Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 16A (Variational Principles)

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation

The integrandis nonnegative, so $I[y]\ge 0$. If $I[y]=0$, both continuous summands vanish everywhere. Hence $y^{\prime }=0$, so $y$ is constant; the common boundary value forces $y(x)=a$. Conversely, $y\equiv a$ has $y^{\prime }=0$ and $U(a)=0$, so $I[y]=0$. Therefore

This agrees with part (a): at $y=a$, the first variation vanishes because $y^{\prime \prime }=U(a)=0$, andThus the constant configuration is a stationary local minimum as well as the global minimum identified directly.

For $U(y)=c^2-y^2$, completing the square gives the Bogomolny boundConsequentlywith equality exactly when the Bogomolny equationholds everywhere. Separating variables, or differentiating the proposed form directly, gives all solutions with the required limits:These are the translated phi-four kinks.

Differentiating the first-order equation yieldswhich is precisely the Euler-Lagrange equation from part (a). Thus every configuration saturating the first-order bound is automatically a stationary solution of the second-order variational equation.