Past exam of the differential equations 2 course of the University of Oxford / 2021 / 2

Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.

For the first Frobenius solution, take the larger indicial root $r=a$ and normalize the first coefficient to one:Substitution and comparison of the coefficient of $x^{n+a}$ giveEvery denominator is nonzero because $a\ge 0$.

For $\beta =1$, the recurrence relation is $c_n=-b{c}_{n-1}/[n(n+2a)]$, soFor $\beta =2$, $c_1=0$ and all odd coefficients vanish; the even coefficients satisfy $c_{2k}=-b{c}_{2k-2}/[2k(2k+2a)]$. HenceFor any positive integer $\beta$, induction in the recurrence relation shows that only indices divisible by $\beta$ survive:The empty product for $k=0$ equals one. Successive terms in the power series factor have ratio of magnitudesThus the ratio test gives convergence for every finite $x$; a nonintegral $a$ still requires a chosen side of zero or a complex branch for the factor $x^a$.

For a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe Frobenius case classification concerns the gap $2a$ between the indicial roots. In the course convention, Case IIIa is the obstructed recurrence requiring a Logarithmic Frobenius solution; Case IIIb permits two pure Frobenius series.

If $a>0$ and $2a\notin \mathbb{Z}$, this is Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is Case II, with a repeated indicial root. There is only one pure Frobenius series with a nonzero leading coefficient. To see the second solution's form, reduction of order givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe logarithm has a nonzero coefficient and cannot be removed by adding a multiple of $y_{+}$.

If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe Frobenius resonance for an Euler equation with monomial potential explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original differential operator to $y_{+}\log x$ gives $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

If $\beta$ does not divide $N$, then $d_{N-\beta }=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is Case IIIb: set $d_N=0$ to select an independent pure Frobenius series starting with $x^{-a}$. Subsequent coefficients again follow the recurrence relation. Changing $d_N$ simply adds a multiple of $y_{+}$, the undetermined coefficient at Frobenius resonance.

The complete answer is thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.

Normalize both leading coefficients to one. For the indicial root $a=1/2$, the recurrence relation from part (b) givesRecognizing the Taylor series of the sine gives the first Frobenius solutionFor the smaller indicial root $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first Frobenius solution. The remaining coefficients areThus the Taylor series of the cosine givesTheir leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the differential equation becomes $x^{3/2}(u^{\prime \prime }+u)=0$, so these two functions solve it on $x>0$. Their Wronskian is $-1/x\ne 0$, proving linear independence. The general solution is $(C_1\sin x+C_2\cos x)/\sqrt{x}$.