Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 17B (Methods)

Let $D=d/dx$. Repeated integration by parts gives the formal adjoint rulesThereforeEquality of the coefficients of $D$ in $L$ and $L^{\ast }$ requiresso $q=p^{\prime }$. This also makes the zeroth-order coefficients equal because $p^{\prime \prime }-q^{\prime }=0$. Hence, under boundary conditions that remove the boundary terms,This is the self-adjoint fourth-order scalar differential operator $D^4+p{D}^2+p^{\prime }D+r$.

With $q=p^{\prime }$ and $r=0$, the equation isMultiply by the real eigenfunction $y$ and integrate over $[a,b]$. The clamped boundary conditions $y=y^{\prime }=0$ at both endpoints eliminate all boundary terms, soSince $p(x)<0$, the right side is a sum of nonnegative terms. It cannot vanish for a nonzero eigenfunction: if $y^{\prime \prime }=y^{\prime }=0$, the boundary conditions force $y=0$. Thus

For $p=0$ and $\lambda =1$, the linear ordinary differential equation is $y^{\prime \prime \prime \prime }=y$, whose general solution isBy the stated parity reduction, an even eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y^{\prime }(c)=0$ have a nonzero solution exactly whenthat is,An odd eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant givesThus $\lambda =1$ is an eigenvalue precisely under one of the two stated conditions.

For the minus sign, division by $\cos c\cosh c$ away from $c=\pi /2$ gives $\tan c=\tanh c$. On $(0,\pi /2)$, $\tan c>\tanh c$ for $c>0$, while on $(\pi /2,\pi )$ their signs differ, so there is no root. For the plus sign the equation isThere is no root in $(0,\pi /2)$. On $(\pi /2,\pi )$, the function $\tan c+\tanh c$ increases strictly from $-\infty$ to $\tanh \pi >0$, so it has exactly one root by the intermediate value theorem. Numerically,Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.