Calculus of Variations Questions and Answers

Q: The functional\(\int_{0}^{1} \left(y'^{2} + (y + 2y')y'' + kxyy' + y^{2}\right) dx\), \(y(0) = 0, y(1) = 1, y'(0) = 2, y'(1) = 3\)is path independent if \(k\) equals

The functional is path independent when \(k = 2\).A functional is path independent when its integrand is an exact derivative \(dG/dx\), since then the integral depends only on the end values. Group the terms:\[ \begin{aligned} y'^{2} + yy'' &= (yy')' \\ 2y'y'' &= (y'^{2})' \\ y^{2} + 2xyy' &= (xy^{2})' \end{aligned} \]So the integrand is \((yy' + y'^{2} + xy^{2})'\) exactly when \(kxyy' = 2xyy'\), that is \(k = 2\). For any other \(k\), the leftover term \((k - 2)xyy'\) is not a total derivative (its Euler-Lagrange expression \(-(k-2)y\) does not vanish), so the value would depend on the path. Hence 1, 3 and 4 are wrong.

Q: Suppose that among all continuously differentiable functions \(y(x),\ x \in \mathbb{R}\), with \(y(0) = 0\) and \(y(1) = \frac{1}{2}\), the function \(y_{0}(x)\) minimizes the functional\[ \int_{0}^{1} \left(e^{-(y' - x)} + (1 + y)y'\right) dx. \]Then \(y_{0}\left(\frac{1}{2}\right)\) is equal to

\(y_{0}\left(\frac{1}{2}\right)\) is equal to \(\frac{1}{8}\).The term \((1 + y)y'\) is the derivative of \(y + \frac{y^{2}}{2}\), so its integral depends only on the fixed end values and does not affect which function minimises. Only \(F = e^{-(y' - x)}\) matters.F does not contain \(y\), so the Euler-Lagrange equation reduces to \(\frac{\partial F}{\partial y'} = \text{constant}\):\[ \begin{aligned} -e^{-(y' - x)} &= \text{const} \Rightarrow y' - x = k \\ y &= \tfrac{x^{2}}{2} + kx \end{aligned} \]\(y(0) = 0\) holds, and \(y(1) = \frac{1}{2} + k = \frac{1}{2}\) gives \(k = 0\). So \(y_{0} = \frac{x^{2}}{2}\) and \(y_{0}(\frac{1}{2}) = \frac{1}{8}\). \(\frac{1}{4}\) is the value on the straight line \(y = \frac{x}{2}\) joining the end points, which is not the extremal.

Q: Consider the functional :\[ I(u(x, y)) = \iint_{G} F(x, y, u, u_{x}, u_{y})\, dx\, dy \]over the region of integration G, where \(u\) is continuous and has continuous derivatives upto second order. Further \(u\) takes prescribed values on the boundary of G. Then the differential equation for the extremization of the functional is :

The equation is \(F_{u} - \partial_{x}F_{u_{x}} - \partial_{y}F_{u_{y}} = 0\), option D.This is the Euler-Ostrogradsky equation. Vary \(u\) to \(u + \epsilon\eta\) with \(\eta = 0\) on the boundary. The first variation is\[ \begin{aligned} \iint (&F_{u}\eta + F_{u_{x}}\eta_{x} \\ &+ F_{u_{y}}\eta_{y})\,dx\,dy \end{aligned} \]Integrating the last two terms by parts (Green's theorem), the boundary terms vanish and\[ \begin{aligned} \iint \eta\,(F_{u} &- \partial_{x}F_{u_{x}} \\ &- \partial_{y}F_{u_{y}}) = 0 \end{aligned} \]for every \(\eta\), which forces the bracket to vanish. Each derivative must be paired with its own variable: \(\partial/\partial x\) with \(F_{u_{x}}\) and \(\partial/\partial y\) with \(F_{u_{y}}\). Options A, B and C each mismatch one pair.

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

The extremal is \(y = x^{2} - 1\).\(F = x^{3}/(y')^{2}\) does not contain \(y\), so the Euler equation reduces to \(F_{y'} = \text{constant}\):\[ \frac{-2x^{3}}{(y')^{3}} = c \;\Rightarrow\; y' = kx \]so \(y = \tfrac{k}{2}x^{2} + d\). From \(y(1) = 0\): \(k/2 + d = 0\). From \(y(2) = 3\): \(2k + d = 3\). Subtracting, \(\tfrac{3k}{2} = 3\), so \(k = 2\), \(d = -1\) and \(y = x^{2} - 1\).Check the others against the boundary values: \((x^{2} - 1)x\) gives 6 at \(x = 2\), \((x - 1)x\) gives 2, and \(x - 1\) gives 1, none of them 3.

Q: The extremal of the isoperimetric problem :\[ I(y(x)) = \int_{1}^{4} (y')^{2}\, dx \]subject to the conditions :\[ \int_{1}^{4} y\, dx = 36 \]and \(y(1) = 3\), \(y(4) = 24\) is :

The extremal is a parabola.For an isoperimetric problem, extremise \(\int (F + \lambda G)\,dx\) with the constraint integrand \(G = y\) and a Lagrange multiplier \(\lambda\). Here \(H = (y')^{2} + \lambda y\), and its Euler equation is\[ \begin{aligned} \lambda - \frac{d}{dx}(2y') &= 0 \\ y'' &= \frac{\lambda}{2} \end{aligned} \]So \(y = \tfrac{\lambda}{4}x^{2} + c_{1}x + c_{2}\), a quadratic, whose graph is a parabola. The three constants are fixed by \(y(1) = 3\), \(y(4) = 24\) and the area condition; they give \(y = x^{2} + 2x\), which indeed has \(\int_{1}^{4} y\,dx = 21 + 15 = 36\).The circle is the answer to the classical isoperimetric problem of maximising area for a fixed arc length, where the functional is \(\int\sqrt{1 + y'^{2}}\); this functional is different.

Q: Any function that gives an extremum of the functional\[ \iint_{D} \{z_{x}^{2} + z_{y}^{2} + 2zf(x, y)\}\,dx\,dy \]must satisfy :

The extremal must satisfy Poisson's equation.For \(F(z, z_{x}, z_{y}) = z_{x}^{2} + z_{y}^{2} + 2zf(x, y)\), the Euler-Ostrogradsky equation is\[ \begin{aligned} F_{z} - \frac{\partial}{\partial x}F_{z_{x}} &- \frac{\partial}{\partial y}F_{z_{y}} \\ &= 0 \end{aligned} \]which gives \(2f - 2z_{xx} - 2z_{yy} = 0\), that is\[ \nabla^{2}z = f(x, y) \]That is Poisson's equation. It reduces to Laplace's equation only if \(f \equiv 0\), and there is no time variable for a heat or wave equation.

Q: The curve which extremizes the functional\[ I[y(x)] = \int_{0}^{\pi/4} (y''^{2} - y^{2} + x^{2})\,dx \]under the conditions \(y(0) = 0\), \(y'(0) = 1\), \(y(\pi/4) = y'(\pi/4) = \dfrac{1}{\sqrt{2}}\) is :

The extremal is \(y(x) = \sin x\).For \(F(x, y, y'') = y''^{2} - y^{2} + x^{2}\), the Euler-Poisson equation is \(F_{y} + \dfrac{d^{2}}{dx^{2}}F_{y''} = 0\):\[ -2y + 2y^{(4)} = 0 \Rightarrow y^{(4)} = y \]So \(y = c_{1}e^{x} + c_{2}e^{-x} + c_{3}\cos x + c_{4}\sin x\). The four conditions pick out \(\sin x\): \(\sin 0 = 0\), \(\cos 0 = 1\), and \(\sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2}\), and the four conditions determine the constants uniquely.\(1 - \cos x\) has \(y'(0) = 0\); \(\cos x\) has \(y(0) = 1\); \(\tan x\) has \(y(\pi/4) = 1\). None meets the conditions.

Q: The shortest arc connecting two points on the surface of a sphere is called :

It is a great circle arc.The geodesics of a sphere are its great circles, the intersections of the sphere with planes through its centre. The shorter arc of the great circle through two points is the shortest path on the surface between them; this is why long-distance flight routes follow great circles.A catenary is the shape of a hanging chain, a catenoid is the minimal surface of revolution spanning two circles, and a cycloid solves the brachistochrone (fastest descent) problem. None is a shortest path on a sphere.

Q: Let \(J(y) = \displaystyle\int_{-1}^{1} (y'e^{y} + xy^{2})\, dx\) be a functional defined on \(C^{1}[-1, 1]\). Then the variation of the integral \(J[y]\) is :

The variation is \(\int_{-1}^{1} [[y'e^{y} + 2xy]\delta y + e^{y}\delta y']\,dx\).For \(J = \int F(x, y, y')\,dx\), the first variation is \(\delta J = \int (F_{y}\,\delta y + F_{y'}\,\delta y')\,dx\). Here \(F = y'e^{y} + xy^{2}\), so\[ \begin{aligned} F_{y} &= y'e^{y} + 2xy \\ F_{y'} &= e^{y} \end{aligned} \]which is option C. Option D adds a spurious \(y^{2}\) to \(F_{y}\), and options A and B are not variations at all: they contain no \(\delta y\) and look like attempts at differentiating the integrand.

The functional

\(\int_{0}^{1} \left(y'^{2} + (y + 2y')y'' + kxyy' + y^{2}\right) dx\), \(y(0) = 0, y(1) = 1, y'(0) = 2, y'(1) = 3\)

is path independent if \(k\) equals

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

The functional is path independent when \(k = 2\).

A functional is path independent when its integrand is an exact derivative \(dG/dx\), since then the integral depends only on the end values. Group the terms:

\[ \begin{aligned} y'^{2} + yy'' &= (yy')' \\ 2y'y'' &= (y'^{2})' \\ y^{2} + 2xyy' &= (xy^{2})' \end{aligned} \]

So the integrand is \((yy' + y'^{2} + xy^{2})'\) exactly when \(kxyy' = 2xyy'\), that is \(k = 2\). For any other \(k\), the leftover term \((k - 2)xyy'\) is not a total derivative (its Euler-Lagrange expression \(-(k-2)y\) does not vanish), so the value would depend on the path. Hence 1, 3 and 4 are wrong.

Discuss in Forum

The functional

\[ \begin{aligned} &\int_{0}^{1}\left(y'^{2} + 4y^{2} + 8ye^{x}\right)dx, \\ &y(0) = -\frac{4}{3},\ y(1) = -\frac{4e}{3} \end{aligned} \]

possesses :

A.strong minima on \(y = -\frac{1}{3}e^{x}\)
B.strong minima on \(y = -\frac{4}{3}e^{x}\)
C.weak maxima on \(y = -\frac{1}{3}e^{x}\)
D.strong maxima on \(y = -\frac{4}{3}e^{x}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

The functional has its extremal at \(y = -\frac{4}{3}e^{x}\), and that extremal gives a strong minimum, so the answer is strong minima on \(y = -\frac{4}{3}e^{x}\).

With \(F = y'^{2} + 4y^{2} + 8ye^{x}\), the Euler equation \(\frac{d}{dx}F_{y'} = F_{y}\) reads

\[ \begin{aligned} 2y'' &= 8y + 8e^{x} \\ y'' - 4y &= 4e^{x} \end{aligned} \]

A particular solution \(Ae^{x}\) needs \(A - 4A = 4\), so \(A = -\frac{4}{3}\), and the general solution is

\[ y = c_{1}e^{2x} + c_{2}e^{-2x} - \frac{4}{3}e^{x}. \]

The condition \(y(0) = -\frac{4}{3}\) gives \(c_{1} + c_{2} = 0\), and \(y(1) = -\frac{4e}{3}\) gives \(c_{1}e^{2} + c_{2}e^{-2} = 0\). Together these force \(c_{1} = c_{2} = 0\), so \(y = -\frac{4}{3}e^{x}\). This also rules out the options with \(-\frac{1}{3}e^{x}\), which has \(y(0) = -\frac{1}{3}\) and so misses the boundary value.

For the type of extremum, put \(\hat{y} = y + \eta\) with \(\eta(0) = \eta(1) = 0\). The terms linear in \(\eta\) vanish by the Euler equation, leaving

\[ \begin{aligned} J[\hat{y}] - J[y] &= \int_{0}^{1}\left(\eta'^{2} + 4\eta^{2}\right)dx \\ &\ge 0. \end{aligned} \]

This holds for every admissible curve, however far it lies from \(y\), so the minimum is strong (and weak too); the integrand is convex in \(y\) and \(y'\). A maximum is impossible, which removes the two options that say maxima.

Discuss in Forum

Suppose that among all continuously differentiable functions \(y(x),\ x \in \mathbb{R}\), with \(y(0) = 0\) and \(y(1) = \frac{1}{2}\), the function \(y_{0}(x)\) minimizes the functional

\[ \int_{0}^{1} \left(e^{-(y' - x)} + (1 + y)y'\right) dx. \]

Then \(y_{0}\left(\frac{1}{2}\right)\) is equal to

A.0
B.\(\frac{1}{8}\)
C.\(\frac{1}{4}\)
D.\(\frac{1}{2}\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: B

\(y_{0}\left(\frac{1}{2}\right)\) is equal to \(\frac{1}{8}\).

The term \((1 + y)y'\) is the derivative of \(y + \frac{y^{2}}{2}\), so its integral depends only on the fixed end values and does not affect which function minimises. Only \(F = e^{-(y' - x)}\) matters.

F does not contain \(y\), so the Euler-Lagrange equation reduces to \(\frac{\partial F}{\partial y'} = \text{constant}\):

\[ \begin{aligned} -e^{-(y' - x)} &= \text{const} \Rightarrow y' - x = k \\ y &= \tfrac{x^{2}}{2} + kx \end{aligned} \]

\(y(0) = 0\) holds, and \(y(1) = \frac{1}{2} + k = \frac{1}{2}\) gives \(k = 0\). So \(y_{0} = \frac{x^{2}}{2}\) and \(y_{0}(\frac{1}{2}) = \frac{1}{8}\). \(\frac{1}{4}\) is the value on the straight line \(y = \frac{x}{2}\) joining the end points, which is not the extremal.

Discuss in Forum

Consider the functional :

\[ I(u(x, y)) = \iint_{G} F(x, y, u, u_{x}, u_{y})\, dx\, dy \]

over the region of integration G, where \(u\) is continuous and has continuous derivatives upto second order. Further \(u\) takes prescribed values on the boundary of G. Then the differential equation for the extremization of the functional is :

A.\(\dfrac{\partial F}{\partial u}\) - \(\dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial u_{x}}\right)\) - \(\dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial u_{y}}\right)\) = 0
B.\(\dfrac{\partial F}{\partial u}\) - \(\dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial u_{y}}\right)\) - \(\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial u_{y}}\right)\) = 0
C.\(\dfrac{\partial F}{\partial u}\) - \(\dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial u_{x}}\right)\) - \(\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial u_{x}}\right)\) = 0
D.\(\dfrac{\partial F}{\partial u}\) - \(\dfrac{\partial}{\partial x}\left(\dfrac{\partial F}{\partial u_{x}}\right)\) - \(\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial u_{y}}\right)\) = 0

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The equation is \(F_{u} - \partial_{x}F_{u_{x}} - \partial_{y}F_{u_{y}} = 0\), option D.

This is the Euler-Ostrogradsky equation. Vary \(u\) to \(u + \epsilon\eta\) with \(\eta = 0\) on the boundary. The first variation is

\[ \begin{aligned} \iint (&F_{u}\eta + F_{u_{x}}\eta_{x} \\ &+ F_{u_{y}}\eta_{y})\,dx\,dy \end{aligned} \]

Integrating the last two terms by parts (Green's theorem), the boundary terms vanish and

\[ \begin{aligned} \iint \eta\,(F_{u} &- \partial_{x}F_{u_{x}} \\ &- \partial_{y}F_{u_{y}}) = 0 \end{aligned} \]

for every \(\eta\), which forces the bracket to vanish. Each derivative must be paired with its own variable: \(\partial/\partial x\) with \(F_{u_{x}}\) and \(\partial/\partial y\) with \(F_{u_{y}}\). Options A, B and C each mismatch one pair.

Discuss in Forum

The extremal of the functional :

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

subject to the conditions \(y(1) = 0\), \(y(2) = 3\) is :

A.\(y = (x^{2} - 1)x\)
B.\(y = (x^{2} - 1)\)
C.\(y = (x - 1)x\)
D.\(y = (x - 1)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The extremal is \(y = x^{2} - 1\).

\(F = x^{3}/(y')^{2}\) does not contain \(y\), so the Euler equation reduces to \(F_{y'} = \text{constant}\):

\[ \frac{-2x^{3}}{(y')^{3}} = c \;\Rightarrow\; y' = kx \]

so \(y = \tfrac{k}{2}x^{2} + d\). From \(y(1) = 0\): \(k/2 + d = 0\). From \(y(2) = 3\): \(2k + d = 3\). Subtracting, \(\tfrac{3k}{2} = 3\), so \(k = 2\), \(d = -1\) and \(y = x^{2} - 1\).

Check the others against the boundary values: \((x^{2} - 1)x\) gives 6 at \(x = 2\), \((x - 1)x\) gives 2, and \(x - 1\) gives 1, none of them 3.

Discuss in Forum

The extremal of the isoperimetric problem :

\[ I(y(x)) = \int_{1}^{4} (y')^{2}\, dx \]

subject to the conditions :

\[ \int_{1}^{4} y\, dx = 36 \]

and \(y(1) = 3\), \(y(4) = 24\) is :

A.a parabola
B.an ellipse
C.a hyperbola
D.a circle

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The extremal is a parabola.

For an isoperimetric problem, extremise \(\int (F + \lambda G)\,dx\) with the constraint integrand \(G = y\) and a Lagrange multiplier \(\lambda\). Here \(H = (y')^{2} + \lambda y\), and its Euler equation is

\[ \begin{aligned} \lambda - \frac{d}{dx}(2y') &= 0 \\ y'' &= \frac{\lambda}{2} \end{aligned} \]

So \(y = \tfrac{\lambda}{4}x^{2} + c_{1}x + c_{2}\), a quadratic, whose graph is a parabola. The three constants are fixed by \(y(1) = 3\), \(y(4) = 24\) and the area condition; they give \(y = x^{2} + 2x\), which indeed has \(\int_{1}^{4} y\,dx = 21 + 15 = 36\).

The circle is the answer to the classical isoperimetric problem of maximising area for a fixed arc length, where the functional is \(\int\sqrt{1 + y'^{2}}\); this functional is different.

Discuss in Forum

Any function that gives an extremum of the functional

\[ \iint_{D} \{z_{x}^{2} + z_{y}^{2} + 2zf(x, y)\}\,dx\,dy \]

must satisfy :

A.Laplace equation
B.Heat equation
C.Wave equation
D.Poisson equation

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The extremal must satisfy Poisson's equation.

For \(F(z, z_{x}, z_{y}) = z_{x}^{2} + z_{y}^{2} + 2zf(x, y)\), the Euler-Ostrogradsky equation is

\[ \begin{aligned} F_{z} - \frac{\partial}{\partial x}F_{z_{x}} &- \frac{\partial}{\partial y}F_{z_{y}} \\ &= 0 \end{aligned} \]

which gives \(2f - 2z_{xx} - 2z_{yy} = 0\), that is

\[ \nabla^{2}z = f(x, y) \]

That is Poisson's equation. It reduces to Laplace's equation only if \(f \equiv 0\), and there is no time variable for a heat or wave equation.

Discuss in Forum

The curve which extremizes the functional

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

under the conditions \(y(0) = 0\), \(y'(0) = 1\), \(y(\pi/4) = y'(\pi/4) = \dfrac{1}{\sqrt{2}}\) is :

A.\(y(x) = 1 - \cos x\)
B.\(y(x) = \tan x\)
C.\(y(x) = \cos x\)
D.\(y(x) = \sin x\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The extremal is \(y(x) = \sin x\).

For \(F(x, y, y'') = y''^{2} - y^{2} + x^{2}\), the Euler-Poisson equation is \(F_{y} + \dfrac{d^{2}}{dx^{2}}F_{y''} = 0\):

\[ -2y + 2y^{(4)} = 0 \Rightarrow y^{(4)} = y \]

So \(y = c_{1}e^{x} + c_{2}e^{-x} + c_{3}\cos x + c_{4}\sin x\). The four conditions pick out \(\sin x\): \(\sin 0 = 0\), \(\cos 0 = 1\), and \(\sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2}\), and the four conditions determine the constants uniquely.

\(1 - \cos x\) has \(y'(0) = 0\); \(\cos x\) has \(y(0) = 1\); \(\tan x\) has \(y(\pi/4) = 1\). None meets the conditions.

Discuss in Forum

The shortest arc connecting two points on the surface of a sphere is called :

A.a great circle arc
B.a catenary
C.a catenoid of revolution
D.a cycloid

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

It is a great circle arc.

The geodesics of a sphere are its great circles, the intersections of the sphere with planes through its centre. The shorter arc of the great circle through two points is the shortest path on the surface between them; this is why long-distance flight routes follow great circles.

A catenary is the shape of a hanging chain, a catenoid is the minimal surface of revolution spanning two circles, and a cycloid solves the brachistochrone (fastest descent) problem. None is a shortest path on a sphere.

Discuss in Forum

10.

Let \(J(y) = \displaystyle\int_{-1}^{1} (y'e^{y} + xy^{2})\, dx\) be a functional defined on \(C^{1}[-1, 1]\). Then the variation of the integral \(J[y]\) is :

A.\(\int_{-1}^{1} (y''e^{y} + e^{y}y' + 2xyy')\, dx\)
B.\(\int_{-1}^{1} (y'^{2}e^{y} + y^{2} + 2xyy')\, dx\)
C.\(\int_{-1}^{1} [[y'e^{y} + 2xy]\delta y + e^{y}\delta y']\, dx\)
D.\(\int_{-1}^{1} [[y'e^{y} + 2xy + y^{2}]\delta y + e^{y}\delta y']\, dx\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The variation is \(\int_{-1}^{1} [[y'e^{y} + 2xy]\delta y + e^{y}\delta y']\,dx\).

For \(J = \int F(x, y, y')\,dx\), the first variation is \(\delta J = \int (F_{y}\,\delta y + F_{y'}\,\delta y')\,dx\). Here \(F = y'e^{y} + xy^{2}\), so

\[ \begin{aligned} F_{y} &= y'e^{y} + 2xy \\ F_{y'} &= e^{y} \end{aligned} \]

which is option C. Option D adds a spurious \(y^{2}\) to \(F_{y}\), and options A and B are not variations at all: they contain no \(\delta y\) and look like attempts at differentiating the integrand.

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