Past exam of the differential equations 2 course of the University of Oxford / 2015 / 2

Divide by $2x$ away from zero to obtain $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda /2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^{2}Q(x)=\lambda x^2/2$ are analytic. The pole in $P$ makes zero singular, so $x=0$ is a regular singular point. Its indicial equation isAt $x=1$, both normalized coefficients are analytic functions, so $x=1$ is an ordinary point. These are the standard normalized-coefficient and indicial tests.^[ref]

For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The chain rule gives $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, so the transformed normalized equation isThe product of the first-derivative coefficient with $t$ still has a second-order pole at zero. Hence the criterion for a regular singular point fails and $x=\infty$ is an irregular singular point, including when $\lambda =0$. The primes on $Y$ here refer to differentiation with respect to $t$.

Seek a Frobenius series $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$, on the positive side of zero. The lowest coefficient gives the indicial equation from part (a); the next gives $(r+1)(2r-1)a_1=0$. For either permitted indicial exponent, $a_1=0$. Matching the subsequent powers givesThe two indicial exponents differ by a noninteger, so there are two independent Frobenius solutions without a logarithmic companion. Normalizing each leading coefficient to one, their forms areThus the general local solution is $C_0{f}_0+C_1{f}_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne 0$, so $y(x)\sim C_1{x}^{3/2}$ and $\alpha =3/2$, with the arbitrary nonzero amplitude understood.

The condition at one selects values of $\lambda$ for which this branch also vanishes there; it does not change the selected local exponent. The zero solution, available for every $\lambda$, has no nonzero leading coefficient and is excluded when asking for a leading exponent. This is a direct application of the Frobenius method and the noninteger-separated-root case.^[ref]

First classify nonzero polynomial solutions of the differential equation itself. If $y$ has degree $d$ and leading coefficient $a_d\ne 0$, the coefficient of $x^{d+1}$ is $(d+\lambda )a_d$. Hence $\lambda =-d$ is necessary.

An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power coefficients vanish, because $a_1=0$ and the recurrence relation preserves parity. For the even-power coefficients,All denominators are nonzero for $j\ge 0$. With $\lambda =-2m$, the coefficients through $a_{2m}$ are nonzero when $a_0\ne 0$, the next coefficient vanishes, and every subsequent coefficient vanishes. This proves sufficiency as well as the restriction to even polynomial degree:For example, setting $a_0=1$ gives $1$ when $\lambda =0$, $1+x^2$ when $\lambda =-2$, and $1+2x^2+x^4/5$ when $\lambda =-4$. A polynomial cannot instead begin with a positive integer power: its lowest power must have an indicial exponent, and the only integer indicial exponent is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a terminating Frobenius series.

If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero polynomial solution: every nonzero polynomial just classified has $y(0)=a_0\ne 0$. The displayed family and degree answer therefore concern the differential equation without those extra endpoint conditions.