Let us consider the inner-product in the Hilbert space \(L_{2} ( -1,1 ) \) as follows:
$$ \langle f,g \rangle=\frac{h_{1}}{p_{1}} \int_{-1}^{0}f(x)\overline{g(x)}\,dx+ \frac{h_{2}}{p_{2}} \int_{0}^{1}f(x)\overline{g(x)}\,dx, $$
(7)
where \(F:=f(x)\), \(G:=g(x)\in L_{2} ( -1,1 ) \). In this Hilbert space we define the operator A with domain
$$D(A):=\left \{\textstyle\begin{array}{l} f:f(x) \mbox{ and } D^{\alpha}f(x), {}^{c}D^{\alpha}f(x) \mbox{ are absolutely continuous } \\ \mbox{on } [-1,0)\cup(0,1], \mbox{ and } f(0\pm),D^{\alpha}f(0\pm),I^{1-\alpha}f(0\pm) \mbox{ have} \\ \mbox{finite limits}, L_{i}f=0,i=1,2,3,4.\end{array}\displaystyle \right \} $$
and action law
$$ Af:=\mathfrak{L}_{\alpha,x}f. $$
(8)
Thus, problem (2)-(6) can be written in the operator form as
Note that by eigenvalues and eigenfunctions of problem (2)-(6) we mean eigenvalues and eigenelements of the operator A, respectively.
Theorem 1
The linear operator
A
is symmetric.
Proof
For each \(f,g\in \operatorname{Dom}(A)\), using (7) we write
$$\begin{aligned} \langle Af,g\rangle =&\frac{h_{1}}{p_{1}} \int_{-1}^{0}Af(x)\overline{g(x)}\,dx+ \frac{h_{2}}{p_{2}} \int_{0}^{1}Af(x)\overline{g(x)}\,dx \\ =&\frac{h_{1}}{p_{1}} \int_{-1}^{0} \bigl( {}^{c}D_{0^{-}}^{\alpha }p_{1}D_{-1^{+}}^{\alpha}f(x) \bigr) \overline{g(x)}\,dx+\frac{h_{1}}{p_{1}}\int_{-1}^{0}q(x)f(x)\overline{g(x)}\,dx \\ &{}+\frac{h_{2}}{p_{2}} \int_{0}^{1} \bigl( {}^{c}D_{1^{-}}^{\alpha }p_{2}D_{0^{+}}^{\alpha}f(x) \bigr) \overline{g(x)}\,dx+\frac{h_{2}}{p_{2}}\int_{0}^{1}q(x)f(x)\overline{g(x)}\,dx. \end{aligned}$$
By applying Property 3, we get
$$\begin{aligned} \langle Af,g\rangle =&h_{1}\biggl\{ \int_{-1}^{0}f(x) {}^{c}D_{0^{-}}^{\alpha }D_{-1^{+}}^{\alpha } \overline{g(x)}\,dx+ D_{-1^{+}}^{\alpha}\overline{g(x)}I_{-1^{+}}^{1-\alpha}f(x)\big\vert _{-1}^{0}- D_{-1^{+}}^{\alpha }f(x)I_{-1^{+}}^{1-\alpha}\overline{g(x)} \big\vert _{-1}^{0}\biggr\} \\ &{}+h_{2}\biggl\{ \int_{0}^{1}f(x) {}^{c}D_{1^{-}}^{\alpha}D_{0^{+}}^{\alpha}\overline{g(x)}\,dx+ D_{0^{+}}^{\alpha}\overline{g(x)}I_{0^{+}}^{1-\alpha}f(x) \big\vert _{0}^{1}- D_{0^{+}}^{\alpha }f(x)I_{0^{+}}^{1-\alpha}\overline{g(x)} \big\vert _{0}^{1}\biggr\} \\ &{}+\frac{h_{1}}{p_{1}} \int_{-1}^{0}q(x)f(x)\overline{g(x)}\,dx+ \frac{h_{2}}{p_{2}} \int_{0}^{1}q(x)f(x)\overline{g(x)}\,dx \\ =&\langle f,Ag\rangle+h_{1}\bigl\{ D_{-1^{+}}^{\alpha} \overline{g(x)}I_{-1^{+}}^{1-\alpha}f(x) \big\vert _{-1}^{0}- D_{-1^{+}}^{\alpha }f ( x ) I_{-1^{+}}^{1-\alpha}\overline{g ( x ) }\big\vert _{-1}^{0}\bigr\} \\ &{}+h_{2}\bigl\{ D_{0^{+}}^{\alpha } \overline{g(x)}I_{0^{+}}^{1-\alpha }f(x) \big\vert _{0}^{1}- D_{0^{+}}^{\alpha}f(x)I_{0^{+}}^{1-\alpha}\overline{g(x)}\big\vert _{0}^{1}\bigr\} . \end{aligned}$$
By considering the fractional transmission conditions (5)-(6), we have
$$\langle Af,g\rangle=\langle f,Ag\rangle $$
that the operator A is symmetric. □
Corollary 1
All eigenvalues of problem (2)-(6) are real.
Corollary 2
Let
\(\lambda_{1}\)
and
\(\lambda_{2}\)
be two different eigenvalues of problem (2)-(6). Then the corresponding eigenfunctions
f
and
g
of this problem satisfy the following equality:
$$\frac{h_{1}}{p_{1}} \int_{-1}^{0}Af(x)\overline{g(x)}\,dx+ \frac{h_{2}}{p_{2}}\int_{0}^{1}Af(x)\overline{g(x)}\,dx=0. $$
As a consequence, the eigenfunctions of problem (2)-(6) corresponding to the different eigenvalues are orthogonal to the inner product (7) in the Hilbert space \(L_{2} ( -1,1 ) \).
Naturally, we can now assume that all eigenfunctions of problem (2)-(6) are real-valued.
Lemma 2
The equivalent integral form of equation (2) with fractional conditions (5)-(6) is given as
$$ u(x)=u_{0}(x)+\frac{1}{p_{2}\Gamma(2\alpha)} \int_{0}^{x}\bigl[N_{u}(y)+(-1)^{1-\alpha}(x-y)^{2\alpha-1} \bigl(\lambda+q(y)\bigr)u(y)\bigr]\,dy, $$
(9)
where
\(u_{0}(x)=\frac{x^{\alpha-1}}{\Gamma(\alpha )}(-h_{1}I_{-1^{+}}^{1-\alpha}u(-0))+I_{0^{+}}^{\alpha}(-\frac{1}{h_{2}}D_{-1^{+}}^{\alpha}u(-0))\).
Proof
Let us consider equation (2)
$${}^{c}D_{1^{-}}^{\alpha}p_{2}D_{0^{+}}^{\alpha}u(x)+ \bigl(\lambda+q(x)\bigr)u(x)=0,\quad x\in( 0,1 ], $$
integral operators \(I_{0^{+}}^{\alpha}\) acting on this equation and by Lemma 1, we obtain
$$ I_{0^{+}}^{\alpha}\bigl( {}^{c}D_{1^{-}}^{\alpha }p_{2}D_{0^{+}}^{\alpha}u(x) \bigr)+I_{0^{+}}^{\alpha}\bigl(\lambda+q(x)\bigr)u(x)=0 $$
(10)
and
$$ p_{2}D_{0^{+}}^{\alpha}u(x)=I_{0^{+}}^{\alpha }N_{u}(x)+p_{2}D_{0^{+}}^{\alpha }u(+0)+(-1)^{1-\alpha}I_{0^{+}}^{\alpha} \bigl(\lambda+q(x)\bigr)u(x). $$
(11)
Applying \(I_{0^{+}}^{\alpha}\) on both sides of (11) and using conditions (5)-(6), we find
$$\begin{aligned} u(x) =&\frac{x^{\alpha-1}}{\Gamma(\alpha )}\bigl(-h_{1}I_{-1^{+}}^{1-\alpha }u(-0) \bigr)+I_{0^{+}}^{\alpha}\biggl(-\frac{1}{h_{2}}D_{-1^{+}}^{\alpha}u(-0)\biggr)+\frac{1}{p_{2}}I_{0^{+}}^{2\alpha}N_{u}(x) \\ &{}+\frac{(-1)^{1-\alpha}}{p_{2}}I_{0^{+}}^{2\alpha}\bigl(\lambda+q(y) \bigr)u(x). \end{aligned}$$
Then we reach
$$ u(x)=u_{0}(x)+\frac{1}{p_{2}}I_{0^{+}}^{2\alpha} \bigl[N_{u}(x)+(-1)^{1-\alpha}\bigl(\lambda+q(x)\bigr)u(x)\bigr], $$
(12)
which completes the proof. □
We next define \(u_{m}(x,\lambda)\) to construct the successive approximations
$$u_{m}(x,\lambda)=u_{0}(x,\lambda)+\frac{1}{p_{2}\Gamma(2\alpha)}\int_{0}^{x}(x-y)^{2\alpha-1} \bigl[N_{u_{m-1}}(y)+(-1)^{1-\alpha}\bigl(\lambda+q(y) \bigr)u_{m-1}(y)\bigr]\,dy. $$
Remark 1
The corresponding classical Sturm-Liouville problem with integer orders is covered by the case \(\alpha=1\).
Lemma 3
Let
\(Q:=\max_{x\in ( 0,1 ] } \vert q ( x ) \vert \), \(P_{R}:=\max_{\lvert\lambda\rvert\leq \mathbb{R} }P ( \lambda ) \)
and
\(P ( \lambda ) :=\max_{x\in ( 0,1 ] } \vert u_{0} ( x,\lambda ) \vert \), \(k_{\alpha}:=1/[(2-\alpha)\Gamma (1-\alpha )] \). Then the following estimate
$$ \bigl\lVert u_{m}(x,\lambda)-u_{m-1}(x, \lambda)\bigr\rVert \leq P_{R}\biggl\{ \frac{\lvert\lambda\rvert +2k_{\alpha}+Q}{p_{2}\Gamma (2\alpha+1 )}\biggr\} ^{m} $$
(13)
holds for all
m.
Proof
Let us apply the mathematical induction for m. In what follows, for convenience we shall use the notation \(K=1/\Gamma (2\alpha+1 ) \).
For \(m=1\), we have
$$\begin{aligned} \bigl\lVert u_{1}(x,\lambda)-u_{0}(x,\lambda)\bigr\rVert =&\biggl\lVert \frac{1}{p_{2}}I_{0^{+}}^{2\alpha} \bigl(N_{u_{0}}(x)+ ( -1 ) ^{1-\alpha}\bigl(\lambda+q(x) \bigr)u_{0}(x,\lambda)\bigr)\biggr\rVert . \end{aligned}$$
By using Lemma 2.1 in [2], we have
$$\begin{aligned} \bigl\lVert u_{1}(x,\lambda)-u_{0}(x,\lambda) \bigr\rVert \leq&\frac{1}{p_{2}}K\bigl\lVert N_{u_{0}}(x)+ ( -1 ) ^{1-\alpha}\bigl(\lambda+q(x)\bigr)u_{0}(x,\lambda)\bigr\rVert \\ \leq& \frac{1}{p_{2}}K\bigl[\bigl\lVert N_{u_{0}}(x)\bigr\rVert + \bigl\lVert \bigl(\lambda+q(x)\bigr) u_{0}(x,\lambda)\bigr\rVert \bigr]. \end{aligned}$$
By using Corollary 2.3 in [2], we have
$$\begin{aligned} \bigl\lVert u_{1}(x,\lambda)-u_{0}(x,\lambda) \bigr\rVert \leq& \frac{K}{p_{2}}\bigl[2k_{\alpha}\bigl\lVert u_{0}(x,\lambda)\bigr\rVert +\bigl(\lvert\lambda\rvert+Q\bigr)\bigl\lVert u_{0}(x,\lambda)\bigr\rVert \bigr] \\ \leq& \frac{KP_{R}}{p_{2}}\bigl(2k_{\alpha}+ \lvert\lambda\rvert +Q\bigr). \end{aligned}$$
Suppose that (13) holds for \(m-1\), i.e.,
$$\bigl\lVert u_{m-1}(x,\lambda)-u_{m-2}(x,\lambda)\bigr\rVert \leq P_{R}\biggl\{ \frac{K}{p_{2}}\bigl(\lvert\lambda \rvert+2k_{\alpha}+Q\bigr)\biggr\} ^{m-1}. $$
Then we have
$$\begin{aligned}& \bigl\lVert u_{m}(x,\lambda)-u_{m-1}(x,\lambda)\bigr\rVert \\& \quad = \biggl\lVert \frac{1}{p_{2}}I_{0^{+}}^{2\alpha} \bigl[N_{u_{m-1}-u_{m-2}}(x)+ ( -1 ) ^{\alpha}\bigl(\lambda+q(x)\bigr) \bigl(u_{m-1}(x,\lambda)-u_{m-2}(x,\lambda)\bigr)\bigr] \biggr\rVert \\& \quad \leq \frac{K}{p_{2}}\bigl[\bigl\lVert N_{u_{m-1}-u_{m-2}}(x)\bigr\rVert +\bigl\lVert \bigl(\lambda+q(x)\bigr) \bigl(u_{m-1}(x,\lambda )-u_{m-2}(x,\lambda)\bigr)\bigr\rVert \bigr] \\& \quad \leq \frac{K}{p_{2}}\bigl[2k_{\alpha}\bigl\lVert \bigl(u_{m-1}(x,\lambda)-u_{m-2}(x,\lambda)\bigr) \bigr\rVert + \bigl(\lvert\lambda\rvert+Q\bigr)\bigl\lVert \bigl(u_{m-1}(x,\lambda )-u_{m-2}(x,\lambda)\bigr) \bigr\rVert \bigr] \\& \begin{gathered} \quad = \frac{K}{p_{2}}\bigl(2k_{\alpha}+\lvert\lambda\rvert+Q\bigr)\bigl\lVert \bigl(u_{m-1}(x,\lambda)-u_{m-2}(x,\lambda)\bigr) \bigr\rVert \\ \quad \leq P_{R}\biggl\{ \frac{K}{p_{2}}\bigl(\lvert\lambda\rvert +2k_{\alpha}+Q\bigr)\biggr\} ^{m}. \end{gathered} \end{aligned}$$
The proof is completed. □
Lemma 4
The following IVP
$$\begin{aligned}& {}^{c}D_{0^{-}}^{\alpha}p_{1}D_{-1^{+}}^{\alpha}u ( x ) + \bigl( q ( x ) +\lambda\bigr) u ( x ) =0,\quad x\in[ -1,0 ], \end{aligned}$$
(14)
$$\begin{aligned}& I_{-1^{+}}^{1-\alpha}u ( -1 ) =c_{2}, \end{aligned}$$
(15)
$$\begin{aligned}& D_{-1^{+}}^{\alpha}u ( -1 ) =-c_{1} \end{aligned}$$
(16)
has a unique solution on
\([ -1,0 ]\)
provided that
$$ \frac{K}{p_{1}}\bigl(\lvert\lambda\rvert+2k_{\alpha}+Q \bigr)< 1. $$
(17)
Proof
If we use a similar way in Lemma 2, we get a corresponding integral equation of the problem as follows:
$$ u(x)=u_{0}(x)+\frac{1}{p_{1}}I_{-1^{+}}^{2\alpha} \bigl[N_{u}(x)+(-1)^{1-\alpha}\bigl(\lambda+q(x)\bigr)u(x)\bigr], $$
(18)
where \(u_{0}(x)=\frac{(x+1)^{\alpha}}{\Gamma(1+\alpha)}(-c_{1})+\frac {(x+1)^{\alpha-1}}{\Gamma(\alpha)}c_{2}\).
Let us construct the integral equation by
where the mapping T is defined as
$$Tf=u_{0}+\frac{1}{p_{1}}I_{-1^{+}}^{2\alpha} \bigl[N_{f}+(-1)^{1-\alpha}(\lambda+q)f\bigr], $$
then we have
$$\lVert Tf-Tg\rVert=\biggl\lVert \frac{1}{p_{1}}I_{-1^{+}}^{2\alpha} \bigl[(N_{f}-N_{g})+(-1)^{1-\alpha}(\lambda+q) (f-g) \bigr]\biggr\rVert . $$
By applying Lemma 2.1 in [2], we get
$$\begin{aligned} \lVert Tf-Tg\rVert \leq&\frac{K}{p_{1}}\bigl\lVert (N_{f}-N_{g})+(-1)^{1-\alpha}(\lambda+q) (f-g)\bigr\rVert \\ \leq&\frac{K}{p_{1}}\bigl\lVert (N_{f}-N_{g})\bigr\rVert +\bigl\lVert (\lambda+q) (f-g)\bigr\rVert . \end{aligned}$$
(20)
By relation (1) we have
$$N_{f}-N_{g}= {}^{c}D_{-1^{+}}^{\alpha}(f-g)+(-1)^{1-\alpha} {}^{c}D_{1^{-}}^{\alpha }(f-g), $$
then
$$\begin{aligned} \lVert N_{f}-N_{g}\rVert \leq&k_{\alpha}\lVert f-g \rVert+k_{\alpha}\lVert f-g\rVert \\ =&2k_{\alpha}\lVert f-g\rVert, \end{aligned}$$
where we have used Corollary 2.3 in [2]. If we substitute the last inequality into (20), we find
$$\lVert Tf-Tg\rVert\leq\frac{K}{p_{1}}\bigl(\lvert\lambda\rvert +2k_{\alpha}+Q\bigr)\lVert f-g\rVert. $$
By condition (17), the mapping T is a contraction on the space \(\langle C[-1,0],\lVert\cdot\rVert\rangle \). Consequently, there exists a unique solution of equation (19). The proof is complete. □
Theorem 2
For any
\(\lambda\in \mathbb{C} \)
satisfying
\(Kp_{i}^{-1}(\lvert\lambda\rvert+2k_{\alpha}+Q)<1\) (\(i=1,2\)), the differential equation (2) has a unique solution which satisfies fractional boundary condition (3) and fractional transmission conditions (5)-(6).
Proof
Consider the following problem for each \(\lambda\in \mathbb{C}\):
$$\begin{aligned}& \mathfrak{L}_{\alpha,x}u ( x ) +\lambda u ( x ) =0,\quad x\in[-1,0 ), \end{aligned}$$
(21)
$$\begin{aligned}& {}^{c}D_{0^{-}}^{\alpha}p_{1}D_{-1^{+}}^{\alpha}u ( x ) + \bigl( q ( x ) +\lambda\bigr) u ( x ) =0,\quad x\in[ -1,0 ), \end{aligned}$$
(22)
$$\begin{aligned}& I_{-1^{+}}^{1-\alpha}u ( -1 ) =c_{2}, \end{aligned}$$
(23)
$$\begin{aligned}& D_{-1^{+}}^{\alpha}u ( -1 ) =-c_{1}. \end{aligned}$$
(24)
By considering Lemma 4, the initial value problem has a unique solution \(\phi_{1} ( x,\lambda ) \).
Next, take into account the differential equation
$$\begin{aligned}& \mathfrak{L}_{\alpha,x}u ( x ) +\lambda u ( x ) =0,\quad x\in(0,1 ], \end{aligned}$$
(25)
$$\begin{aligned}& {}^{c}D_{1^{-}}^{\alpha}p_{2}D_{0^{+}}^{\alpha}u ( x ) + \bigl( q ( x ) +\lambda\bigr) u ( x ) =0,\quad x\in( 0,1 ], \end{aligned}$$
(26)
$$\begin{aligned}& I_{0^{+}}^{1-\alpha}u ( +0 ) =-h_{1}I_{-1^{+}}^{1-\alpha } \phi_{1} ( -0 ), \end{aligned}$$
(27)
$$\begin{aligned}& D_{0^{+}}^{\alpha}u ( +0 ) =-\frac{1}{h_{2}}D_{-1^{+}}^{\alpha} \phi_{1} ( -0 ). \end{aligned}$$
(28)
We establish the sequence \(\{ u_{n} ( x,\lambda ) \} \) for \(x\in ( 0,1 ] \) and \(n=1,2,\ldots \) such that
$$ \begin{aligned}[b] u_{n} ( x,\lambda) &=u_{0}(x,\lambda)\\ &\quad {}+\frac{1}{p_{2}\Gamma(2\alpha)} \int_{0}^{x}(x-y)^{2\alpha-1} \bigl[N_{u_{n-1}}(y)+(-1)^{1-\alpha}\bigl(\lambda+q(y)\bigr) \bigr]u_{n-1}(y,\lambda)\,dy,\end{aligned} $$
(29)
where
$$ u_{0}(x,\lambda)=I_{-1^{+}}^{\alpha}\bigl(kD_{-1^{+}}^{\alpha} \phi_{1}(-0,\lambda)\bigr),\quad x\in( 0,1 ]. $$
(30)
Obviously, each of the functions \(u_{n} ( x,\lambda ) \) is an entire function of λ for each \(x\in (0,1 ] \).
Now let us consider the series
$$ u^{\ast} ( x,\lambda) =\lim_{n\rightarrow\infty } \bigl(u_{n} ( x,\lambda) -u_{0}(x,\lambda) \bigr) =\sum _{j=1}^{\infty} \bigl( u_{j} ( x, \lambda) -u_{j-1}(x,\lambda) \bigr). $$
(31)
According to estimate (13) in Lemma 3, for \(0< x\leq1\), the absolute value of its terms is less than the corresponding terms of the convergent numeric series
$$P_{R}\sum_{j=1}^{\infty}\biggl\{ \frac{K}{p_{2}}\bigl(\lvert\lambda\rvert+2k_{\alpha}+Q\bigr)\biggr\} ^{j}. $$
Hence, series (31) converges uniformly. Obviously, each term \(( u_{j} ( x,\lambda ) -u_{j-1}(x,\lambda) ) \) of series (31) is continuous on \(x\in(0,1]\). Therefore, the sum of series (31) is continuous on \(x\in(0,1]\) and
$$\phi_{2} ( x,\lambda) =\lim_{n\rightarrow\infty }u_{n} ( x,\lambda) =u_{0}(x,\lambda)+u^{\ast} ( x,\lambda) $$
is continuous on \(x\in(0,1]\).
The uniform convergency of the sequence \(u_{n} ( x,\lambda ) \) allows us to take \(n\rightarrow\infty\) in the relation (29). This gives equations (12) showing that \(\phi_{2} ( x,\lambda ) \), the limit function of the process defined by (30) and (29), is the solution of (12). Furthermore, it is trivial that \(\phi _{2} ( x,\lambda ) \) satisfies the initial conditions (27)-(28). Finally, the function \(\phi ( x,\lambda ) \) given by
$$ \phi( x,\lambda) = \textstyle\begin{cases} \phi_{1} ( x,\lambda ), & x\in [ -1,0 ), \\ \phi_{2} ( x,\lambda ), & x\in ( 0,1 ]\end{cases} $$
(32)
satisfies the differential equation (2), fractional boundary condition (3) and fractional transmission conditions (5) and (6). □
In a similar manner, we can prove the following theorem.
Theorem 3
For any
\(\lambda\in \mathbb{C} \), the differential equation
$$\mathfrak{L}_{\alpha,x}u ( x ) +\lambda u ( x ) =0,\quad x\in[ -1,0 ) \cup ( 0,1 ] $$
has a unique solution
$$\chi( x,\lambda) = \textstyle\begin{cases} \chi_{1} ( x,\lambda ), & x\in [ -1,0 ), \\ \chi_{2} ( x,\lambda ), & x\in ( 0,1 ]\end{cases} $$
satisfying fractional boundary condition (4) and fractional transmission conditions (5) and (6) for each
\(x\in [ -1,0 ) \cup ( 0,1 ] \).
Remark 2
If λ is not eigenvalue, then \(\phi_{1}\) and \(\chi_{1}\) are linearly independent solutions of equation (2) in the interval \([ -1,0 )\). Similarly, \(\phi_{2}\) and \(\chi_{2}\) are linearly independent solutions of equation (2) in the interval \((0,1 ]\). Then it is obvious that the four functions \(\tilde{u_{1}}\), \(\tilde {u_{2}}\), \(\tilde{u_{3}}\), \(\tilde{u_{4}}\) which are defined by
$$\begin{gathered} \tilde{u_{1}} = \textstyle\begin{cases} \phi_{1} ( x,\lambda ), & x\in [ -1,0 ), \\ 0, & x\in ( 0,1 ],\end{cases}\displaystyle \qquad \tilde{u_{2}} = \textstyle\begin{cases} \chi_{1} ( x,\lambda ), & x\in [ -1,0 ), \\ 0, & x\in ( 0,1 ],\end{cases}\displaystyle \\ \tilde{u_{3}} = \textstyle\begin{cases} 0, & x\in [ -1,0 ), \\ \phi_{2} ( x,\lambda ), & x\in ( 0,1 ],\end{cases}\displaystyle \qquad\tilde{u_{4}} = \textstyle\begin{cases} 0, & x\in [ -1,0 ), \\ \chi_{2} ( x,\lambda ), & x\in ( 0,1 ]\end{cases}\displaystyle \end{gathered} $$
are linearly independent solutions of equation (2) in whole \([ -1,0 ) \cup ( 0,1 ]\).
To prove this fact, suppose if possible that \(\lambda=\lambda_{0} \) is not an eigenvalue but the corresponding solutions \(\phi_{1} ( x,\lambda_{0} ) \) and \(\chi_{1} ( x,\lambda_{0} ) \) are linearly dependent. Then there is a constant \(\alpha_{0}\neq0 \) such that
$$\chi_{1} ( x,\lambda_{0} )= \alpha_{0} \phi _{1} ( x,\lambda_{0} ). $$
From this equality obviously follows that the solution \(\chi_{1} (x,\lambda_{0} ) \) also satisfies the first boundary condition. Consequently, the solution
$$\chi( x,\lambda) = \textstyle\begin{cases} \chi_{1} ( x,\lambda_{0} ), & x\in [ -1,0 ), \\ \chi_{2} ( x,\lambda_{0} ), & x\in ( 0,1 ]\end{cases} $$
satisfies also the first boundary condition. Therefore, \(\chi (x,\lambda_{0} ) \) satisfies all boundary and transmission conditions, that is, \(\chi ( x,\lambda_{0} ) \) is an eigenfunction for the considered problem (2)-(6), and consequently \(\lambda _{0} \) is an eigenvalue. Thus we have a contradiction which completes the proof.
Let us consider the fractional Wronskians
$$\begin{aligned} \omega_{i} ( \lambda) := &W_{F} \bigl( \phi _{i} ( x,\lambda),\chi_{i} ( x,\lambda) \bigr),\quad i=1,2 \\ := &I_{-1^{+}}^{1-\alpha}\phi_{i} ( x,\lambda) D_{0^{+}}^{\alpha}\chi_{i} ( x,\lambda) -I_{0^{+}}^{1-\alpha}\chi_{i} ( x,\lambda) D_{-1^{+}}^{\alpha}\phi_{i} ( x,\lambda) \end{aligned}$$
which are independent of x and are entire functions. The short calculation gives
$$\omega_{1} ( \lambda) =\omega_{2} ( \lambda). $$
Now we may introduce to the consideration the characteristic function
$$ \omega( \lambda) :=\omega_{1} ( \lambda) =\omega_{2} ( \lambda). $$
(33)
Lemma 5
The fractional Wronskian
\(W_{F} \)
satisfies the following relation:
$$W_{F} ( \lambda) =-h_{1}\omega^{3} ( \lambda ), $$
where
$$W_{F} ( \lambda) =\left \vert \begin{matrix} L_{1}(\phi_{1}) & L_{1}(\chi_{1}) & L_{1}(\phi_{2}) & L_{1}(\chi _{2}) \\ L_{2}(\phi_{1}) & L_{2}(\chi_{1}) & L_{2}(\phi_{2}) & L_{2}(\chi _{2}) \\ L_{3}(\phi_{1}) & L_{3}(\chi_{1}) & L_{3}(\phi_{2}) & L_{3}(\chi _{2}) \\ L_{4}(\phi_{1}) & L_{4}(\chi_{1}) & L_{4}(\phi_{2}) & L_{4}(\chi _{2})\end{matrix} \right \vert . $$
Proof
Employing the definitions of the functions \(\phi_{i} (x,\lambda ) \) and \(\chi_{i} ( x,\lambda )\), \(i=1,2\), we obtain
$$\begin{aligned} W_{F} ( \lambda) =&\left \vert \begin{matrix} 0 & \omega_{1} ( \lambda ) & 0 & 0 \\ 0 & 0 & \omega_{2} ( \lambda ) & 0 \\ h_{1}I_{-1^{+}}^{1-\alpha}\phi_{1} ( -0,\lambda ) & h_{1}I_{-1^{+}}^{1-\alpha}\chi_{1} ( -0,\lambda ) & I_{0^{+}}^{1-\alpha}\phi_{2} ( +0,\lambda ) & I_{0^{+}}^{1-\alpha}\chi_{2} ( +0,\lambda ) \\ D_{-1^{+}}^{\alpha}\phi_{1} ( -0,\lambda ) & D_{-1^{+}}^{\alpha}\chi_{1} ( -0,\lambda ) & h_{2}D_{0^{+}}^{\alpha}\phi_{2} ( +0,\lambda ) & h_{2}D_{0^{+}}^{\alpha}\chi_{2} (+0,\lambda )\end{matrix} \right \vert \\ =&\omega_{1} ( \lambda) \omega_{2} ( \lambda) \left \vert \begin{matrix} h_{1}I_{-1^{+}}^{1-\alpha}\phi_{1} ( -0,\lambda ) & I_{0^{+}}^{1-\alpha}\chi_{2} ( +0,\lambda ) \\ D_{-1^{+}}^{\alpha}\phi_{1} ( -0,\lambda ) & h_{2}D_{0^{+}}^{\alpha}\chi_{2} ( +0,\lambda )\end{matrix} \right \vert \\ =&-h_{1}\omega_{1}^{2} ( \lambda) \omega _{2} ( \lambda) \\ =&-h_{1}\omega^{3} ( \lambda). \end{aligned}$$
□
Corollary 3
The zeros of the function
\(W_{F} ( \lambda ) \)
consist of the zeros of the characteristic function
\(\omega ( \lambda ) \).
Theorem 4
The eigenvalues of fractional boundary value problem (2)-(6) are the same as the roots of the characteristic function
\(\omega ( \lambda ) \).
Proof
Let \(\lambda=\lambda_{0}\) be a root of the characteristic function \(\omega ( \lambda ) \), hence \(\omega _{2} ( \lambda_{0} ) =0\). It follows that \(\phi_{2}\) and \(\chi_{2}\) are linearly dependent, that is,
$$\phi_{2} ( x,\lambda_{0} ) =c\chi_{2} ( x, \lambda);\quad x\in( 0,1 ] $$
for some \(c\neq0\). As a result, the function \(\phi_{2} (x,\lambda_{0} ) \) satisfies fractional boundary condition (4). So, \(\phi ( x,\lambda _{0} ) \) which is given by
$$\phi( x,\lambda_{0} ) = \textstyle\begin{cases} \phi_{1} ( x,\lambda_{0} ), & x\in [ -1,0 ), \\ \phi_{2} ( x,\lambda_{0} ), & x\in ( 0,1 ]\end{cases} $$
satisfies the main problem (2)-(6). So, the function \(\phi ( x,\lambda _{0} ) \) is an eigenfunction of problem (2)-(6) corresponding to the eigenvalue \(\lambda_{0}\).
Let \(\lambda=\lambda_{0}\) be an eigenvalue and \(u_{0} (x,\lambda _{0} ) \) be any corresponding eigenfunction. It must be proved that \(\omega ( \lambda_{0} ) =0\). Let us suppose that \(\omega ( \lambda_{0} ) \neq0\). Then, since \(\omega_{1} (\lambda_{0} ) \neq0\) and \(\omega_{2} ( \lambda _{0} ) \neq0\), there exist constants \(c_{i}\), \(i=1,2,3,4\), at least one of which is not zero, such that
$$u_{0} ( x,\lambda_{0} ) = \textstyle\begin{cases} c_{1}\phi_{1} ( x,\lambda_{0} ) +c_{2}\chi_{1} (x,\lambda _{0} ), & x\in [ -1,0 ), \\ c_{3}\phi_{2} ( x,\lambda_{0} ) +c_{4}\chi_{2} (x,\lambda _{0} ), & x\in ( 0,1 ]\end{cases} $$
since \(\omega_{1} ( \lambda_{0} ) \neq0\) and \(\omega _{2} ( \lambda_{0} ) \neq0\).
Since the eigenfunction \(u_{0} ( x,\lambda_{0} ) \) satisfies both fractional boundary and fractional transmission conditions (3)-(6), we have
$$L_{i}u_{0} ( \cdot,\lambda_{0} ) =0, \quad \mbox{for } i=1,2,3,4. $$
Also since at least one of the constants \(c_{i}\), \(i=1,2,3,4\), is not zero,
$$\det\bigl( L_{i}u_{0} ( \cdot,\lambda_{0} ) \bigr) =0, $$
that is, \(W_{F} ( \lambda ) =0\). But, by Lemma 5, \(W_{F} ( \lambda ) \neq0\). This contradiction completes the proof. □