Solution. We note that (using the chain rule and the product rule)

\begin{gather*} \frac {d\Phi }{d\phi }=\frac {d\Phi }{dx}\frac {dx}{d\phi }, \\ \frac {d^2\Phi }{d\phi ^2}=\frac {d}{d\phi }\left (\frac {d\Phi }{dx}\frac {dx}{d\phi }\right ) =\frac {dx}{d\phi }\frac {d}{d\phi }\frac {d\Phi }{dx}+\frac {d\Phi }{dx}\frac {d}{d\phi }\frac {dx}{d\phi } =\frac {dx}{d\phi }\frac {dx}{d\phi }\frac {d}{dx}\frac {d\Phi }{dx} +\frac {d\Phi }{dx}\frac {d^2x}{d\phi ^2} =\left (\frac {dx}{d\phi }\right )^2\frac {d^2\Phi }{dx^2} +\frac {d\Phi }{dx}\frac {d^2x}{d\phi ^2}. \end{gather*} We have \(\frac {dx}{d\phi }=-\sin (\phi )\) and \(\frac {d^2x}{d\phi ^2}=-\cos (\phi )\) so that the equation becomes (slightly inappropriately mixing variables for clarity)

\[ \sin ^2(\phi )\frac {d^2}{dx^2}\Phi -\cos (\phi )\frac {d}{dx}\Phi -\cot (\phi )\sin (\phi )\frac {d}{dx}\Phi +\lambda ^\Theta \csc ^2(\phi )\Phi =\lambda ^\Phi \Phi . \]

Using \(\cot (\phi )\sin (\phi )=\cos (\phi )=x\), \(\sin ^2(\phi )=1-x^2\) and \(\csc ^2(\phi )=\frac {1}{1-x^2}\), this is

\[ (1-x^2)\frac {d^2}{dx^2}\Phi -2x\frac {d}{dx}\Phi +\frac {\lambda ^\Theta }{1-x^2}\Phi =\lambda ^\Phi \Phi , \]

as desired.

We further have (using \(\frac {dx}{d\phi }=-\sin (\phi )\) and that as \(\phi \) goes from \(0\) to \(\pi \), then \(x=\cos (\phi )\) goes from 1 to \(-1\)):

\[ \int _0^\pi \Phi (\phi )^2\,\sin (\phi )\,d\phi =-\int _{1}^{-1} \Phi (x)^2\,dx =\int _{-1}^1 \Phi (x)^2\,dx, \]

as desired. □

Solution. From the solution space we see that \(\a =-1\), \(\b =1\) and \(w=1\). The right-hand side of the differential equation therefore is already of the form \(\lambda wu\) and the left-hand side of the differential equation should then equal \(pu''+p'u'+qu\). From the coefficients of \(u''\) and \(u\) we then read off that \(p=1-x^2\) and \(q=-\frac {\sigma ^2}{1-x^2}\). Since then \(p'=-2x\), this is consistent with the coefficient of \(u'\). In summary:

\[ \a =-1,~\b =1, \quad w=1,\quad p=1-x^2,\quad q=\frac {-\sigma ^2}{1-x^2}. \]

The smoothness assumptions on \(w,p,q\) and the sign conditions on \(w,p\) on the interval \((\a ,\b )\) are clearly satisfied. □

Solution. From the solution space we see that \(\a =0\), \(\b =b\) and \(w=x^2\). The left-hand side of the differential equation should then equal \(pu''+p'u'+qu\). From the coefficients of \(u''\) and \(u\) we then read off that \(p=x^2\) and \(q=-\sigma (\sigma +1)\). Since then \(p'=2x\), this is consistent with the coefficient of \(u'\). In summary:

\[ \a =0,~\b =b, \quad w=x^2,\quad p=x^2,\quad q=-\sigma (\sigma +1). \]

The smoothness assumptions on \(w,p,q\) and the sign conditions on \(w,p\) on the interval \((\a ,\b )\) are clearly satisfied. □

Solution. From the solution space we see that \(\a =-1\), \(\b =1\) and \(w=\frac {1}{\sqrt {1-x^2}}\). Re-writing the differential equation so that the right-hand side equals \(\lambda wu\) gives

\[ \sqrt {1-x^2}\,u''-\frac {x}{\sqrt {1-x^2}}u'=\lambda \frac {1}{\sqrt {1-x^2}}u. \]

The left-hand side of this equation should then equal \(pu''+p'u'+qu\). From the coefficients of \(u''\) and \(u\) we then read off that \(p=\sqrt {1-x^2}\) and \(q=0\). Since then \(p'=\frac {-x}{\sqrt {1-x^2}}\), this is consistent with the coefficient of \(u'\). In summary:

\[ \a =-1,~\b =1, \quad w=\frac {1}{\sqrt {1-x^2}},\quad p=\sqrt {1-x^2},\quad q=0. \]

The smoothness assumptions on \(w,p,q\) and the sign conditions on \(w,p\) on the interval \((\a ,\b )\) are clearly satisfied. □

Solution. We have by the chain rule

\[ \frac {dy}{d\theta }=\frac {dx}{d\theta }~\frac {du}{dx}=-\sin (\theta )\frac {du}{dx}. \]

Using the product rule and the chain rule we obtain

\[ \frac {d^2y}{d\theta ^2}=-\cos (\theta )\frac {du}{dx}-\sin (\theta )\frac {dx}{d\theta }\frac {d^2u}{dx^2} =-\cos (\theta )\frac {du}{dx}+\sin ^2(\theta )\frac {d^2u}{dx^2}. \]

The original differential equation is

\[ \sin (\theta )^2\frac {du^2}{dx^2}-\cos (\theta )\frac {du}{dx}=\lambda u, \]

and we see that this equals \(\frac {d^2y}{d\theta ^2}=\lambda y\).

Alternatively, we have \(u(x)=y(\arccos (x))\) and (using \(x=\cos (\theta )\), \(\theta =\arccos (x)\))

\[ \frac {du}{dx}=\frac {dy}{d\theta }\,\frac {d\theta }{dx}=\frac {dy}{d\theta }\,\frac {-1}{\sqrt {1-x^2}}, \]

so that

\[ \sqrt {1-x^2}\frac {du}{dx}=-\frac {dy}{d\theta }, \]

and therefore

\[ \frac {d}{dx}\left (\sqrt {1-x^2}\frac {du}{dx}\right ) =-\frac {d\theta }{dx}\,\frac {d}{d\theta }\frac {dy}{d\theta } =\frac {1}{\sqrt {1-x^2}}\frac {d^2y}{d\theta ^2}. \]

Hence the differential equation

\[ \left (\sqrt {1-x^2}\,u'\right )'=\lambda \frac {1}{\sqrt {1-x^2}}u, \]

is equivalent to

\[ y''=\lambda y. \]

From our coordinate transformation we have \(\theta \in (0,\pi )\) and we have

\[ d\theta =\frac {-1}{\sqrt {1-x^2}}\,dx, \]

so the solution space \(L^2\left (-1,1;\frac {1}{\sqrt {1-x^2}}\right )\) for \(u\) transforms into the solution space \(L^2(0,\pi )\) for \(y\). □

Solution. From the solution space we see that \(\a =0\), \(\b =\infty \) and \(w=\e ^{-x}\). Re-writing the differential equation so that the right-hand side equals \(\lambda wu\) gives

\[ x\e ^{-x}u''+(1-x)\e ^{-x}u'=\lambda \e ^{-x}u. \]

The left-hand side of this equation should then equal \(pu''+p'u'+qu\). From the coefficients of \(u''\) and \(u\) we then read off that \(p=x\e ^{-x}\) and \(q=0\). Since then \(p'=\e ^{-x}-x\e ^{-x}\), this is consistent with the coefficient of \(u'\). In summary:

\[ \a =0,~\b =\infty , \quad w=\e ^{-x},\quad p=x\e ^{-x},\quad q=0. \]

The smoothness assumptions on \(w,p,q\) and the sign conditions on \(w,p\) on the interval \((\a ,\b )\) are clearly satisfied. □

Solution. From the solution space we see that \(\a =-\infty \), \(\b =\infty \) and \(w=\e ^{-x^2}\). Re-writing the differential equation so that the right-hand side equals \(\lambda wu\) gives

\[ \e ^{-x^2}u''-2\e ^{-x^2}u'=\lambda \e ^{-x^2}u. \]

The left-hand side of this equation should then equal \(pu''+p'u'+qu\). From the coefficients of \(u''\) and \(u\) we then read off that \(p=\e ^{-x^2}\) and \(q=0\). Since then \(p'=-2x\e ^{-x^2}\), this is consistent with the coefficient of \(u'\). In summary:

\[ \a =-\infty ,~\b =\infty , \quad w=\e ^{-x^2},\quad p=\e ^{-x^2},\quad q=0. \]

The smoothness assumptions on \(w,p,q\) and the sign conditions on \(w,p\) on the interval \((\a ,\b )\) are clearly satisfied. □

Solution. (a) From the solution space we obtain that \(w=1\). Therefore the right-hand side is of the form \(\lambda wu\) so that the left-hand side should be of the form \(pu''+p'u'+qu\). From the coefficient of \(u''\) we have that \(p=1\). Therefore \(p'=0\), but this is inconsistent with the coefficient of \(u'\). Therefore this is not a Sturm–Liouville problem.

(b) From the solution space we have \(w=1\). Therefore the right-hand side is of the form \(\lambda wu\) so that the left-hand side should be of the form \(pu''+p'u'+qu\). From the coefficient of \(u''\) we have that \(p=x\). Therefore \(p'=1\), which is consistent with the coefficient of \(u'\). We have \(p(0)=0\) and \(0\in (-1,1)\) so that the sign condition on \(p\) is not satisfied. Therefore this is not a Sturm–Liouville problem.

(c) From the solution space we have \(w=x\). Since \(0\in (-1,1)\) and \(w(0)=0\) we do not have that \(w>0\) on \((\a ,\b )\). Therefore this is not a Sturm–Liouville problem.

Writing the equation in standard form as \(xu''=\lambda xu\), we can also read off \(p=x\) (which also doesn’t satisfy the sign condition) and \(p'=1\) which is not consistent with the coefficient of \(u'\). □

Solution. By symmetry we only have to consider one boundary point (the other boundary point will be of the same type). We solve \(Du=0\) which is

\begin{equation} \label {eq:leg}\tag {$\dagger $} [(1-x^2)u']'=\frac {\sigma ^2}{1-x^2}u. \end{equation}

We first consider the case \(\sigma =0\) in which case the above reduces to

\[ [(1-x^2)u']'=0, \]

from which we obtain for \(C_1\) constant

\[ u'=\frac {C_1}{1-x^2}, \]

which gives for \(C_2\) constant

\[ u=\frac {C_1}{2}\ln \left (\frac {1+x}{1-x}\right )+C_2. \]

We have \(\ln \left (\frac {1+x}{1-x}\right )=\ln (1+x)-\ln (1-x)\) and at \(\b =1\) the first of these summands is continuous. Therefore we consider only the second summand and we have (using standard limits for the “evaluation” of the anti-derivative at zero):

\[ \int _0^1 |\ln (1-x)|^2\,dx=\int _0^1 \ln ^2(y)\,dy=\left [y\ln ^2(y)-2y\ln (y)+2y\right ]_0^1=2<\infty . \]

Clearly we also have that \(1\) belongs to \(L^2(0,1)\). This shows that we have two linearly independent solutions in \(L^2(0,1)\), so that \(\b =1\) is quasi-regular. By symmetry, both boundary points are quasi-regular.

We now consider the case \(\sigma >0\). We make the Ansatz that (for some \(r\in \mR \))

\[ u(x)=\left (\frac {1+x}{1-x}\right )^r. \]

We then have (using the chain and quotient rules)

\[ u'(x)=2r\frac {(1+x)^{r-1}}{(1-x)^{r+1}}, \]

so that

\[ (1-x^2)u'(x)=(1-x)(1+x)u'(x)=2r\left (\frac {1+x}{1-x}\right )^r, \]

and therefore

\[ [(1-x^2)u']'=4r^2\frac {(1+x)^{r-1}}{(1-x)^{r+1}}, \]

which gives

\[ (1-x^2)[(1-x^2)u']'=(1-x)(1+x)[(1-x^2)u']'=4r^2\left (\frac {1+x}{1-x}\right )^r=4r^2u(x). \]

Since by (\(\dagger \)) this should equal \(\sigma ^2u\), we obtain \(\sigma ^2=4r^2\) which gives

\[ r=\pm \frac {\sigma }{2}. \]

We therefore obtain the solutions \(\left (\frac {1+x}{1-x}\right )^{\sigma /2}\) and \(\left (\frac {1+x}{1-x}\right )^{-\sigma /2}\). By linearity, any linear combination of these is also a solution and we obtain the general solution (\(\dagger \)).

It remains to determine when all these solutions are in \(L^2(0,1)\) (considering quasi-regularity of the right boundary point). We have that \(\left (\frac {1+x}{1-x}\right )^{-\sigma /2}=\left (\frac {1-x}{1+x}\right )^{\sigma /2}\) is continous on \([0,1]\) and therefore in \(L^2(0,1)\), we therefore only have to consider \(\left (\frac {1+x}{1-x}\right )^{\sigma /2}\). The numerator is bounded from above and below on \([0,1]\), so it suffices to consider \((1-x)^{-\sigma /2}\). We have

\[ \int _0^1 \left |(1-x)^{-\sigma /2}\right |^2\,dx =\int _0^1 (1-x)^{-\sigma }\,dx =\int _0^1 y^{-\sigma }\,dy, \]

which is finite if and only if \(\sigma <1\). Therefore if \(\sigma <1\) both boundary points are quasi-regular and if \(\sigma \geq 1\) both boundary points are not quasi-regular. □

Solution. The right boundary point is regular (since the functions \(w,p,q\) satisfy the smoothness assumptions at that point and \(w(\b )>0\) and \(p(\b )>0\)) and therefore quasi-regular.

The left boundary point is not regular, so we have to investigate further. The equation to solve (\(Du=0\)) is

\[ \left (x^2u'\right )'-\sigma (\sigma +1)u=0. \]

We make the Ansatz \(u(x)=x^r\). Then \(u'=rx^{r-1}\) so that \(x^2u'=rx^{r+1}\), which gives \((x^2u')'=r(r+1)x^r\). Substituting this in the differential equation gives

\[ r(r+1)x^r-\sigma (\sigma +1)x^r=0. \]

We conclude that \(r(r+1)=\sigma (\sigma +1)\), so that \(r=\sigma \) or \(r=-(1+\sigma )\) and we obtain the general solution \(u(x)=C_1x^\sigma +C_2x^{-(\sigma +1)}\).

We choose \(\delta =b/2\). We have that \(x^\sigma \) is continuous and therefore belongs to \(L^2(0,b/2;x^2)\) (as the weight \(x^2\) is also continuous and the interval is bounded). We have

\[ \int _0^{b/2}|x^{-(\sigma +1)}|^2\,x^2\,dx=\int _0^{b/2} x^{-2\sigma }\,dx, \]

and from standard integrals we know that this is finite if and only if \(-2\sigma >-1\), i.e. \(\sigma <\frac {1}{2}\). Therefore we obtain quasi-regularity of the left boundary point if and only if \(\sigma <\frac {1}{2}\). □

Solution. By symmetry \(-1\) is quasi-regular if and only if \(1\) is quasi-regular, so we only consider \(1\).

The equation to solve (\(Du=0\)) is

\[ \left (\sqrt {1-x^2}\,u'\right )'=0. \]

Integrating this gives

\[ \sqrt {1-x^2}\,u'=C_1, \]

where \(C_1\) is a constant. This gives

\[ u'(x)=\frac {C_1}{\sqrt {1-x^2}}, \]

and integrating this gives

\[ u(x)=C_1\arcsin (x)+C_2, \]

where \(C_2\) is a constant. The solution space is \(L^2\left (-1,1;\frac {1}{\sqrt {1-x^2}}\right )\), so for the boundary point \(1\) we consider \(L^2\left (0,1;\frac {1}{\sqrt {1-x^2}}\right )\). We have that the constant function 1 is in this solution space since

\[ \int _0^1 |1|^2\,\frac {1}{\sqrt {1-x^2}}\,dx=\left [\arcsin (x)\right ]_{x=0}^1=\frac {\pi }{2}. \]

We also have that \(\arcsin \) is in the solution space since

\[ \int _0^1 \arcsin ^2(x)\,\frac {1}{\sqrt {1-x^2}}\,dx =\left [\frac {1}{3}\arcsin ^3(x)\right ]_{x=0}^1 =\frac {\pi ^3}{24}. \]

Hence (using symmetry) both boundary points are quasi-regular. □

Solution. The equation to solve (\(Du=0\)) is

\[ \left (x\e ^{-x}\,u'\right )'=0. \]

Integrating this gives

\[ x\e ^{-x}\,u'=C_1, \]

where \(C_1\) is a constant. This gives

\[ u'(x)=\frac {C_1}{x}\e ^x, \]

and integrating this gives

\[ u(x)=C_1F(x)+C_2, \]

where \(C_2\) is a constant and

\[ F(x):=\int _1^x \frac {\e ^t}{t}\,dt. \]

The solution space is \(L^2(0,\infty ;\e ^{-x})\).

We first consider the boundary point \(\infty \) and therefore the space \(L^2(1,\infty ;\e ^{-x})\). We have \(1\in L^2(1,\infty ;\e ^{-x})\) since

\[ \int _1^\infty |1|^2\e ^{-x}\,dx<\infty . \]

We consider the behavior of \(F\) near infinity. We have by looking at the power series (noting that the terms we omit are nonnegative) for \(z\geq 0\)

\[ \e ^z\geq 1+z+\frac {1}{2}z^2. \]

From this we conclude that for \(t\geq 0\)

\[ \e ^{t/2}\geq 1+\frac {t}{2}+\frac {1}{2}\left (\frac {t}{2}\right )^2. \]

We have for all \(t\in \mR \) that

\[ 1+\frac {t}{2}+\frac {1}{2}\left (\frac {t}{2}\right )^2\geq t, \]

since

\[ 1-\frac {t}{2}+\frac {1}{2}\left (\frac {t}{2}\right )^2=\frac {1}{2}\left (\frac {t}{2}-1\right )^2+\frac {1}{2}\geq 0. \]

It follows that for \(t\geq 0\)

\[ \e ^{t/2}\geq t, \]

so that for \(t>0\)

\[ \frac {\e ^t}{t}\geq \e ^{t/2}, \]

which is the hint given (which may also be used without proof). Integrating this gives for \(x\geq 1\)

\[ F(x)=\int _1^x \frac {\e ^t}{t}\,dt\geq \int _1^x \e ^{t/2}\,dt =\left [2\e ^{t/2}\right ]_{t=1}^x =2\e ^{x/2}-2\sqrt {\e }. \]

Hence

\[ \int _1^\infty F(x)^2 \e ^{-x}\,dx\geq 4\int _1^\infty (\e ^{x/2}-\sqrt {\e })^2\e ^{-x}\,dx =4\int _1^\infty 1-2\sqrt {\e }\,\e ^{-x/2}+\e \,\e ^{-x}\,dx =\infty . \]

We conclude that \(F\notin L^2(1,\infty ;\e ^{-x})\). Hence the boundary point \(\infty \) is not quasi-regular.

We now consider the boundary point \(0\) and therefore the space \(L^2(0,1;\e ^{-x})\). We have \(1\in L^2(0,1;\e ^{-x})\) since

\[ \int _0^1 |1|^2\e ^{-x}\,dx<\infty . \]

For \(t\in [0,1]\) we have \(1\leq \e ^t\leq \e \) and therefore for \(t\in (0,1)\)

\[ \frac {1}{t}\leq \frac {\e ^t}{t}\leq \frac {\e }{t}. \]

Integrating gives for \(x\in (0,1)\)

\[ \int _{x}^1 \frac {1}{t}\,dt\leq \int _x^1 \frac {\e ^t}{t}\,dt\leq \e \int _{x}^1 \frac {1}{t}\,dt, \]

which reversing signs gives (noting that reversing the integration limits reverse signs)

\[ \int _1^x \frac {1}{t}\,dt\geq \int _1^x \frac {\e ^t}{t}\,dt\geq \e \int _1^x \frac {1}{t}\,dt. \]

This gives

\[ \ln (x)\geq F(x) \geq \e \ln (x). \]

Hence

\[ |F(x)|^2\leq \e ^2|\ln (x)|^2. \]

It follows that

\[ \int _0^1 |F(x)|^2\e ^{-x}\,dx\leq \e ^2 \int _0^1 |\ln (x)|^2\e ^{-x}\,dx. \]

Using \(\e ^{-x}\leq 1\), this gives

\[ \int _0^1 |F(x)|^2\e ^{-x}\,dx\leq \e ^2\int _0^1 |\ln (x)|^2\,dx =\e ^2\left [x\ln ^2(x)-2x\ln (x)+2x\right ]_{x=0}^1, \]

which is finite by standard limits. Therefore we have two linearly independent solutions in \(L^2(0,1;\e ^{-x})\) and therefore the boundary point \(0\) is quasi-regular. □

Solution. By symmetry, \(\infty \) is quasi-regular if and only if \(-\infty \) is quasi-regular, so we only consider \(\infty \).

The equation to solve (\(Du=0\)) is

\[ \left (\e ^{-x^2}u'\right )'=0. \]

Integrating this gives

\[ \e ^{-x^2}u'=C_1, \]

where \(C_1\) is a constant. This gives

\[ u'(x)=C_1\e ^{x^2}, \]

and integrating this gives

\[ u(x)=C_1 G(x)+C_2, \]

where \(G\) is the anti-derivative of \(\e ^{x^2}\) which satisfies \(G(0)=0\) and \(C_2\) is a constant. The solution space is \(L^2(-\infty ,\infty ;\e ^{-x^2})\), so for the boundary point \(\infty \) we consider \(L^2(0,\infty ;\e ^{-x^2})\). We have \(1\in L^2(0,\infty ;\e ^{-x^2})\) since

\[ \int _0^\infty |1|^2\e ^{-x^2}\,dx<\infty . \]

We now consider whether or not \(G\) belongs to \(L^2(0,\infty ;\e ^{-x^2})\). Noting that \(G(x)=\int _0^x \e ^{t^2}\,dt\), so that \(\e ^{-x^2}G(x)=D_+(x)\) we have

\[ \int _0^\infty |G(x)|^2\e ^{-x^2}\,dx=\int _0^\infty G(x)D_+(x)\,dx. \]

Since \(\e ^{t^2}\geq 1\), we have \(G(x)\geq x\). Using that \(D_+(x)\) behaves like \(\frac {1}{2x}\) near infinity (as given in the question), we then have that the integrand \(GD_+\) is bounded from below by a positive constant near infinity. It follows that \(\int _0^\infty G(x)D_+(x)\,dx=\infty \). Hence \(G\notin L^2(0,\infty ;\e ^{-x^2})\). It follows that \(\infty \) is not quasi-regular. By symmetry, neither boundary point is then quasi-regular. □