Selfadjoint Restrictions of Legendre Operator $-\frac{d}{dx}(1-x^{2})\frac{d}{dx}$ Problem: Let $Lf =-((1-x^{2})f')'$ be the Legendre differential operator defined on the domain $\mathcal{D}(L)$ consisting of twice absolutely continuous functions on $(-1,1)$ for which $f, Lf \in L^{2}[-1,1]$. Find all of the possible selfadjoint restrictions of $L$. Show that the restriction to functions $f \in \mathcal{D}(L)$ that are bounded near the ends of $[-1,1]$ is selfadjoint (this is the classical selfadjoint restriction for which the Legendre polynomials form a complete orthogonal basis of eigenfunctions.)
Background: There are many selfadjoint restrictions other than the classical one. In solving this problem it is useful to notice that $e=\ln(1+x)-\ln(1-x)$ satisfies $Le=0$; of course $L1=0$ also holds. In classical language, the Legendre equation is in the limit circle case at both endpoints $x=\pm 1$. There is confusion and folklore concerning how to pose this classical problem; so I thought this was a useful problem to post. I've seen questions of this type pop up just recently, and I cannot find any full treatment of this equation in regard to the possible conditions leading to valid selfadjoint problems for the Legendre operator. Perhaps this is somehow duplicated, but I cannot find a general discussion about endpoint conditions. I'll post a full solution if nobody else does, just for the record.
Outline: For each $f \in \mathcal{D}(L)$ show that the following limits exists
$$
\begin{align}
          A_{\pm}(f) & =  \lim_{x\rightarrow \pm 1}\left[(1-x^{2})f'(x)\right], \\
          B_{\pm}(f) & = \lim_{x\rightarrow\pm 1}\left[f(x)-A_{\pm}(f)\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right].
\end{align}
$$
Establish the more precise asymptotics
$$
\begin{align}
      (1-x^{2})f'(x) & = A_{\pm}(f) + o\left(\sqrt{1-x^{2}}\ln\left(\frac{1+x}{1-x}\right)\right), \\
       f(x) & = A_{\pm}(f)\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)+B_{\pm}(f)+o\left(\sqrt{1-x^{2}}\ln\left(\frac{1+x}{1-x}\right)\right).
\end{align}
$$
Use this to show that the following holds for all $f,g\in\mathcal{D}(L)$:
$$
      (Lf,g)-(f,Lg) = (B_{+}\wedge A_{+})(f,g)-(B_{-}\wedge A_{-})(f,g),
$$
where $(C\wedge D)(f,g)$ means $C(f)D(g)-C(g)D(f)$. To finish the problem, you may assume that $L^{\star}$ is the restriction of $L$ to the functions $f\in\mathcal{D}(L)$ for which $A_{\pm}(f)=0$ and $B_{\pm}(f)=0$.
 A: Let $\mathcal{D}(L)$ be as stated in the problem, meaning that $f \in \mathcal{D}(L)$ iff


*

*$f$ is continuously differentiable on $(-1,1)$,

*$f'$ is absolutely continuous on $(-1,1)$,

*$f$ and $Lf =-((1-x^{2})f')'$ are in $L^{2}(-1,1)$.


Aymptotics: The two classical solutions of $Lf=0$, both of which are in $L^{2}(-1,1)$:
$$
         e_{1}(x)=\frac{1}{2}\ln\left(\frac{1-x}{1+x}\right),\;\;\; e_{2}(x)\equiv 1.
$$
If $f \in \mathcal{D}(L)$, then $Lf=0$ iff $f = Ae_{1}+Be_{2}$ for unique constants $A$, $B$. Let $(f,g)$ denote the inner-product of $f,g \in L^{2}(-1,1)$ and let $(f,g)_{a}^{b}$ denote the inner-product in $L^{2}(a,b)$ for $-1 \le a < b \le 1$. For $f, g \in \mathcal{D}(L)$ integration by parts gives
$$
             (Lf,g)_{a}^{b}-(f,Lg)_{a}^{b} = (1-x^{2})(fg'-f'g)|_{a}^{b}.
$$
This seems innocent at first, but notice that the left side has limits as $a$, $b$ approach $-1$, $1$, which guarantees the existence of limits on the right side. For $g=e_{2}$, one has $e_{2}'=0$, which leads to
$$
         (1-x^{2})f'(x) = f'(0)-(Lf,e_{2})_{0}^{x}
$$
So the following limit exists:
$$
     A_{+}(f) = \lim_{x\rightarrow\pm 1}(1-x^{2})f'(x) 
              = f'(0)-\lim_{x\rightarrow 1}(Lf,e_{2})_{0}^{x}=f'(0)-(Lf,e_{2})_{0}^{1}.
$$
Furthermore, one has the asymptotic
$$
       |(1-x^{2})f'(x)-A_{+}(f)|
                = |(Lf,e_{2})_{x}^{1}| \le \|Lf\|_{x}^{1}\|e_{2}\|_{x}^{1}=o(\sqrt{1-x}).
$$
To obtain the second asymptotic, let $g=e_{1}$. Notice that $e_{1}'=-1/(1-x^{2})$. Therefore
$$
\begin{align}
    (Lf,e_{2})_{0}^{x}
        & =(1-x)^{2}(fe_{2}'-f'e_{2})|_{0}^{x} \\
        & =-f(x)-(1-x^{2})f'(x)e_{2}(x)-f(0).
\end{align}
$$
Therefore,
$$
\begin{align}
         f(x) & = -(Lf,e_{2})_{0}^{x}-(1-x^{2})f'(x)e_{2}(x)-f(0) \\
              & = -(Lf,e_{2})_{0}^{x}+A_{+}(f)e_{2}(x)+o(\sqrt{1-x})e_{2}(x)-f(0).
\end{align}
$$
This gives
$$
         f(x)-A_{+}(f)e_{2}(x)-B_{+}(f)  = -(Lf,e_{2})_{0}^{x}+o(\sqrt{1-x})e_{2}(x)-f(0).
$$
So the stated limit $B_{\pm}(f)=\lim_{x\rightarrow +1}(f(x)-A_{+}e_{2}(x))$ exists and
$$
       f(x)-A_{+}(f)e_{2}(x)-B_{+}(f) = o(\sqrt{1-x}e_{2})+O((Lf,e_{2})_{x}^{1})=o(\sqrt{1-x}e_{2}).
$$
Substituting back into the original identity,
$$
   (Lf,g)-(f,Lg) = (1-x^{2})(fg'-f'g)|_{-1}^{1}.
$$
Near $x=1$, one has
$$
    (1-x^{2})g'f-(1-x^{2})f'g \\
   = (A_{+}(g)+o(\sqrt{1-x}))(B_{+}(f)+A_{+}(f)e_{2}+o(\sqrt{1-x}e_{2}))
   - (A_{+}(f)+o(\sqrt{1-x}))(B_{+}(g)+A_{+}(g)e_{2}+o(\sqrt{1-x}e_{2})) \\
   = A_{+}(g)B_{+}(f)-A_{+}(f)B_{+}(g)+o(\sqrt{1-x}e_{2}^{2}).
$$
Therefore,
$$
\begin{align}
    (Lf,g)-(f,Lg) & = 
 -\left|\begin{array}{cc} A_{+}(f) & B_{+}(f) \\ A_{+}(g) & B_{+}(g)\end{array}\right|
 +\left|\begin{array}{cc} A_{-}(f) & B_{-}(f) \\ B_{-}(g) & B_{-}(g)\end{array}\right|.
\end{align}
$$
Separated Conditions: We are told to assume that the unconstrained operator $L$ has adjoint $L^{\star} = L_{0}$, where $L_{0}$ is the restriction of $L$ to the functions $f \in \mathcal{D}(L)$ for which $A_{\pm}(f)=0$ and $B_{\pm}(f)=0$. It follows that $L_{0}$ is closed, and the graph of $L_{0}$ is of finite co-dimmension in the graph of $L$, which means that $L$ is also closed. So $L_{0}^{\star}=L$ also holds.
Let $\alpha,\beta \in [0,\pi)$ and define $L_{\alpha,\beta}$ as the restriction of $L$ to the $f \in \mathcal{D}(L)$ satisfying
$$
          \cos\alpha A_{-}(f)+\sin\alpha B_{-}(f) = 0,\\
          \cos\beta A_{+}(f) + \sin\beta B_{+}(f) = 0.
$$
We show that $L_{\alpha,\beta}$ is selfadjoint.
This operator $L_{\alpha,\beta}$ is symmetric on its domain because the two evaluation determinants are $0$ for the symmetric difference $(Lf,g)-(f,Lg)$, which is proved by noticing that the associated matrices have non-trivial homogenous solutions:
$$
   \left[\begin{array}{cc} A_{-}(f) & B_{-}(f) \\ A_{-}(g) & B_{-}(g)\end{array}\right]
   \left[\begin{array}{c} \cos\alpha \\ \sin\alpha\end{array}\right]=0,\;\;\;
   \left[\begin{array}{cc} A_{+}(f) & B_{+}(f) \\ A_{+}(g) & B_{+}(g)\end{array}\right]
   \left[\begin{array}{c} \cos\beta \\ \sin\beta\end{array}\right]=0.
$$
Therefore $L_{\alpha,\beta}\preceq L_{\alpha,\beta}^{\star}$.
Suppose $(L_{\alpha,\beta}f,g)=(f,h)$ for some $g,h\in L^{2}$ and all $f \in \mathcal{D}(L_{\alpha,\beta})$. Then this is true for all $f \in \mathcal{D}(L_{0})$ which implies that $h \in \mathcal{D}(L_{0}^{\star})=\mathcal{D}(L)$ and $Lg=h$. Because $(L_{\alpha,\beta}f,g)=(f,Lg)$ for all $f \in \mathcal{D}(L_{\alpha,\beta})$ then the left and right evaluation determinants must separately vanish. Define
$$
         \Phi_{\alpha}(f) = +\cos\alpha A_{-}(f)+\sin\alpha B_{-}(f),\\
         \Psi_{\alpha}(f) = -\sin\alpha A_{-}(f)+\cos\alpha B_{-}(f).
$$
Using muliplicative properties of determinant and the fact that $\Phi_{\alpha}(f)=0$ gives
$$
    0=  \left|\begin{array}{cc}
               A_{-}(g) & B_{-}(g) \\
               A_{-}(f) & B_{-}(f)
        \end{array}\right|
      = \left|\begin{array}{cc}
               \Phi_{\alpha}(g) & \Psi_{\alpha}(g) \\
               \Phi_{\alpha}(f) & \Psi_{\alpha}(f)
        \end{array}\right| = \Phi_{\alpha}(g)\Psi_{\alpha}(f).
$$
This is true for all $f \in \mathcal{D}(L_{\alpha,\beta})$ which implies $\Phi_{\alpha}(g)=0$ for all $g \in \mathcal{D}(L_{\alpha,\beta}^{\star})$. The analogous result holds at the right endpoint as well. So  $L_{\alpha,\beta}^{\star} \preceq L_{\alpha,\beta}$, which finishes the proof that $L_{\alpha,\beta}^{\star}=L_{\alpha,\beta}$.
Classical Conditions: The classical Legendre operator is $L_{0,0}$ which is equivalent to requiring $A_{\pm}(f)=0$ for $f \in \mathcal{D}(L)$. Or, because of the asymptotics, this is also equivalent to requiring the functions in the domain are bounded near $x=\pm 1$. Automatically such functions will have limits at $x=\pm 1$ because of the asymptotics derived for a general $f \in \mathcal{D}(L)$. So boundedness is equivalent:
$$
    \mathcal{D}(L_{0,0})= \{ f \in \mathcal{D}(L) : f \mbox{ is bounded on } (-1,1) \}.
$$
This is why only eigenfunctions which are bounded near $\pm 1$ are considered valid. The only $f \in\mathcal{D}(L_{0,0})$ for which $Lf = \lambda f$ are the Legendre polynomials. The eigenvalues are $\lambda =n(n+1)$ for $n=0,1,2,3,\cdots$, and this set of eigenfunctions is a complete orthogonal basis of $L^{2}(-1,1)$. All kinds of other separted conditions are possible, but none of the others appears to be appropriate for applications to Physics and Engineering. The non-physical selfadjoint formulations do lead to orthogonal types of expansions.
Periodic Condition: I don't believe there is any reason that periodic types of conditions do not lead to selfadjoint restrictions. Conditions such as
$$
               A_{+}(f) = A_{-}(f),\;\; B_{+}(f) = B_{-}(f)
$$
appear to be okay. However, $f(-1)=f(1)$, $f'(-1)=f'(1)$ does not make sense because $f  \in \mathcal{D}(L)$ does not necessarily have limiting values for $f$ or for $f'$. And a single condition is not enough to determine a selfadjoint problem.
