Discussion

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.