Calculus of Variations Questions and Answers

Answer: C

The extremals are \(y(x) = \sinh(Ax + B)\).

F does not contain \(x\), so the Euler-Lagrange equation has the first integral \(F - y'F_{y'} = c\):

\[ \begin{aligned} F_{y'} &= -\frac{2(1 + y^{2})}{y'^{3}} \\ F - y'F_{y'} &= \frac{3(1 + y^{2})}{y'^{2}} = c \end{aligned} \]

So \(y'^{2} = A^{2}(1 + y^{2})\), that is \(dy/\sqrt{1 + y^{2}} = A\,dx\), giving \(\sinh^{-1} y = Ax + B\) and \(y = \sinh(Ax + B)\).

The trigonometric families solve \(y'' = -y\), and \(A\sinh x + B\cosh x\) solves \(y'' = y\); neither satisfies \(y'^{2} = A^{2}(1 + y^{2})\) in general.

Answer: C

The shortest distance is \(\dfrac{19\sqrt{2}}{8}\).

The distance from \((x, x^{2})\) to the line \(x - y - 5 = 0\) is \(|x - x^{2} - 5|/\sqrt{2}\). Since \(x^{2} - x + 5 \gt 0\) always, the line and parabola do not meet, and we minimise \(x^{2} - x + 5\):

\[ \begin{aligned} x &= \tfrac{1}{2}: \ \tfrac{1}{4} - \tfrac{1}{2} + 5 = \tfrac{19}{4} \\ d &= \frac{19/4}{\sqrt{2}} = \frac{19\sqrt{2}}{8} \end{aligned} \]

(At \(x = 1/2\) the tangent slope \(2x = 1\) is parallel to the line.) The distance is not 0 because the curves do not intersect, and 5 and \(\sqrt{5}\) do not come from the minimisation.

Answer: D

There are infinitely many extremals if \(\lambda = -1\).

The Euler-Lagrange equation for \(F = y'^{2} - y^{2}\) is \(y'' + y = 0\), with solutions \(y = A\cos x + B\sin x\). The condition \(y(0) = 1\) gives \(A = 1\). Then

\[ y(\pi) = \cos\pi + B\sin\pi = -1 \]

whatever \(B\) is. So if \(\lambda = -1\), every \(y = \cos x + B\sin x\) is an extremal, infinitely many; if \(\lambda \neq -1\), there is none.

In particular there is no extremal at all when \(\lambda = 1\), so A and B fail, and C fails because the extremal is not unique.

Answer: B

The first integral is option B:

\[ \begin{aligned} &\frac{\partial f}{\partial y'} - \frac{d}{dx}\left(\frac{\partial f}{\partial y''}\right) \\ &= \text{constant} \end{aligned} \]

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

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

If \(f\) does not depend on \(y\), then \(f_{y} = 0\) and the rest is an exact derivative:

\[ \frac{d}{dx}\left(f_{y'} - \frac{d}{dx}f_{y''}\right) = 0 \]

so \(f_{y'} - \frac{d}{dx}f_{y''}\) is constant, option B. The other options pair the derivatives wrongly: the term differentiated by \(x\) must be \(f_{y''}\), and the undifferentiated term must be \(f_{y'}\).

Answer: B

The extremal is \(y(x) = -x^{3} + 2x\).

For \(F = y'^{2} - 12xy\), the Euler-Lagrange equation \(F_{y} - \frac{d}{dx}F_{y'} = 0\) gives

\[ \begin{aligned} -12x - 2y'' = 0 &\Rightarrow y'' = -6x \\ &\Rightarrow y = -x^{3} + c_{1}x + c_{2} \end{aligned} \]

\(y(0) = 0\) gives \(c_{2} = 0\), and \(y(1) = 1\) gives \(-1 + c_{1} = 1\), so \(c_{1} = 2\).

\(y = x^{3}\) satisfies the end conditions but not \(y'' = -6x\) (its \(y''\) is \(+6x\)). \(x^{3} - 2x\) gives \(y(1) = -1\), and \(x^{2} - x^{3}\) gives \(y(1) = 0\).

Answer: D

The curve is the cycloid of which one arc contains both points A and B.

This is the brachistochrone problem posed by Johann Bernoulli in 1696. Minimising the descent time \(\int ds / \sqrt{2gy}\) by the calculus of variations gives a cycloid, the curve traced by a point on a circle rolling along a horizontal line, inverted with its cusp at A.

The straight line is the shortest path but not the fastest: the cycloid drops steeply at first, so the particle gains speed early. A parabola is not the extremal, and a minimising curve does exist, so 'no curve' is wrong.

Answer: A

The brachistochrone satisfies \(y(1 + y'^{2}) = c\).

Measure \(y\) downward from the starting point. A particle released from rest has speed \(\sqrt{2gy}\), so the descent time is

\[ T = \int \frac{\sqrt{1 + y'^{2}}}{\sqrt{2gy}}\, dx \]

The integrand \(F = \sqrt{(1 + y'^{2})/y}\) does not contain \(x\), so the Beltrami identity \(F - y' F_{y'} = \text{const}\) applies and simplifies to

\[ \begin{aligned} \frac{1}{\sqrt{y(1 + y'^{2})}} &= \text{const} \\ y(1 + y'^{2}) &= c \end{aligned} \]

Its solutions are cycloids. The equations involving \(\sqrt{a^{2} - x^{2}}\) belong to other problems (curves related to circles), and the square-root form in option B is not the first integral of this functional.

Answer: D

The curve is a circle.

This is the isoperimetric problem: among all closed plane curves of a given length \(L\), the one enclosing the greatest area is the circle. The isoperimetric inequality states \(4\pi A \le L^{2}\), with equality only for a circle. In the calculus of variations it is solved by maximising the area integral with the length as a constraint (a Lagrange multiplier), and the Euler equation says the curvature is constant.

For perimeter 4, the square has area 1, a 1.5 by 0.5 rectangle has 0.75, while the circle has area \(4/\pi \approx 1.27\). An ellipse with unequal axes also encloses less than the circle of the same perimeter.

Answer: D

The extremal is \(4y = x(x - 1)\).

The integrand \(F = xy' - y'^{2}\) does not contain \(y\), so the Euler-Lagrange equation reduces to \(F_{y'} = \text{const}\):

\[ \begin{aligned} x - 2y' &= c \\ y' &= \frac{x - c}{2} \\ y &= \frac{x^{2}}{4} - \frac{cx}{2} + d \end{aligned} \]

The condition \(y(0) = 0\) gives \(d = 0\), and \(y(4) = 3\) gives \(4 - 2c = 3\), so \(c = 1/2\). Then \(y = x^{2}/4 - x/4\), that is \(4y = x(x - 1)\).

Checking the others: \(y = x^{2}/4\) gives \(y(4) = 4\), \(x^{2}/2 + 3x\) gives 20, and \(x^{2}/4 - x + 1\) gives \(y(0) = 1\). Each fails a boundary condition.

Answer: B

The expanded Euler-Lagrange equation is \(f_{y'y'} \dfrac{d^{2}y}{dx^{2}} + f_{y'y} \dfrac{dy}{dx} + (f_{y'x} - f_{y}) = 0\).

The Euler-Lagrange equation is \(\dfrac{d}{dx} f_{y'} - f_{y} = 0\). Since \(f_{y'}\) depends on \(x\), \(y\) and \(y'\), the chain rule gives

\[ \begin{aligned} \frac{d}{dx} f_{y'} &= f_{y'x} + f_{y'y}\, y' \\ &\quad + f_{y'y'}\, y'' \end{aligned} \]

so the equation becomes \(f_{y'y'} y'' + f_{y'y} y' + f_{y'x} - f_{y} = 0\), the second option. It is a second-order equation whenever \(f_{y'y'} \neq 0\).

The first option drops the \(f_{y'y}\,y'\) term, which comes from \(y\) varying with \(x\). The last two put \(f_{y'x}\) as the coefficient of \(y''\), which mixes up the partial derivatives.