Put $a(y)=(1-y^2)^2$. Multiply $u_{xx}=\partial_y(au_y)$ by $2u_x$ and integrate over $-1<y<1$. An <integration by parts> gives
$$
\begin{aligned}
\frac d{dx}\int_{-1}^1u_x^2\,dy
&=2\bigl[au_xu_y\bigr]_{-1}^1-2\int_{-1}^1a u_{xy}u_y\,dy\\
&=-\frac d{dx}\int_{-1}^1a u_y^2\,dy,
\end{aligned}
$$
because $a(1)=a(-1)=0$. Therefore the <energy estimate> is in fact the conservation law
$$
\boxed{\frac d{dx}\int_{-1}^1\left(u_x^2+(1-y^2)^2u_y^2\right)dy=0.}
$$
If $u(0,y)=u_x(0,y)=0$, 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
$$
\boxed{u(x,y)=0\qquad((x,y)\in\mathbb R\times(-1,1)).}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-105/3/d/solution