Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 5A (Differential Equations)

After multiplying either solution by $-1$ if necessary, suppose both are positive on $(x_1,x_2)$. SetThe equations giveSince ${\phi }_1$ is positive between consecutive zeros,ThereforeBut $W$ is nondecreasing, so both endpoint values coincide at zero and $W^{\prime }\equiv 0$. Since both solutions are strictly positive inside,This is the equality case in the wronskian proof of Sturm comparison.

Suppose ${\phi }_2$ has no zero in $(x_1,x_2)$. It has a constant sign there, so replace it by its negative if needed and apply part (i). The conclusion is $q_1\equiv q_2$ on that interval. Taking the contrapositive proves the Sturm comparison theorem: unless the coefficients agree identically, ${\phi }_2$ has a zero between the consecutive zeros of ${\phi }_1$.

TakeIts consecutive zeros are $-1$ and $\pi -1$. For the stated equation,and $q_2$ is not identically equal to $1$ on the interval. The Sturm comparison theorem therefore shows that every nontrivial solution has at least one zero in

Rewrite the equation asOn $[2,6]$, one has $q(x)\le 1/3$. If a nontrivial solution had two zeros, choose two consecutive ones $\alpha <\beta$ in that interval. Then $\beta -\alpha \le 4<\pi \sqrt{3}$.

The solutionof ${\phi }^{\prime \prime }+\frac{1}{3}\phi =0$ is strictly positive on $[\alpha ,\beta ]$. Since $q\le 1/3$ and the coefficients are not identical on this interval, the Sturm comparison theorem requires $\phi$ to have a zero between $\alpha$ and $\beta$, a contradiction. Hence every nontrivial solution has