z′′=0\int[(y')^{2} + (z')^{2} + y'z']: 2y'' + z'' = 0 and 2z″ + y″ = 0 ⇒ y″ = z″ = 0.
  • Isoperimetric constraint ∫G=c\int{}G = c: extremise F+λGF + \lambda{}G; the multiplier is fixed by the constraint.
  • Higher derivatives: ∂F/∂y−d/\partial{}F/\partial{}y - d/dx(∂F/∂y′)+d2/(\partial{}F/\partial{}y') + d^{2}/dx2(∂F/∂y′′)=0^{2}(\partial{}F/\partial{}y'') = 0.

Weak vs strong extrema

A weak minimum compares y′ too (C1−(C^{1}-neighbourhood); a strong minimum only compares y(C0−y (C^{0}-neighbourhood).∫(1−(y′)2)y2). \int(1 - (y')^{2})y^{2} has y ≡ 0 as a weak minimum but not a strong one — in aC0−a C^{0}-ball, y′ can be huge and make J negative.

Key takeaways

  • No x ⇒ Beltrami; no y ⇒ momentum constant.
  • Exact-derivative terms drop out of the Euler–Lagrange equation.
  • Weak vs strong extremum is decided by which norm the neighbourhood uses.

Euler–Lagrange, and the two shortcuts worth knowinginteractive

This visual is part of Trap Analysis

When F has no y, and when F has no x — each collapses the equation by one order.

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Exam focus: Write the Euler–Lagrange equation; use the Beltrami identity when F has no explicit x. Terms that are exact derivatives (null Lagrangians) do not change the extremals.

Practice

Part C2025 Decembershow ▾
Suppose y(x) is the extremal of the variational problem J(y)=∫01((y′)2sin⁡x+(2cosx)y)J(y) = \int_{0}^{1} ((y')^{2}\sin x + (2cos x)y) dx subject to y(0)=0, y(1)=1. Then which of the following statements are true?
  1. A.y(1/2) = 1
  2. B.y′(0) = 1✓ Correct
  3. C.y(1/4) = 2
  4. D.y′(1/2) = 1✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾
For α≥0\alpha \ge 0, consider the functional Jα[y]=∫12(y′)2/xαJ_\alpha[y] = \int_{1}^{2} (y')^{2}/x^\alpha dx defined for all continuously differentiable functions defined on the interval [1, 2] satisfying the conditions y(1) = 1, y(2) = 2. Then which of the following statements are true?
  1. A.y(x)=(1/15)(x4+14)y(x) = (1/15)(x^{4} + 14) is an extremal for J3J_{3}✓ Correct
  2. B.y(x)=(1/3)(x2+2)y(x) = (1/3)(x^{2} + 2) is an extremal for J1J_{1}✓ Correct
  3. C.y(x) = x is an extremal for J0J_{0}✓ Correct
  4. D.y(x)=(1/2)(x2−x+2)y(x) = (1/2)(x^{2} - x + 2) is an extremal for J1J_{1}

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾
Suppose y(x) is the extremal of the variational problem J(y)=∫04(y′)2/y2J(y) = \int_{0}^{4} (y')^{2}/y^{2} dx subject to y(0)=1,y(4)=e8y(0) = 1, y(4) = e^{8}. Then which of the following statements is true?
  1. A.y(loge2)=2y(log_{e} 2) = 2.
  2. B.y(loge3)=9y(log_{e} 3) = 9.✓ Correct
  3. C.y(loge4)=4y(log_{e} 4) = 4.
  4. D.y(loge5)=5y(log_{e} 5) = 5.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾
For a,b,c∈Ra, b, c \in \mathbb{R}, consider the variational problem: Minimize J[y]=∫02[a(y′)2+2J[y] = \int_{0}^{2} [a(y')^{2} + 2byy′ + cy2]^{2}] dx, subject to y(0) = 10, y(1) = 100. Then which of the following statements are true?
  1. A.If (a, b, c) = (−2, 1, −2), then every admissible extremal is a minimizer✓ Correct
  2. B.If (a, b, c) = (1, 0, 2), then every admissible extremal is a minimizer✓ Correct
  3. C.If (a, b, c) = (2, −1, 1), then every admissible extremal is a minimizer✓ Correct
  4. D.If (a, b, c) = (1, −2, 5), then every admissible extremal is a minimizer✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾
Let y(x) be the extremal of the functional J[y]=∫0π/4J[y] = \int_{0}^{\pi/4} ((y′)2−4y2+2((y')^{2} - 4y^{2} + 2xy) dx subject to y(0)=0,y(π/4)=1y(0) = 0, y(\pi/4) = 1. Then y(x) is equal to
  1. A.(1−π/4)sin⁡(2x)+x(1 - \pi/4)\sin(2x) + x
  2. B.(1−π/16)sin⁡(2x)+x/4(1 - \pi/16)\sin(2x) + x/4✓ Correct
  3. C.(1+π/4)sin⁡(2x)−x(1 + \pi/4)\sin(2x) - x
  4. D.(1+π/16)sin⁡(2x)−x/4(1 + \pi/16)\sin(2x) - x/4

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Consider the variational problem minimize J[y]=∫01[(y′)2+2025J[y] = \int_{0}^{1}[(y')^{2} + 2025 yy′]dx, where y(0) and y(1) are free. Then which of the following statements are true?
  1. A.There are infinitely many extremals
  2. B.There are more than one but only finitely many extremals
  3. C.There is no extremal
  4. D.There is a unique extremal✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
For any b∈Rb \in \mathbb{R}, let S(b) denote the set of all broken extremals with one corner of the variational problem: minimize J[y]=∫01((y′)4−3(y′)2)J[y] = \int_{0}^{1}((y')^{4} - 3(y')^{2})dx, subject to y(0) = 0, y(1) = b. Then which of the following statements are true?
  1. A.S(2) has exactly two elements
  2. B.S(1/2) has exactly one element
  3. C.S(2) is empty✓ Correct
  4. D.S(1/2) has exactly two elements✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Define S := {y∈C1[0,1]:y(0)=y(1)=0y \in C^{1}[0, 1] : y(0) = y(1) = 0}, ‖y‖(∞):=max⁡(x∈[0,1])(\infty) := \max(x\in[0,1])|y(x)| for every y∈S,B0(0,ε):=y \in S, B_{0}(0, \varepsilon) := {y ∈ S : ‖y‖(∞)<ε(\infty) < \varepsilon}, B1(0,ε):=B_{1}(0, \varepsilon) := {y ∈ S : ‖y‖(∞)+(\infty) + ‖y′‖(∞)<ε(\infty) < \varepsilon}. Consider the functional J:S→RJ : S \to \mathbb{R} given by J[y]=∫01[(y′)2−2x(y′)4]J[y] = \int_{0}^{1}[(y')^{2} - 2x(y')^{4}]dx, then there exists an ε>0\varepsilon > 0 such that
  1. A.J[y] ≤ J[0] for every y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  2. B.J[y] ≤ J[0] for every y∈B1(0,ε)y \in B_{1}(0, \varepsilon)
  3. C.J[y] ≥ J[0] for every y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  4. D.J[y] ≥ J[0] for every y∈B1(0,ε)y \in B_{1}(0, \varepsilon)✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾
Let B(0,1) = {(x,y)∈R2(x, y) \in \mathbb{R}^{2} | x2+y2<1x^{2} + y^{2} < 1} be the open unit disc in R2,∂B(0,1)\mathbb{R}^{2}, \partial{}B(0,1) denote the boundary of B(0,1), and ν\nu denote the unit outward normal to ∂B(0,1)\partial{}B(0,1). Let f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R} be a given continuous function. The Euler–Lagrange equation of the minimization problem min { ½∬_B(0,1) |∇u\nabla{}u|2^{2} dxdy + ½∬B(0,1)e(u2)_B(0,1) e^(u^{2}) dxdy +∫∂B(0,1)fu+ \int_\partial{}B(0,1) f u ds } subject to u∈C1(u \in C^{1}(closure of B(0,1)) is
  1. A.Δu=−\Delta{}u = -ue(u2)^(u^{2}) in B(0,1),∂u/∂ν=fB(0,1), \partial{}u/\partial\nu = f on ∂B(0,1)\partial{}B(0,1)
  2. B.Δu=\Delta{}u = ue(u2)+f^(u^{2}) + f in B(0,1), u = 0 on ∂B(0,1)\partial{}B(0,1)
  3. C.Δu=\Delta{}u = ue(u2)^(u^{2}) in B(0,1),∂u/∂ν=−fB(0,1), \partial{}u/\partial\nu = -f on ∂B(0,1)\partial{}B(0,1)✓ Correct
  4. D.Δu=\Delta{}u = ue(u2)^(u^{2}) in B(0,1),∂u/∂ν+u=fB(0,1), \partial{}u/\partial\nu + u = f on ∂B(0,1)\partial{}B(0,1)

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Define S := {y∈C1[−1,1]:y(−1)=−1,y(1)=3y \in C^{1}[-1, 1] : y(-1) = -1, y(1) = 3}. Let φ\varphi be the extremal of the functional J:S→RJ : S \to \mathbb{R} given by J[y]=∫J[y] = \int₋11[(y′)3+(y′)2]_{1}^{1}[(y')^{3} + (y')^{2}]dx. Define ‖y‖(∞):=max⁡(x∈[−1,1])(\infty) := \max(x\in[-1,1])|y(x)| for every y ∈ S and let B0(φ,ε):=B_{0}(\varphi, \varepsilon) := {y ∈ S : ‖y−φy - \varphi‖(∞)<ε(\infty) < \varepsilon}, B1(φ,ε):=B_{1}(\varphi, \varepsilon) := {y ∈ S : ‖y−φy - \varphi‖(∞)+(\infty) + ‖y′−φ′y' - \varphi'‖(∞)<ε(\infty) < \varepsilon}. Then which of the following statements are true?
  1. A.φ(x)=2x+1\varphi(x) = 2x + 1 for every x ∈ [−1, 1]✓ Correct
  2. B.There exists ε>0\varepsilon > 0 such that J[y]≥J[φ]J[y] \ge J[\varphi] for every y∈B0(φ,ε)y \in B_{0}(\varphi, \varepsilon)
  3. C.There exists ε>0\varepsilon > 0 such that J[y]≥J[φ]J[y] \ge J[\varphi] for every y∈B1(φ,ε)y \in B_{1}(\varphi, \varepsilon)✓ Correct
  4. D.There exists ε>0\varepsilon > 0 such that J[y]≤J[φ]J[y] \le J[\varphi] for every y∈B1(φ,ε)y \in B_{1}(\varphi, \varepsilon)

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾
The infimum of the set {∫a\int_{a}ᵇ 1+(y′(t))2\sqrt{1 + (y'(t))^{2}}dt : y∈C1[a,b],y(a)=a2,y(b)=b−5y \in C^{1}[a, b], y(a) = a^{2}, y(b) = b - 5} is
  1. A.192/819\sqrt{2}/8✓ Correct
  2. B.19219\sqrt{2}
  3. C.19/8
  4. D.19/(22)19/(2\sqrt{2})

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾
If φ∈C4[0,1]\varphi \in C^{4}[0, 1] is the extremal of the variational problem minimize J[y]=∫01(J[y] = \int_{0}^{1}(yy′+(y′′)2)' + (y'')^{2})dx, subject to y(0) = 0, y′(0) = 1, y(1) = 2, y′(1) = 4, then φ(1/2)\varphi(1/2) is equal to
  1. A.5/8✓ Correct
  2. B.3/4
  3. C.3/8
  4. D.5/4

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Juneshow ▾
Let y(x) and z(x) be the stationary functions (extremals) of J(y,z)=∫01[(y′)2+(z′)2+y′z′]J(y, z) = \int_{0}^{1} [(y')^{2} + (z')^{2} + y'z'] dx subject to y(0) = 1, y(1) = 0, z(0) = −1, z(1) = 2. Which of the following statements are correct?
  1. A.z(x) + 3y(x) = 2 for x ∈ [0,1].✓ Correct
  2. B.3z(x) + y(x) = 2 for x ∈ [0,1].
  3. C.y(x) + z(x) = 2x for x ∈ [0,1].✓ Correct
  4. D.y(x) + z(x) = x for x ∈ [0,1].

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾
Consider the variational problem (P):J(y)=∫01[(y′)2−y(P): J(y) = \int_{0}^{1} [(y')^{2} - y|y|·y′ + xy] dx, y(0) = 0, y(1) = 0. Which of the following statements is correct?
  1. A.(P) has no stationary function (extremal).
  2. B.y ≡ 0 is the only stationary function (extremal) for (P).
  3. C.(P) has a unique stationary function (extremal) y not identically equal to 0.✓ Correct
  4. D.(P) has infinitely many stationary functions (extremals).

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾
Define S = {y∈C1[0,π]:y(0)=y(π)=0y \in C^{1}[0, \pi] : y(0) = y(\pi) = 0}, ‖f‖∞=max⁡_\infty = \max |f|, B0(0,ε)=B_{0}(0, \varepsilon) = {f ∈ S : ‖f‖∞<ε_\infty < \varepsilon} and B1(0,ε)=B_{1}(0, \varepsilon) = {f ∈ S : ‖f‖∞+_\infty + ‖f′‖∞<ε_\infty < \varepsilon}. For the functional J[y]=∫0π(1−(y′)2)y2J[y] = \int_{0}^\pi (1 - (y')^{2}) y^{2} dx, there exists ε>0\varepsilon > 0 such that
  1. A.J[y] ≤ J[0] for all y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  2. B.J[y] ≤ J[0] for all y∈B1(0,ε)y \in B_{1}(0, \varepsilon)
  3. C.J[y] ≥ J[0] for all y∈B0(0,ε)y \in B_{0}(0, \varepsilon)
  4. D.J[y] ≥ J[0] for all y∈B1(0,ε)y \in B_{1}(0, \varepsilon)✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾
Among the curves connecting the points (1, 2) and (2, 8), let γ\gamma be the curve on which an extremal of J[y]=∫12(1+x3y′)y′J[y] = \int_{1}^{2} (1 + x^{3}y') y' dx can be attained. Which of the following points lie on the curve γ\gamma?
  1. A.(2,3)(\sqrt{2}, 3)
  2. B.(2,6)(\sqrt{2}, 6)✓ Correct
  3. C.(3,22/3)(\sqrt{3}, 22/3)✓ Correct
  4. D.(3,23/3)(\sqrt{3}, 23/3)

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾
The extremals of the arc-length functional J[y]=∫aJ[y] = \int_{a}ᵇ 1+(y′)2\sqrt{1 + (y')^{2}} dx are
  1. A.straight lines✓ Correct
  2. B.circular arcs
  3. C.parabolas
  4. D.catenaries

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Consider J[y]=∫01(y′)2J[y] = \int_{0}^{1} (y')^{2} dx over smooth y with y(0) = 0, y(1) = 1. Which of the following are true?
  1. A.The Euler–Lagrange equation is y″ = 0.✓ Correct
  2. B.The extremal is y(x) = x.✓ Correct
  3. C.The minimum value of J is 1.✓ Correct
  4. D.The infimum of J is not attained by any admissible function.

The worked solution and the per-option analysis are in Trap Analysis.

2. Isoperimetric problems

Read this as a chapter →

Lec-08 Calculus of Variations and Integral Equations

NPTEL · Calculus of Variations and Integral Equations

Constrained variational problems and Lagrange multipliers.

Constrained variational problems free

In plain words

Extremise one integral while holding another one fixed. It is the Lagrange multiplier you already know, with the multiplier found at the end from the constraint rather than chosen up front.

The method

To extremise J[y]=∫FJ[y] = \int{}F subject to K[y]=∫G=cK[y] = \int{}G = c, extremise ∫(F+λG)\int(F + \lambda{}G). The extremal and λ\lambda are determined together by the boundary conditions and the constraint.

Counting constants: an n-th order Euler–Lagrange equation gives n constants, plus λ\lambda. Boundary conditions plus the constraint must supply exactly that many equations — this is how "how many extremals are there?" questions are answered.

Example. Extremise ∫01(y′)2\int_{0}^{1}(y')^{2} with y(0) = 1, y(1) = 6 and ∫01y=3\int_{0}^{1} y = 3. The modified equation 2y′′=λ2y'' = \lambda gives a quadratic with three unknown coefficients, and the three conditions fix them uniquely: exactly one extremal.

Classical results

Problem Answer
Maximum area for fixed perimeter circle
Minimum surface of revolution catenoid (from the catenary)
Shortest path on a sphere great circle
Fastest descent under gravity cycloid
Fixed length, maximum area under a curve circular arc
  • Holonomic constraint g(x, y, z) = 0: use λ(x)\lambda(x) as a function.
  • Variable endpoints: add the transversality condition [F − y′Fy′]δx+y'F_{y'}]\delta{}x + Fy′δy=0F_{y'}\delta{}y = 0 at the free end. For a fully free endpoint this reduces to Fy′F_{y'} = 0 there.
  • Eigenvalue problems: minimising ∫(y′)2\int(y')^{2} subject to ∫y2=1\int{}y^{2} = 1 gives the Sturm–Liouville problem, with λ\lambda the smallest eigenvalue (Rayleigh quotient).

Key takeaways

  • One multiplier per integral constraint; count constants against conditions.
  • Free endpoints bring transversality conditions, not extra constants.
  • The Rayleigh quotient links isoperimetric problems to Sturm–Liouville eigenvalues.

Constrained extremals: the multiplier methodinteractive

This visual is part of Trap Analysis

Fixed perimeter, maximum area — set up with a Lagrange multiplier and solved.

Unlock interactive visuals

Exam focus: Introduce a Lagrange multiplier, solve the modified Euler–Lagrange equation, then use the constraint plus boundary conditions to pin every constant.

Practice

Part C2025 Decembershow ▾
Suppose y(x) is the extremal of the variational problem J(y)=∫01x2(y′)2J(y) = \int_{0}^{1} x^{2}(y')^{2} dx subject to y(0)=0,y(1)=1,∫01y2y(0)=0, y(1)=1, \int_{0}^{1}y^{2}dx = 1/7. Then which of the following statements are true?
  1. A.y′(1/2) = 3/4✓ Correct
  2. B.y′(1/3)=1y'(1/\sqrt{3}) = 1✓ Correct
  3. C.y′(1/3) = 1/3✓ Correct
  4. D.y′(1/4) = 1/2

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾
The extremizer of the problem min⁡[(1/2)∫\min[(1/2)\int₋11((y′(x))2+(y(x))2)_{1}^{1} ((y'(x))^{2} + (y(x))^{2})dx] subject to y∈C1[−1,1],∫y \in C^{1}[-1, 1], \int₋11_{1}^{1} xy(x)dx = 0 and y(−1) = y(1) = 1 is
  1. A.(e/(1+e2))(e(e/(1 + e^{2}))(eˣ + e⁻ˣ)+x2−1) + x^{2} - 1
  2. B.(e/(1+e2))(e(e/(1 + e^{2}))(eˣ + e⁻ˣ)+1−x2) + 1 - x^{2}
  3. C.(e/(1+e2))(e(e/(1 + e^{2}))(eˣ + e⁻ˣ)✓ Correct
  4. D.(e/(1+e2))(e(e/(1 + e^{2}))(eˣ + e⁻ˣ)+sin⁡(2πx)) + \sin(2\pi{}x)

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾
Let S denote the set of all solutions of the Euler-Lagrange equation of the variational problem: minimize J[y]=∫01(y2+(y′)2)J[y] = \int_{0}^{1}(y^{2} + (y')^{2})dx, subject to y(0)=0,y(1)=0,∫01y2y(0) = 0, y(1) = 0, \int_{0}^{1}y^{2}dx = 1. Then the set {φ(1/2):φ∈S\varphi(1/2) : \varphi \in S} is equal to
  1. A.{−2,2-\sqrt{2}, \sqrt{2}}
  2. B.{2/k:k∈Z,k≠0\sqrt{2}/k : k \in \mathbb{Z}, k \ne 0}
  3. C.{2/k:k∈N\sqrt{2/k} : k \in \mathbb{N}}
  4. D.{−2,0,2-\sqrt{2}, 0, \sqrt{2}}✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾
The cardinality of the set of extremals of J[y]=∫01(y′)2J[y] = \int_{0}^{1} (y')^{2} dx subject to y(0)=1,y(1)=6,∫01yy(0) = 1, y(1) = 6, \int_{0}^{1} y dx = 3 is
  1. A.0
  2. B.1✓ Correct
  3. C.2
  4. D.countably infinite

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Consider extremizing J[y]=∫01y′2J[y] = \int_{0}^{1} y'^{2} dx subject to ∫01y\int_{0}^{1} y dx = 1, with y(0) = y(1) = 0. Which of the following are true?
  1. A.The Euler–Lagrange equation with a Lagrange multiplier λ\lambda gives y″ = constant, so the extremal is a parabola.✓ Correct
  2. B.The extremal function is y(x) = 6x(1 − x).✓ Correct
  3. C.The extremal value of J is 12.✓ Correct
  4. D.The Lagrange multiplier λ\lambda used to enforce an isoperimetric constraint ∫y\int{}y dx = k must always be positive.

Solution

Extremizing ∫(y′2−λy)\int(y'^{2} - \lambda{}y) dx gives 2y′′+λ=02y'' + \lambda = 0, so y is quadratic — (1). Solving with the boundary conditions gives y=(λ/4)x(1−x)y = (\lambda/4)x(1-x); imposing ∫01y\int_{0}^{1}y dx = 1 fixes λ=24\lambda = 24, giving y(x) = 6x(1−x) — (2). Then J=∫01(6−12x)2J = \int_{0}^{1}(6-12x)^{2} dx = 12 — (3). (4) is false: the sign of λ\lambda tracks the sign of the constraint value k, not a universal law — a constraint ∫y\int{}y dx = −1 would produce a negative λ\lambda by the same computation.

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Part Cshow ▾
Consider the classical isoperimetric problem: among closed curves of fixed perimeter L in the plane, find the one enclosing maximum area. Which of the following are true?
  1. A.The circle uniquely maximizes enclosed area among closed curves of perimeter L.✓ Correct
  2. B.The isoperimetric inequality states 4πA≤L24\pi{}A \le L^{2}, with equality iff the curve is a circle.✓ Correct
  3. C.Using a Lagrange multiplier on the constrained problem, the Euler–Lagrange equation forces the extremal curve to have constant curvature.✓ Correct
  4. D.Among closed curves of a fixed enclosed area A, the perimeter-minimizing curve is a square.

Solution

(1)–(2) are the isoperimetric theorem. (3) constant curvature is exactly what the constrained variational problem yields, consistent with the circle being the extremal. (4) is false: the perimeter-minimizing curve for fixed area is, again, the circle — the same inequality read the other way round, not a square.

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Part Cshow ▾
Among smooth simple closed plane curves of perimeter L, which of the following are true?
  1. A.The enclosed area satisfies A≤L2/(4π)A \le L^{2}/(4\pi).✓ Correct
  2. B.Equality holds exactly for the circle.✓ Correct
  3. C.Among rectangles of perimeter L, the square encloses the largest area.✓ Correct
  4. D.Some closed curve of perimeter L encloses area strictly greater than L2/(4π)L^{2}/(4\pi).

The worked solution and the per-option analysis are in Trap Analysis.