Calculus of Variations Questions and Answers

Q: The Euler Lagrange equation associated with the functional\[ \begin{aligned} I(w) &= \frac{1}{2} \int_{u} |Dw|^{2}\, dx \\ &\quad + \int_{u} f w\, dx \end{aligned} \]where \(u\) is an open set is :

The Euler-Lagrange equation is \(\Delta w = f\) in \(u\).Perturb \(w\) by \(\varepsilon v\) with \(v\) vanishing on the boundary of the open set \(u\), and set the first variation to zero:\[ \begin{aligned} \frac{d}{d\varepsilon} I(w + \varepsilon v)\Big|_{0} &= \int_{u} Dw \cdot Dv + f v\, dx \\ &= \int_{u} (-\Delta w + f)\, v\, dx = 0 \end{aligned} \]after integrating by parts. Since \(v\) is arbitrary, \(-\Delta w + f = 0\), that is \(\Delta w = f\).The sign matters: Evans's textbook form \(-\Delta w = f\) arises from the functional with \(-\int f w\). Here the linear term is \(+\int f w\), so the sign flips. The options with \(w\) on the right-hand side would need a term \(\frac{1}{2}\int w^{2}\) in the functional.

Q: The curve for which the area of surface of revolution is minimum when revolved about x-axis is the solution of the equation :

The curve satisfies a y' = √(y2 − a2).The surface area is \(S = 2\pi\int y\sqrt{1 + y'^{2}}\, dx\). The integrand \(F = y\sqrt{1 + y'^{2}}\) does not contain x, so the Euler-Lagrange equation has the first integral \(F - y'F_{y'} = \text{constant}\):\[ \begin{aligned} y\sqrt{1 + y'^{2}} - \frac{yy'^{2}}{\sqrt{1 + y'^{2}}} &= a \\ \frac{y}{\sqrt{1 + y'^{2}}} &= a \end{aligned} \]Squaring, \(y^{2} = a^{2}(1 + y'^{2})\), so \(ay' = \sqrt{y^{2} - a^{2}}\), whose solution is the catenary \(y = a\cosh\frac{x - b}{a}\). The options with \(x^{2} - a^{2}\) put the wrong variable under the root, and \(\sqrt{y^{2} - a^{2}}\,y' = a\) inverts the relation.

Q: The Euler-Lagrange's differential equation for extremization of the functional\[ I(y(x)) = \int_{0}^{1} (y'^{2} - y^{2})\, dx \]subject to the condition that \(\displaystyle\int_{0}^{\pi} y\, dx = 1\) is :

The Euler-Lagrange equation is 2y'' + 2y = λ.For an isoperimetric problem, extremise \(\int (F + \mu G)\, dx\) with \(F = y'^{2} - y^{2}\) and the constraint integrand \(G = y\). With \(H = y'^{2} - y^{2} + \mu y\):\[ \begin{aligned} H_{y} - \frac{d}{dx}H_{y'} &= 0 \\ -2y + \mu - 2y'' &= 0 \\ 2y'' + 2y &= \mu \end{aligned} \]Renaming the multiplier as \(\lambda\) gives option B. (The paper's two integrals have different upper limits, 1 and \(\pi\), but the equation does not depend on the limits.) \(y'' + y = 0\) forgets the constraint, and the options with \(y'' - y\) get the sign of the \(y^{2}\) term wrong.

Q: The first integral of Euler-Lagrange's differential equation of the functional\[ I(y(x)) = \int_{x_{1}}^{x_{2}} f(x, y', y'')\, dx \]is :

The first integral is ∂f/∂y' − d/dx(∂f/∂y'') = constant.For \(f(x, y, y', y'')\) the Euler-Poisson equation is\[ f_{y} - \frac{d}{dx}f_{y'} + \frac{d^{2}}{dx^{2}}f_{y''} = 0 \]Here f does not contain y, so \(f_{y} = 0\) and the equation becomes \(\frac{d}{dx}\left(f_{y'} - \frac{d}{dx}f_{y''}\right) = 0\). Integrating once gives \(f_{y'} - \frac{d}{dx}f_{y''} = \text{constant}\).The option with a plus sign has the wrong sign. \(f - y'f_{y'} - y''f_{y''}\) is not the first integral for this case, and \(f_{y} - \frac{d}{dx}f_{y'}\) is the ordinary first order Euler-Lagrange expression.

Q: The extremal of the functional\[ I(y(x)) = \int_{0}^{1} \frac{\dot{y}^{2}}{x^{3}}\, dx, \]when \(x(0) = 0\), \(x(1) = 1\) is :

The extremal is y = x4.The boundary conditions are printed as x(0) = 0 and x(1) = 1; they are the conditions \(y(0) = 0\), \(y(1) = 1\) on the unknown function. The integrand \(F = y'^{2}/x^{3}\) does not contain y, so the Euler-Lagrange equation gives\[ \begin{aligned} \frac{\partial F}{\partial y'} = \frac{2y'}{x^{3}} &= c \\ y' &= \frac{c}{2}x^{3} \\ y &= Ax^{4} + B \end{aligned} \]\(y(0) = 0\) gives \(B = 0\) and \(y(1) = 1\) gives \(A = 1\), so \(y = x^{4}\). The line \(y = x\) and the constants c and 0 do not satisfy the Euler-Lagrange equation with these end values.

Q: The solution of the functional \(I(y(x)) = \displaystyle\int_{x_{1}}^{x_{2}} f(x, y, y'')\, dx\) is a :

The extremals form a four parameter family of curves.When the integrand depends on y'', the Euler-Poisson equation\[ f_{y} - \frac{d}{dx}f_{y'} + \frac{d^{2}}{dx^{2}}f_{y''} = 0 \]is in general a differential equation of order four in y. Its general solution contains four arbitrary constants, fixed by four boundary conditions (y and y' at each end).A two parameter family is what a functional depending only on y' gives, since the Euler-Lagrange equation is then of second order.

Q: The extremal of the functional\[ \begin{aligned} I(r(\theta)) &= \int_{\theta_{0}}^{\theta} \sqrt{r^{2} + r'^{2}}\, d\theta, \\ r' &= \frac{dr}{d\theta} \end{aligned} \]is :

The extremal is a straight line.In polar coordinates the element of arc length is\[ \begin{aligned} ds &= \sqrt{dr^{2} + r^{2}\,d\theta^{2}} \\ &= \sqrt{r^{2} + r'^{2}}\,d\theta \end{aligned} \]so the functional is the length of the curve between two points. The curve of least length in the plane is a straight line; in polar form the extremals are \(r\cos(\theta - \alpha) = p\).A catenary minimises the area of a surface of revolution, a cycloid solves the brachistochrone problem, and great-circle arcs are the shortest paths on a sphere, not in the plane.

Q: The shortest distance between two points in a 3-dimensional Euclidean space is a curve of intersection of :

The shortest path is the curve of intersection of two planes.In Euclidean 3-space the shortest distance between two points is the straight segment joining them, and a straight line is exactly the intersection of two non-parallel planes. (In the calculus-of-variations treatment, minimising \(\int \sqrt{1 + y'^{2} + z'^{2}}\,dx\) gives \(y\) and \(z\) linear in \(x\), each equation being a plane.)'Two surfaces' or 'a surface by a plane' generally give curves that are not straight. A sphere cut by a plane through its centre gives a great circle, which is the answer for distance measured on a sphere, not in space.

Q: The extremals of the functional\[ I(y(x)) = \int_{x_{1}}^{x_{2}} f(x, y')\, dx \]is obtained as the solution of the equation :

The extremals satisfy \(f_{y'y'}\, y'' + f_{y'x} = 0\).The Euler-Lagrange equation is \(f_{y} - \frac{d}{dx} f_{y'} = 0\). Here f does not contain y, so \(f_{y} = 0\) and\[ \frac{d}{dx} f_{y'}(x, y') = 0 \]Differentiating \(f_{y'}\) along the curve by the chain rule gives \(f_{y'x} + f_{y'y'}\, y'' = 0\), which is option B. Integrated once, it says \(f_{y'}\) is constant.\(y' f_{y'} - f = \) constant is the Beltrami identity, which applies when f does not contain x. \(f_{y'x} = 0\) drops the \(y''\) term. Option A has the wrong derivatives.

Q: The functional\[ I(y(x)) = \int_{0}^{\pi/4} (y^{2} - y'^{2})\, dx \]satisfying y(0) = 0 attains its extremal on the curve :

The extremal is \(y = 0\).The Euler-Lagrange equation for \(F = y^{2} - y'^{2}\) is\[ F_{y} - \frac{d}{dx}F_{y'} = 2y + 2y'' = 0 \]so \(y = A\cos x + B\sin x\), and \(y(0) = 0\) gives \(y = B\sin x\).The right end \(x = \pi/4\) has no prescribed value, so it carries the natural boundary condition \(F_{y'} = 0\) there, that is \(-2y'(\pi/4) = 0\). Then \(B\cos(\pi/4) = 0\), so \(B = 0\) and the extremal is \(y = 0\).\(y = b\sin x\) is the family that satisfies only the Euler equation and \(y(0) = 0\); it ignores the free-end condition. \(e^{x}\) and \(\cos x\) fail \(y(0) = 0\) or the Euler equation.

21.

The Euler Lagrange equation associated with the functional

\[ \begin{aligned} I(w) &= \frac{1}{2} \int_{u} |Dw|^{2}\, dx \\ &\quad + \int_{u} f w\, dx \end{aligned} \]

where \(u\) is an open set is :

A.\(-\Delta w = w\) in \(u\)
B.\(-\Delta w = f\) in \(u\)
C.\(\Delta w = w\) in \(u\)
D.\(\Delta w = f\) in \(u\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The Euler-Lagrange equation is \(\Delta w = f\) in \(u\).

Perturb \(w\) by \(\varepsilon v\) with \(v\) vanishing on the boundary of the open set \(u\), and set the first variation to zero:

\[ \begin{aligned} \frac{d}{d\varepsilon} I(w + \varepsilon v)\Big|_{0} &= \int_{u} Dw \cdot Dv + f v\, dx \\ &= \int_{u} (-\Delta w + f)\, v\, dx = 0 \end{aligned} \]

after integrating by parts. Since \(v\) is arbitrary, \(-\Delta w + f = 0\), that is \(\Delta w = f\).

The sign matters: Evans's textbook form \(-\Delta w = f\) arises from the functional with \(-\int f w\). Here the linear term is \(+\int f w\), so the sign flips. The options with \(w\) on the right-hand side would need a term \(\frac{1}{2}\int w^{2}\) in the functional.

Discuss in Forum

22.

The curve for which the area of surface of revolution is minimum when revolved about x-axis is the solution of the equation :

A.\(ay' = \sqrt{y^{2} - a^{2}}\), a is a constant
B.\(\sqrt{y^{2} - a^{2}}\ y' = a\)
C.\(\sqrt{x^{2} - a^{2}}\ y' = a\)
D.\(ay' = \sqrt{x^{2} - a^{2}}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The curve satisfies a y' = √(y² − a²).

The surface area is \(S = 2\pi\int y\sqrt{1 + y'^{2}}\, dx\). The integrand \(F = y\sqrt{1 + y'^{2}}\) does not contain x, so the Euler-Lagrange equation has the first integral \(F - y'F_{y'} = \text{constant}\):

\[ \begin{aligned} y\sqrt{1 + y'^{2}} - \frac{yy'^{2}}{\sqrt{1 + y'^{2}}} &= a \\ \frac{y}{\sqrt{1 + y'^{2}}} &= a \end{aligned} \]

Squaring, \(y^{2} = a^{2}(1 + y'^{2})\), so \(ay' = \sqrt{y^{2} - a^{2}}\), whose solution is the catenary \(y = a\cosh\frac{x - b}{a}\). The options with \(x^{2} - a^{2}\) put the wrong variable under the root, and \(\sqrt{y^{2} - a^{2}}\,y' = a\) inverts the relation.

Discuss in Forum

23.

The Euler-Lagrange's differential equation for extremization of the functional

\[ I(y(x)) = \int_{0}^{1} (y'^{2} - y^{2})\, dx \]

subject to the condition that \(\displaystyle\int_{0}^{\pi} y\, dx = 1\) is :

A.\(y'' - y = 0\)
B.\(2y'' + 2y = \lambda\)
C.\(y'' + y = 0\)
D.\(y'' - y = \lambda\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The Euler-Lagrange equation is 2y'' + 2y = λ.

For an isoperimetric problem, extremise \(\int (F + \mu G)\, dx\) with \(F = y'^{2} - y^{2}\) and the constraint integrand \(G = y\). With \(H = y'^{2} - y^{2} + \mu y\):

\[ \begin{aligned} H_{y} - \frac{d}{dx}H_{y'} &= 0 \\ -2y + \mu - 2y'' &= 0 \\ 2y'' + 2y &= \mu \end{aligned} \]

Renaming the multiplier as \(\lambda\) gives option B. (The paper's two integrals have different upper limits, 1 and \(\pi\), but the equation does not depend on the limits.) \(y'' + y = 0\) forgets the constraint, and the options with \(y'' - y\) get the sign of the \(y^{2}\) term wrong.

Discuss in Forum

24.

The first integral of Euler-Lagrange's differential equation of the functional

\[ I(y(x)) = \int_{x_{1}}^{x_{2}} f(x, y', y'')\, dx \]

is :

A.\(f-y'\frac{\partial f}{\partial y'}-y''\frac{\partial f}{\partial y''}\) = constant
B.\(\frac{\partial f}{\partial y'}+\frac{d}{dx}(\frac{\partial f}{\partial y''})\) = constant
C.\(\frac{\partial f}{\partial y'}-\frac{d}{dx}(\frac{\partial f}{\partial y''})\) = constant
D.\(\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y'})\) = constant

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The first integral is ∂f/∂y' − d/dx(∂f/∂y'') = constant.

For \(f(x, y, y', y'')\) the Euler-Poisson equation is

\[ f_{y} - \frac{d}{dx}f_{y'} + \frac{d^{2}}{dx^{2}}f_{y''} = 0 \]

Here f does not contain y, so \(f_{y} = 0\) and the equation becomes \(\frac{d}{dx}\left(f_{y'} - \frac{d}{dx}f_{y''}\right) = 0\). Integrating once gives \(f_{y'} - \frac{d}{dx}f_{y''} = \text{constant}\).

The option with a plus sign has the wrong sign. \(f - y'f_{y'} - y''f_{y''}\) is not the first integral for this case, and \(f_{y} - \frac{d}{dx}f_{y'}\) is the ordinary first order Euler-Lagrange expression.

Discuss in Forum

25.

The extremal of the functional

\[ I(y(x)) = \int_{0}^{1} \frac{\dot{y}^{2}}{x^{3}}\, dx, \]

when \(x(0) = 0\), \(x(1) = 1\) is :

A.\(y = x\)
B.\(y = c\)
C.\(y = x^{4}\)
D.\(y = 0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The extremal is y = x⁴.

The boundary conditions are printed as x(0) = 0 and x(1) = 1; they are the conditions \(y(0) = 0\), \(y(1) = 1\) on the unknown function. The integrand \(F = y'^{2}/x^{3}\) does not contain y, so the Euler-Lagrange equation gives

\[ \begin{aligned} \frac{\partial F}{\partial y'} = \frac{2y'}{x^{3}} &= c \\ y' &= \frac{c}{2}x^{3} \\ y &= Ax^{4} + B \end{aligned} \]

\(y(0) = 0\) gives \(B = 0\) and \(y(1) = 1\) gives \(A = 1\), so \(y = x^{4}\). The line \(y = x\) and the constants c and 0 do not satisfy the Euler-Lagrange equation with these end values.

Discuss in Forum

26.

The solution of the functional \(I(y(x)) = \displaystyle\int_{x_{1}}^{x_{2}} f(x, y, y'')\, dx\) is a :

A.one parameter family of curves
B.two parameter family of curves
C.three parameter family of curves
D.four parameter family of curves

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The extremals form a four parameter family of curves.

When the integrand depends on y'', the Euler-Poisson equation

\[ f_{y} - \frac{d}{dx}f_{y'} + \frac{d^{2}}{dx^{2}}f_{y''} = 0 \]

is in general a differential equation of order four in y. Its general solution contains four arbitrary constants, fixed by four boundary conditions (y and y' at each end).

A two parameter family is what a functional depending only on y' gives, since the Euler-Lagrange equation is then of second order.

Discuss in Forum

27.

The extremal of the functional

\[ \begin{aligned} I(r(\theta)) &= \int_{\theta_{0}}^{\theta} \sqrt{r^{2} + r'^{2}}\, d\theta, \\ r' &= \frac{dr}{d\theta} \end{aligned} \]

is :

A.a catenary
B.a cycloid
C.a straight line
D.an arc of the great circle

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The extremal is a straight line.

In polar coordinates the element of arc length is

\[ \begin{aligned} ds &= \sqrt{dr^{2} + r^{2}\,d\theta^{2}} \\ &= \sqrt{r^{2} + r'^{2}}\,d\theta \end{aligned} \]

so the functional is the length of the curve between two points. The curve of least length in the plane is a straight line; in polar form the extremals are \(r\cos(\theta - \alpha) = p\).

A catenary minimises the area of a surface of revolution, a cycloid solves the brachistochrone problem, and great-circle arcs are the shortest paths on a sphere, not in the plane.

Discuss in Forum

28.

The shortest distance between two points in a 3-dimensional Euclidean space is a curve of intersection of :

A.two surfaces
B.two planes
C.a surface by a plane
D.a sphere of radius r by the plane through the centre of the sphere

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The shortest path is the curve of intersection of two planes.

In Euclidean 3-space the shortest distance between two points is the straight segment joining them, and a straight line is exactly the intersection of two non-parallel planes. (In the calculus-of-variations treatment, minimising \(\int \sqrt{1 + y'^{2} + z'^{2}}\,dx\) gives \(y\) and \(z\) linear in \(x\), each equation being a plane.)

'Two surfaces' or 'a surface by a plane' generally give curves that are not straight. A sphere cut by a plane through its centre gives a great circle, which is the answer for distance measured on a sphere, not in space.

Discuss in Forum

29.

The extremals of the functional

\[ I(y(x)) = \int_{x_{1}}^{x_{2}} f(x, y')\, dx \]

is obtained as the solution of the equation :

A.\(f_{y'x}\, y' + f_{y'} = 0\)
B.\(f_{y'y'}\, y'' + f_{y'x} = 0\)
C.\(y' f_{y'} - f = \text{constant}\)
D.\(f_{y'x} = 0\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The extremals satisfy \(f_{y'y'}\, y'' + f_{y'x} = 0\).

The Euler-Lagrange equation is \(f_{y} - \frac{d}{dx} f_{y'} = 0\). Here f does not contain y, so \(f_{y} = 0\) and

\[ \frac{d}{dx} f_{y'}(x, y') = 0 \]

Differentiating \(f_{y'}\) along the curve by the chain rule gives \(f_{y'x} + f_{y'y'}\, y'' = 0\), which is option B. Integrated once, it says \(f_{y'}\) is constant.

\(y' f_{y'} - f = \) constant is the Beltrami identity, which applies when f does not contain x. \(f_{y'x} = 0\) drops the \(y''\) term. Option A has the wrong derivatives.

Discuss in Forum

30.

The functional

\[ I(y(x)) = \int_{0}^{\pi/4} (y^{2} - y'^{2})\, dx \]

satisfying y(0) = 0 attains its extremal on the curve :

A.\(y = b \sin x\)
B.\(y = e^{x}\)
C.\(y = \cos x\)
D.\(y = 0\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The extremal is \(y = 0\).

The Euler-Lagrange equation for \(F = y^{2} - y'^{2}\) is

\[ F_{y} - \frac{d}{dx}F_{y'} = 2y + 2y'' = 0 \]

so \(y = A\cos x + B\sin x\), and \(y(0) = 0\) gives \(y = B\sin x\).

The right end \(x = \pi/4\) has no prescribed value, so it carries the natural boundary condition \(F_{y'} = 0\) there, that is \(-2y'(\pi/4) = 0\). Then \(B\cos(\pi/4) = 0\), so \(B = 0\) and the extremal is \(y = 0\).

\(y = b\sin x\) is the family that satisfies only the Euler equation and \(y(0) = 0\); it ignores the free-end condition. \(e^{x}\) and \(\cos x\) fail \(y(0) = 0\) or the Euler equation.

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