Discussion
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
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.
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