Solution. We have

\[ W(u,v;p)=(1-x^2)\left [u'v-uv'\right ], \]

which with \(v=1\) gives

\[ W(u,1;p)=(1-x^2)u'(x). \]

The boundary conditions therefore are (note that we have to take limits since the boundary points are singular):

\[ \lim _{x\downarrow -1}(1-x^2)u'(x)=0,\qquad \lim _{x\uparrow 1}(1-x^2)u'(x)=0. \]

If \(u'\) is continuous at the boundary points, then since \(1-x^2\to 0\) at the boundary points, we have by algebra of limits that the boundary conditions are indeed satisfied. □

Solution. For \(\sigma \geq 1\) (so in particular for all \(\sigma \in \mN \)) we have by Problem C.9 that both boundary points are not quasi-regular. It follows from Remark 9.4 that there are no admissible boundary conditions. □

Solution. (i) The Wronskian for this problem is

\[ W(u,v;p)(x)=x^2\left (u'(x)v(x)-u(x)v'(x)\right ). \]

To obtain from \(W(u,v;p)(\b )=0\) the boundary condition \(u(\b )=0\), we choose a \(v\) with (\(Dv=0\) and) \(v(\b )=0\) and then check that this indeed works. The general solution of \(Dv=0\) is \(v(x)=C_1x^\sigma +C_2x^{-(\sigma +1)}\). We pick \(C_1=1\) and to obtain \(v(\b )=0\) we then choose \(C_2=-\b ^{2\sigma +1}\). Therefore

\[ v(x)=x^\sigma -\b ^{2\sigma +1}x^{-\sigma -1}. \]

We check that this indeed works. We have

\[ v'(x)=\sigma x^{\sigma -1}+\b ^{2\sigma +1}(\sigma +1)x^{-\sigma -2}, \]

so that

\[ v'(\b )=\sigma \b ^{\sigma -1}+\b ^{\sigma -1}(\sigma +1)=(2\sigma +1)\b ^{\sigma -1}. \]

Therefore

\[ W(u,v;p)(\b )=-\b ^2u(\b )(2\sigma +1)\b ^{\sigma -1}, \]

which is indeed zero if and only if \(u(\b )=0\).

(ii) From Problem C.10 we have that \(0\) is quasi-regular if and only if \(\sigma <\frac {1}{2}\). Therefore, we need a boundary condition at \(0\) if and only if \(\sigma <\frac {1}{2}\).

(iii) By (ii), we restrict to \(\sigma <\frac {1}{2}\). We obtain a suitable boundary condition from any non-zero solution of \(Dv=0\). We choose \(v(x)=x^\sigma \) (other choices are possible). The Wronskian then is

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

The corresponding boundary condition therefore is

\[ \lim _{x\downarrow 0} u'(x)x^{2+\sigma }-\sigma u(x)x^{1+\sigma }=0. \]

□

Solution. The Wronskian for this problem is

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

With \(v=1\), this gives

\[ W(u,1;p)(x)=\sqrt {1-x^2}\,u'(x). \]

This gives the boundary conditions (we need limits because the boundary points are singular)

\[ \lim _{x\downarrow -1}\sqrt {1-x^2}\,u'(x)=0,\qquad \lim _{x\uparrow 1}\sqrt {1-x^2}\,u'(x)=0. \]

If \(u\) is continuously differentiable on \([-1,1]\) (so that \(\lim _{x\downarrow -1}u'(x)\) and \(\lim _{x\uparrow 1}u'(x)\) exist), then since \(\lim _{x\to \pm 1}\sqrt {1-x^2}=0\) we have by algebra of limits that the boundary conditions are satisfied. □

Solution. The Wronskian for this problem is

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

With \(v=\arcsin \), this gives (using that \(v'(x)=\frac {1}{\sqrt {1-x^2}}\))

\[ W(u,v;p)(x)=\sqrt {1-x^2}\,u'(x)\arcsin (x)-u(x). \]

This gives the boundary conditions (we need limits because the boundary points are singular)

\[ \lim _{x\downarrow -1}\sqrt {1-x^2}\,u'(x)\frac {\pi }{2}+u(x)=0,\qquad \lim _{x\uparrow 1}\sqrt {1-x^2}\,u'(x)\frac {\pi }{2}-u(x)=0. \]

□

Solution. The Wronskian for this problem is

\[ W(u,v;x\e ^{-x})(x)=x\e ^{-x}\left (u'(x)v(x)-u(x)v'(x)\right ), \]

which with \(v=1\), gives

\[ W(u,1;x\e ^{-x})(x)=x\e ^{-x}\,u'(x). \]

The boundary condition therefore is (note that we have to take a limit since the boundary point is singular):

\[ \lim _{x\downarrow 0}x\e ^{-x}\,u'(x)=0. \]

Since \(\lim _{x\to 0}\e ^{-x}=1\), we can simplify this to

\[ \lim _{x\downarrow 0}x\,u'(x)=0. \]

If \(u'\) is continuous at \(0\), then since \(x\downarrow 0\), we have by algebra of limits that the boundary condition is indeed satisfied. □

Solution. We have by Problem C.13 that both boundary points are not quasi-regular. It follows from Remark 9.4 that there are no admissible boundary conditions. □