Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 3 / d / Solution

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give