In this section, we use the q-calculus of variations we developed in Section 4 to investigate the existence of solutions of the qFSLP
$$ D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y(x) +r(x)y(x)=\lambda w_{\alpha }y(x),\quad x\in A_{q,qa}^{*}, $$
(5.1)
under the boundary condition
The proof of the main result of this section depends on the Arzelà-Ascoli theorem [64], p.156. The setting of this theorem is a compact metric space X. Let \(C(X)\) denote the space of all continuous functions on X with values in \(\mathbb{C}\) or \(\mathbb{R}\). \(C(X)\) is associated with the metric function
$$d(f, g) = \max \bigl\{ \bigl\vert f(x)- g(x)\bigr\vert : x\in X \bigr\} . $$
Theorem 5.1
(Arzelà-Ascoli theorem)
If a sequence
\(\{f_{n} \}_{n}\)
in
\(C(X)\)
is bounded and equicontinuous then it has a uniformly convergent subsequence.
In our q-setting, we take \(X=A_{q,a}^{*}\). Hence \(f\in C(A_{q,a}^{*}) \) if and only if f is q-regular at zero, i.e.,
$$f(0):=\lim_{n\to\infty}f\bigl(aq^{n}\bigr). $$
Remark 5.2
A question may be raised as to why in (5.1) we have only \(x\in A_{q,qa}^{*}\) instead of \(A_{q,a}^{*}\). The reason for that is that the qFSLP (5.1)-(5.2) will be solved by using the q-fractional isoperimetric problem developed in Theorem 4.7, and its q-Euler-Lagrange equation (4.9) holds only for \(x\in A_{q,qa}^{*}\). Also, in order for (5.1) to hold at \(x=a\), we should have \(D_{q,a^{-}}^{\alpha}(p(\cdot){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }y(\cdot))(a)=0\) and this holds only if \(p(a){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y(a)=0\), which may not hold.
Theorem 5.3
Let
\(\frac{1}{2}<\alpha<1\). Assume that the functions
p, r, \(w_{\alpha}\)
are defined on
\(A_{q,a}^{*}\)
and satisfying the conditions:
-
(i)
\(w_{\alpha}\)
is a positive continuous function on
\([0,a]\)
such that
\(D_{q}^{k}\frac{1}{w_{\alpha}}\) (\(k=0,1,2\)) are bounded functions on
\(A_{q,a}\),
-
(ii)
r
is a bounded function on
\(A_{q,a}\),
-
(iii)
\(p\in C(A_{q,a}^{*})\)
such that
\(\inf_{x\in A_{q,a}}p(x)>0\), and
\(\sup_{x\in A_{q,a}}\vert \frac {r(x)}{w_{\alpha}(x)}\vert <\infty\).
The
q-fractional Sturm-Liouville problem (5.1)-(5.2) has an infinite number of eigenvalues
\(\lambda^{(1)}\), \(\lambda ^{(2)}\), …, and to each eigenvalue
\(\lambda^{(n)}\)
there is a corresponding eigenfunction
\(y^{(n)}\), which is unique up to a constant factor. Furthermore, the eigenfunctions
\(y^{(n)}\)
form an orthogonal set of solutions in the Hilbert space
\(L_{q}^{2}(A_{q,a}^{*},w_{\alpha})\).
Proof
As we mentioned in Remark 5.2, the qFSLP (5.1)-(5.2) can be recast as the q-fractional variational isoperimetric problem: Find the extremal of the functional
$$ J(y):= \int_{0}^{a} \bigl[p(x) \bigl({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y \bigr)^{2}+r(x)y^{2} \bigr] \,d_{q}x $$
(5.3)
subject to the boundary condition
and the isoperimetric constraint
$$ I(y)= \int_{0}^{a}w_{\alpha}(x)y^{2} \,d_{q}x=1. $$
(5.5)
The q-fractional Euler-Lagrange equation for the functional I is
$$2 w_{\alpha}(x) y(x)=0\quad \mbox{for all } x\in A_{q,a}, $$
which is satisfied only for the trivial solution \(y=0\), because \(w_{\alpha}\) is positive on \(A_{q,a}\). So, no extremals for I can satisfy the q-isoperimetric condition. If y is an extremal for the q-fractional isoperimetric problem, then from Theorem 4.7, there exists a constant λ such that y satisfies the q-fractional Euler-Lagrange equation (4.9) in \(A_{q,qa}^{*}\) but this is equivalent to the qFSLP (5.1).
In the following, we shall derive a method for approximating the eigenvalues and the eigenfunctions at the same time similar to the technique in [5, 11]. The proof follows in six steps.
Step 1. First let us point out that functional J defined in (5.3) is bounded from below. Indeed, since p, \(w_{\alpha}\) are positive on \(A_{q,a}\),
$$\begin{aligned} J(y)&= \int_{0}^{a} \bigl[p(x) \bigl({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y \bigr)^{2}+r(x)y^{2} \bigr] \\ &\geq\inf_{x\in A_{q,a}}\frac{r(x)}{w_{\alpha}(x)} \int _{0}^{a}w_{\alpha}(x)y^{2}(x) \,d_{q}x \\ &=\inf_{x\in A_{q,a}}\frac {r(x)}{w_{\alpha}(x)}=:M>-\infty. \end{aligned}$$
According to the Ritz method [11], p.201, we approximate a solution of (5.3)-(5.4) using the following q-trigonometric functions with the coefficients depending on \(w_{\alpha}\):
$$ y_{m}(x)=\frac{1}{\sqrt{w_{\alpha}}}\sum _{k=1}^{m}\frac{\beta _{k}}{\sqrt{\mu_{k}}}S_{q}\biggl( \frac{w_{k}x}{a}\biggr). $$
(5.6)
Observe that \(y_{m}(0)=y_{m}(a)=0\). By substituting (5.6) into (5.3) and (5.5) we obtain
$$\begin{aligned} {J_{m}}(\beta_{1},\ldots,\beta_{m}) =&{J_{m}} \bigl([\beta]\bigr) \\ =&\sum_{k,j=1}^{m}\frac {\beta_{j}\beta_{k}}{\sqrt{\mu_{j}}\sqrt{\mu_{k}}} \int_{0}^{a} \biggl[p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} \frac {S_{q}(\frac {w_{k}x}{a})}{\sqrt{w_{\alpha}}}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} \frac{S_{q}(\frac {w_{j}x}{a})}{\sqrt{w_{\alpha}}} \\ &{}+\frac{r(x)}{w_{\alpha}(x)}S_{q}\biggl(\frac {w_{k}x}{a} \biggr)S_{q}\biggl(\frac{w_{j}x}{a}\biggr) \biggr] \,d_{q}x \end{aligned}$$
(5.7)
subject to the condition
$$ {I_{m}}(\beta_{1},\beta_{2},\ldots, \beta _{m})={I_{m}}\bigl([\beta ]\bigr)=\frac{a\sqrt{q}}{2}\sum _{k=1}^{m}\beta_{k}^{2}=1. $$
(5.8)
The functions defined in (5.7) and (5.8) are functions of the m variables \(\beta_{1}, \beta_{2},\ldots,\beta_{m}\). Thus, in terms of the variables \(\beta_{1},\ldots, \beta_{m}\), our problem is to minimize \({J_{m}}(\beta_{1},\beta_{2},\ldots,\beta_{m})\) on the surface \(\sigma_{m}\) of the m dimensional sphere defined in (5.8). Since \(\sigma_{m}\) is a compact set and \({J_{m}}(\beta_{1},\beta_{2},\ldots,\beta_{m})\) is continuous on \(\sigma _{m}\), \({J_{m}}(\beta_{1},\beta_{2},\ldots,\beta_{m})\) has a minimum \(\lambda _{m}^{{1}}\) at some point \((\beta_{1}^{(1)},\ldots,\beta_{m}^{(1)} )\) of \(\sigma_{m}\). Let
$$y_{m}^{(1)}=\frac{1}{\sqrt{w_{\alpha}}}\sum _{k=1}^{m}\frac{\beta _{k}^{(1)}}{\mu_{k}}S_{q}\biggl( \frac{w_{k}x}{a}\biggr). $$
If this procedure is carried out for \(m=1,2,\ldots\) , we obtain a sequence of numbers \(\lambda_{1}^{(1)}, \lambda_{2}^{(1)}, \ldots\) , and a corresponding sequence of functions
$$y_{1}^{(1)}(x),\qquad y_{2}^{(1)}(x), \qquad y_{3}^{(1)}(x),\qquad \ldots. $$
Noting that \(\sigma_{m}\) is the subset of \(\sigma_{m+1}\) obtained by setting \(\beta_{m+1}=0\), while
$$J_{m}(\beta_{1},\ldots,\beta_{m})=J_{m+1}( \beta_{1},\ldots,\beta_{m},0), $$
consequently,
$$ \lambda_{m+1}^{(1)}\leq\lambda_{m}^{(1)}. $$
(5.9)
Increasing the domain of definition of a function can only decrease its minimum. It follows from (5.9) and the fact that \(J(y)\) is bounded from below that its limit
$$\lambda^{(1)}=\lim_{m\to\infty}\lambda_{m}^{(1)} $$
exists.
Step 2. We shall prove that the sequence \((y_{m}^{(1)})_{m\in \mathbb{N}}\) contains a uniformly convergent subsequence. From now on, for simplicity, we shall write \(y_{m}\) instead of \(y_{m}^{(1)}\). Recall that
$$\lambda_{m}^{(1)}= \int_{0}^{a} \bigl[p(x) \bigl({}^{\mathrm {c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2}+r(x)y_{m}^{2} \bigr] \,d_{q}x $$
is convergent, so it must be bounded, i.e., there exists a constant \(M_{0}>0\) such that
$$\int_{0}^{a} \bigl[p(x) \bigl({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2}+r(x)y_{m}^{2} \bigr] \,d_{q}x\leq M_{0},\quad m\in\mathbb{N}. $$
Therefore, for all \(m\in\mathbb{N}\) we have the inequality
$$\begin{aligned} \int_{0}^{a}p(x) \bigl({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2} \,d_{q}x \leq & M_{0}+ \biggl\vert \int _{0}^{a}r(x)y_{m}^{2}(x) \,d_{q}x\biggr\vert \\ \leq& M_{0}+\sup_{x\in A_{q,a}}\biggl\vert \frac{r(x)}{w_{\alpha }(x)}\biggr\vert \int_{0}^{a}w_{\alpha}(x)y_{m}^{2}(x) \,d_{q}x \\ :=&M_{0}+\sup_{x\in A_{q,a}}\biggl\vert \frac{r(x)}{w_{\alpha}(x)}\biggr\vert =:M_{1}. \end{aligned}$$
Moreover, since \(\inf_{x\in A_{q,a}}p(x)>0\) we have
$$\Bigl(\inf_{x\in A_{q,a}}p(x) \Bigr) \int_{0}^{a}\bigl({}^{\mathrm {c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2} \,d_{q}x\leq \int_{0}^{a}p(x) \bigl({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2} \,d_{q}x\leq M_{1}, $$
and hence
$$ \int_{0}^{a}\bigl({}^{\mathrm {c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr)^{2} \,d_{q}x \leq\frac {M_{1}}{\inf_{x\in A_{q,a}}p(x)}=:M_{2}^{2}. $$
(5.10)
Since \(y_{m}(0)=0\), from (2.15) and (5.10)
$$\begin{aligned} \Vert y_{m}\Vert =&\bigl\Vert I_{q,0^{+}}^{\alpha }{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}\bigr\Vert \leq\widetilde {M}_{\alpha}\bigl\Vert {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m} \bigr\Vert _{2} \\ \leq&\widetilde{M}_{\alpha} {M_{2}} \end{aligned}$$
for \(\alpha>1/2\). Hence, \((y_{m})_{m}\) is uniformly bounded on \(A_{q,a}^{*}\). Now we prove that the sequence \((y_{m})_{m}\) is equicontinuous. Let \(x_{1},x_{2}\in A_{q,a}\). Assume that \(x_{1}< x_{2}\). Applying the Schwarz inequality and (2.9)
$$\begin{aligned}& \Gamma_{q}(\alpha)\bigl\vert y_{m}(x_{2})-y_{m}(x_{1}) \bigr\vert \\& \quad =\Gamma _{q}(\alpha)\bigl\vert I_{q,0^{+}}^{\alpha}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}(x_{2})-I_{q,0^{+}}^{\alpha}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m}(x_{2})\bigr\vert \\& \quad = \biggl\vert x_{2}^{\alpha-1} \int_{0}^{x_{2}}(qt/x_{2};q)_{\alpha -1}{}^{\mathrm{c}}D_{q,0^{+}} ^{\alpha}y_{m}(t) \,d_{q}t-x_{1}^{\alpha-1} \int_{0}^{x_{1}}(qt/x_{1};q)_{\alpha-1}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }y_{m}(t) \,d_{q}t\biggr\vert \\& \quad \leq \biggl\vert x_{2}^{\alpha-1} \int_{x_{1}}^{x_{2}}(qt/x_{2};q)_{\alpha -1}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y_{m}(t) \,d_{q}t\biggr\vert \\& \qquad {} +\biggl\vert \int_{0}^{x_{1}} \bigl\{ x_{2}^{\alpha-1}(qt/x_{2};q)_{\alpha -1}-x_{2}^{\alpha-1}(qt/x_{1};q)_{\alpha-1} \bigr\} {}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }y_{m}(t) \,d_{q}t\biggr\vert \\& \quad \leq M_{2} \biggl(x_{2}^{2\alpha-2} \int_{x_{1}}^{x_{2}}(qt/x_{2};q)_{\alpha -1}^{2} \,d_{q}t \biggr)^{1/2} \\& \qquad {} + M_{2} \biggl( \int_{0}^{x_{1}} \bigl(x_{2}^{\alpha-1}(qt/x_{2};q)_{\alpha -1}-x_{1}^{\alpha-1}(qt/x_{1};q)_{\alpha-1} \bigr)^{2} \,d_{q}t \biggr)^{1/2}. \end{aligned}$$
Since \(x_{1}< x_{2}\), we have
$$x_{2}^{\alpha-1}({qt}/{x_{2}};q)_{\alpha-1}\leq x_{1}^{\alpha-1} ({qt}/{x_{1}};q)_{\alpha-1}\quad \mbox{for all } t< x_{1}< x_{2}. $$
Using the inequality
$$t_{1}\geq t_{2} \geq0\rightarrow(t_{1}-t_{2})^{2} \leq t_{1}^{2}-t_{2}^{2}, $$
we obtain
$$\begin{aligned}& \int_{0}^{x_{1}} \bigl(x_{2}^{\alpha-1}(qt/x_{2};q)_{\alpha -1}-x_{1}^{\alpha-1}(qt/x_{1};q)_{\alpha-1} \bigr)^{2} \,d_{q}t \\& \quad \leq \int_{0}^{x_{1}}x_{1}^{2\alpha-2}(qt/x_{1};q)_{\alpha-1}^{2} \,d_{q}t- \int_{0}^{x_{1}}x_{2}^{2\alpha-2}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t \\& \quad = \int_{x_{1}}^{x_{2}}x_{2}^{2\alpha-2}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t+ \int _{0}^{x_{1}}x_{1}^{2\alpha-2}(qt/x_{1};q)_{\alpha-1}^{2} \,d_{q}t \\& \qquad {} - \int_{0}^{x_{2}}x_{2}^{2\alpha-2}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t \\& \quad = \int_{x_{1}}^{x_{2}}x_{2}^{2\alpha-2}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t+ \bigl(x_{1}^{2\alpha-1}-x_{2}^{2\alpha-1} \bigr) \int_{0}^{1}(q\xi ;q)^{2}_{\alpha-1} \,d_{q}\xi \\& \quad \leq \int_{x_{1}}^{x_{2}}x_{2}^{2\alpha-2}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t \end{aligned}$$
for \(\alpha>\frac{1}{2}\). Hence, we have
$$\begin{aligned} \begin{aligned} \bigl\vert y_{m}(x_{2})-y_{m}(x_{1}) \bigr\vert &\leq \frac {2M_{2}}{\Gamma_{q}(\alpha)}x_{2}^{\alpha-1} \biggl( \int _{x_{1}}^{x_{2}}(qt/x_{2};q)_{\alpha-1}^{2} \,d_{q}t \biggr)^{1/2} \\ &\leq\frac{2M_{2}}{\Gamma_{q}(\alpha)(q^{\alpha};q)_{\infty }^{2}}x_{2}^{\alpha-1}\sqrt{x_{2}-x_{1}} \leq\frac{2M_{2}}{\Gamma_{q}(\alpha )^{2}(q^{\alpha};q)_{\infty}^{2}}(x_{2}-x_{1})^{\alpha-\frac{1}{2}}. \end{aligned} \end{aligned}$$
Hence \(\{y_{m} \}\) is equicontinuous. Therefore, from the Arzelà-Ascoli theorem for metric spaces, a uniformly convergent subsequence \((y_{m_{n}})_{n\in\mathbb{N}}\) exists. It means that we can find \(y^{(1)}\in C(A_{q,a}^{*})\) such that
$$ y^{(1)}=\lim_{n\to\infty}y_{m_{n}}. $$
(5.11)
Step 3. From the Lagrange multiplier at \([\beta]=(\beta_{1}^{(1)},\ldots,\beta _{m}^{(1)})\), we have
$$\frac{\delta}{\delta\beta_{j}} \bigl[ J_{m}\bigl([\beta]\bigr)-\lambda _{m}^{(1)}I_{m}\bigl([\beta]\bigr) \bigr] \big|_{[\beta]=[\beta^{(1)}]},\quad j=1,2,\ldots,m. $$
By multiplying each equation by an arbitrary constant \(c_{j}\) and summing from 1 to m, we obtain
$$ 0=\sum_{j=1}^{m}c_{j} \frac{\delta}{\delta\beta_{j}} \bigl[ J_{m}\bigl([\beta ]\bigr)- \lambda_{m}^{(1)}I_{m}\bigl([\beta]\bigr) \bigr] \big|_{[\beta]=[\beta^{(1)}]}. $$
(5.12)
For \(m\in\mathbb{N}\), set
$$h_{m}(x):=\frac{1}{\sqrt{w_{\alpha}}}g_{m}(x);\qquad g_{m}(x):=\sum_{j=1}^{m} \frac{c_{j}}{\sqrt{\mu_{j}}}S_{q}\biggl(\frac{w_{j}x}{a}\biggr). $$
According to Proposition 3.10, we can choose the coefficients \(c_{j}\) such that there exists a function g satisfying
$$\lim_{m\to\infty}D_{q}^{k} g_{m}= D_{q}^{k} g\quad (k=0,1,2) $$
and the convergence is in \(L_{q}^{2}(A_{q,a}^{*})\) norm. Hence
$$ \lim_{m\to\infty}\bigl\Vert D_{q}^{k} h_{m}-D_{q}^{k} h\bigr\Vert _{2}=0 \quad (k=0,1,2). $$
(5.13)
We can write (5.12) in the form
$$ 0= \int_{0}^{a} \bigl[p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} y_{m} {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h_{m}+ \bigl(r(x)-\lambda_{m}^{1} w_{\alpha}(x) \bigr)y_{m} h_{m} \bigr] \,d_{q}x. $$
(5.14)
Since \(y_{m}(0)=0\), from (2.11)
$${}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y_{m}= D_{q,0^{+}}^{\alpha}y_{m}=D_{q} I_{q,0^{+}}^{1-\alpha}y_{m}. $$
Then replacing \({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y_{m}\) by \(D_{q,0^{+}}^{\alpha}y_{m}\) in (5.14) and applying the q-integration by parts rule (1.10), we obtain
$$\begin{aligned} 0 =& I_{m} \\ :=&- \int_{0}^{a}D_{q} p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx) \,d_{q}x \\ &{}- \int_{0}^{a}p(qx)D_{q} {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha }y_{m} \bigr) (qx) \,d_{q}x \\ &{}+ \int_{0}^{a} \bigl[r(x)-\lambda_{m}^{(1)}w_{\alpha}(x) \bigr]y_{m} h_{m} \,d_{q}x \\ &{}+p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h_{m}(x)I_{q,0^{+}}^{1-\alpha }y_{m}(x) \big|_{x=0}^{x=a}. \end{aligned}$$
In the following we shall prove that
$$\begin{aligned} I :=&\lim_{m\to\infty}I_{m} \\ =& \int_{0}^{a}-D_{q} p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha} h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y^{(1)} \bigr) (qx) \,d_{q}x \\ &{}- \int_{0}^{a}p(qx)D_{q} {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h(x) \bigl(I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr) (qx) \,d_{q}x \\ &{}+p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h(x)I_{q,0^{+}}^{1-\alpha }y(x) \big|_{x=0}^{x=a}+ \int_{0}^{a} \bigl[r(x)-\lambda^{(1)}w_{\alpha }(x) \bigr]y^{(1)}h \,d_{q}x. \end{aligned}$$
(5.15)
Indeed,
$$\begin{aligned}& |I_{m}-I| \\& \quad \leq \int_{0}^{a}\bigl\vert D_{q} p(x) \bigl[{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)-{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha } h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y^{(1)} \bigr) (qx) \bigr] \bigr\vert \,d_{q}x \\& \qquad {}+ \int_{0}^{a}\bigl\vert p(qx) \bigl[D_{q}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)-D_{q}{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y^{(1)} \bigr) (qx) \bigr]\bigr\vert \,d_{q}x \\& \qquad {}+\bigl\vert p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} h_{m}(x)I_{q,0^{+}}^{1-\alpha}y_{m}(x)-p(x){}^{\mathrm{c}}D_{q,0^{+}} ^{\alpha} h(x)I_{q,0^{+}}^{1-\alpha}y^{(1)}(x)\bigr\vert _{x=0}^{x=a} \\& \qquad {}+ \int_{0}^{a}\bigl\vert \bigl[r(x)- \lambda_{m}^{(1)}w_{\alpha}(x) \bigr]y_{m} h_{m}- \bigl[r(x)-\lambda^{(1)}w_{\alpha}(x) \bigr]y^{(1)}h \bigr\vert \,d_{q}x. \end{aligned}$$
(5.16)
For the first q-integral in (5.16), by adding and subtracting the term
$$D_{q}p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx) $$
to the integrand, we obtain
$$\begin{aligned}& \int_{0}^{a}\bigl\vert D_{q} p(x) \bigl[{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)-{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha} h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y^{(1)} \bigr) (qx) \bigr] \bigr\vert \,d_{q}x \\& \quad \leq \Vert D_{q} p\Vert \bigl\Vert {}^{\mathrm {c}}D_{q,0^{+}}h \bigr\Vert _{\infty}\bigl\Vert \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)- \bigl(I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr) (qx) \bigr\Vert _{1} \\& \qquad {} +\Vert D_{q} p\Vert M_{3}K_{1-\alpha} \bigl\Vert {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}(h_{m}-h) \bigr\Vert _{2} \\& \quad \leq \frac{\Vert D_{q} p\Vert }{q} \bigl\{ \bigl\Vert {}^{\mathrm{c}}D_{q,0^{+}}h \bigr\Vert _{\infty}\bigl\Vert I_{q,0^{+}}^{1-\alpha}y_{m}-I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr\Vert _{1}+M_{3}K_{1-\alpha}\bigl\Vert {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }(h_{m}-h)\bigr\Vert _{2} \bigr\} , \end{aligned}$$
where \(K_{1-\alpha}\) is the constant defined in (2.13) and \(M_{3}:=\sup_{m\in\mathbb{N}}\Vert y_{m}\Vert _{\infty}\). From (5.11) and (5.13)
$$\lim_{m\to\infty}\bigl\Vert y_{m}-y^{(1)}\bigr\Vert =\lim_{m\to \infty} \Vert D_{q} h_{m}-D_{q}h\Vert _{2}=0, $$
then applying (2.12)-(2.14), we obtain
$$\lim_{m\to\infty}\bigl\Vert I_{q,0^{+}}^{1-\alpha }y_{m}-I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr\Vert _{1}=\lim_{m\to\infty }\bigl\Vert {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}(h_{m}-h)\bigr\Vert _{2}=0 $$
and the first q-integral vanishes as \(m\to\infty\). As for the second q-integral, we add and subtract the term \(p(qx)D_{q}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}h(x) (I_{q,0^{+}}^{1-\alpha}y_{m} )(qx)\). This gives
$$\begin{aligned}& \int_{0}^{a}\bigl\vert p(qx) \bigl[D_{q}{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }h_{m}(x) \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)-D_{q}{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }h(x) \bigl(I_{q,0^{+}}^{1-\alpha}y^{(1)} \bigr) (qx) \bigr]\bigr\vert \,d_{q}x \\& \quad \leq \Vert p\Vert \bigl\Vert D_{q}{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}h \bigr\Vert _{2}\bigl\Vert \bigl(I_{q,0^{+}}^{1-\alpha}y_{m} \bigr) (qx)- \bigl(I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr) (qx) \bigr\Vert _{2} \\& \qquad {} +\Vert p\Vert M_{3}K_{1-\alpha}\bigl\Vert D_{q}{}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}(h_{m}-h) \bigr\Vert _{2} \\& \quad \leq \frac{\Vert p\Vert }{q} \bigl\{ \bigl\Vert D_{q}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }h \bigr\Vert _{2}\bigl\Vert I_{q,0^{+}}^{1-\alpha}y_{m}-I_{q,0^{+}}^{1-\alpha }y^{(1)} \bigr\Vert _{2}+M_{3}K_{1-\alpha}\bigl\Vert D_{q}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }(h_{m}-h) \bigr\Vert _{2} \bigr\} . \end{aligned}$$
Since \(D_{q}{}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}f(x)=I_{q}^{1-\alpha}D_{q}^{2} f\) if \(D_{q}f(0)=0\), and since \(\lim_{m\to\infty} \Vert D_{q}^{2}h_{m}-D_{q}^{2}h\Vert _{2}=0\), from (2.14), the second q-integral tends to zero as m tends to ∞. For the next two terms, we have for \(x=0, a\),
$$\bigl(I_{q,0^{+}}^{1-\alpha}y_{m}\bigr) (qx)=q^{1-\alpha}I_{q,0^{+}}^{\alpha }y_{m}(qx) \to q^{1-\alpha}I_{q,0^{+}}^{\alpha}y^{(1)}(qx) $$
resulting from the convergence of the sequence \(\Vert y_{m}-y \Vert \to0\), and at the points \(x=0\), \(x=a\), we obtain
$$D_{q,0^{+}}^{\alpha}h_{m}(0)\to D_{q,0^{+}}^{\alpha}h(0), \qquad D_{q,0^{+}}^{\alpha }h_{m}(a)\to D_{q,0^{+}}^{\alpha}h(a). $$
Therefore,
$$\bigl\vert p(x)D_{q,0^{+}}^{\alpha}h_{m}(x)I_{q,0^{+}}^{1-\alpha }y_{m}(x)-p(x)D_{q,0^{+}}^{\alpha} h(x)I_{q,0^{+}}^{1-\alpha}y^{(1)}(x)\bigr\vert _{x=0}^{x=a}=0. $$
Similarly, the last term in the estimation (5.16) vanishes as \(m\to\infty\).
Step 4. We have
$$ I= \int_{0}^{a} p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}y(x){}^{\mathrm{c}}D_{q,0^{+}} ^{\alpha}h(x)+\bigl(r(x)-\lambda w_{\alpha}\bigr)y(x)h(x) \,d_{q}x=0. $$
(5.17)
Set
$$ \begin{aligned} &\gamma_{1}(x) := p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y(x), \\ &\gamma_{2}(x) := \bigl(r(x)-\lambda w_{\alpha}\bigr)y(x). \end{aligned} $$
(5.18)
Thus, since \(h(0)=h(a)=0\),
$$I= \int_{0}^{a} \bigl[ \bigl( I_{q,a^{-}}^{1-\alpha} \gamma_{1} \bigr) (x)-(I_{q,0^{+}}\gamma_{2}) (qx) \bigr]D_{q}h(x) \,d_{q}x=0. $$
Hence, from Lemma 4.4 there is a constant c such that
$$ \bigl(I_{q,a^{-}}^{1-\alpha}\gamma_{1} \bigr) (x)-(I_{q,0^{+}}\gamma_{2}) (qx)=c,\quad \forall x\in A_{q,a}^{*}. $$
(5.19)
Acting on the two sides of (5.19) by \(-\frac{1}{q}D_{q^{-1}}\), we obtain
$$D_{q,a^{-}}^{\alpha}\gamma_{1}(x)+\gamma_{2}(x)=0, \quad x\in A_{q,qa}^{*}. $$
Hence, y is a solution of the qFSLP.
Step 5. In the following, we show that \((y_{m}^{(1)})_{m\in \mathbb{N}}\) itself converges to \(y^{(1)}\). First, from Theorem 3.12 of [34], for a given λ the solution of
$$ \bigl[D_{q,a^{-}}^{\alpha} p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}y +r(x) \bigr]y(x)=\lambda w_{\alpha}(x)y(x), $$
(5.20)
subject to the boundary conditions
and the normalization condition
$$ \int_{0}^{a}w_{\alpha}(x)y^{2}(x) \,d_{q}x=1, $$
(5.22)
is unique except for a sign. Let us assume that \(y^{(1)}\) solves (5.20) and the corresponding eigenvalue is \(\lambda= \lambda^{(1)}\). Suppose that \(y^{(1)}\) is nontrivial, i.e., there exists \(x_{0}\in A_{q,qa}^{*}\) such that \(y(x_{0})\neq0\) and choose the sign so that \(y^{(1)}(x_{0})>0\). Similarly, for all \(m\in\mathbb{N}\), let \(y_{m}^{(1)}\) solve (5.20) with corresponding eigenvalue \(\lambda=\lambda_{m}^{(1)}\), and let us choose the sign so that \(y_{m}^{(1)}(x_{0})\geq0\). Now, suppose that \((y_{m}^{(1)})\) does not converge to \(y^{(1)}\). It means that we can find another subsequence of \(y_{m}^{(1)}\) such that it converges to another solution \(\tilde{y}^{(1)}\). But for \(\lambda=\lambda^{(1)}\), the solution of (5.20)-(5.22) is unique except for a sign, hence
$$\tilde{y}^{(1)}=-y^{(1)} $$
and we must have \(\tilde{y}^{(1)}(x_{0})<0\). However, this is impossible because for all \(m\in\mathbb{N}\), \(y_{m}^{(1)}(x_{0})\geq0\). This is a contradiction, hence the solution is unique.
Step 6. In order to find the eigenfunction \(y^{(2)}\) and the corresponding eigenvalue \(\lambda^{(2)}\), we minimize the functional (5.3) subject to (5.4) and (5.5) but now with an extra orthogonality condition,
$$\int_{0}^{a}y(x)y^{(1)}(x)w_{\alpha}(x) \,d_{q}x=0. $$
If we approximate the solution by
$$y_{m}(x)=\frac{1}{\sqrt{w_{\alpha}}}\sum_{k=1}^{m} \frac{\beta _{k}}{\sqrt{\mu_{k}}}S_{q}\biggl(\frac{w_{k} x}{a}\biggr),\qquad y_{m}(0)=y_{m}(a)=0, $$
then we again obtain the quadratic form (5.7). However, admissible solutions are satisfying (5.8) together with
$$ \frac{a\sqrt{q}}{2}\sum_{k=1}^{m} \beta_{k} \beta_{k}^{(1)}=0, $$
(5.23)
i.e., they lie in the \((m-1)\)-dimensional sphere. As before, we find that the function \(\widetilde{J}([\beta])\) has a minimum \(\lambda_{m}^{(2)}\) and there exists \(\lambda^{(2)}\) such that
$$\lambda^{(2)}=\lim_{m\to\infty} \lambda_{m}^{(2)}, $$
because \(J(y)\) is bounded from below. Moreover, it is clear that
$$ \lambda^{(1)}\leq\lambda^{(2)}. $$
(5.24)
The function \(y_{m}^{(2)}\) defined by
$$y_{m}^{(2)}(x):=\frac{1}{\sqrt{w_{\alpha}}}\sum _{k=1}^{m}\frac{\beta _{k}^{(2)}}{\sqrt{\mu_{k}}}S_{q}\biggl( \frac{w_{k} x}{a} \biggr), $$
achieves its minimum \(\lambda_{m}^{(2)}\), where \(\beta^{(2)}= (\beta_{1}^{(2)},\ldots,\beta_{m}^{(2)} )\) is the point satisfying (5.8) and (5.23). By the same argument as before, we can prove that the sequence \((y_{m}^{(2)})\) converges uniformly to a limit function \(y^{(2)}\), which satisfies the qFSLP (5.1) with \(\lambda^{(2)}\), boundary conditions (5.4) and orthogonality condition (5.5). Therefore, the solution \(y^{(2)}\) of the qFSLP corresponding to the eigenvalue \(\lambda^{(2)}\) exists. Furthermore, because orthogonal functions cannot be linearly dependent, and since only one eigenfunction corresponds to each eigenvalue (except for a constant factor) we have the strict inequality
$$\lambda^{(1)}< \lambda^{(2)} $$
instead of (5.24). Finally, if we repeat the above procedure with similar modifications, we can obtain the eigenvalues \(\lambda^{(3)}, \lambda^{(4)},\ldots\) and the corresponding eigenvectors \(y^{(3)}, y^{(4)},\ldots\) . □
5.1 The first eigenvalue
Definition 5.4
The Rayleigh quotient for the q-fractional Sturm-Liouville problem (5.1)-(5.2) is defined by
$$R(y):=\frac{J(y)}{I(y)}, $$
where \(J(y)\) and \(I(y)\) are given by (5.3) and (5.5), respectively.
Theorem 5.5
Let
y
be a non-zero function satisfying
y
and
\({}^{\mathrm{c}}D_{q,0^{+}} ^{\alpha}y\)
be in
\(C(A_{q,a}*)\)
and
\(y(0)=y(a)=0\). Then
y
is a minimizer of
\(R(y)\)
and
\(R(y)=\lambda\)
if and only if
y
is an eigenfunction of problem (5.1)-(5.2) associated with
λ. That is, the minimum value of
R
at
y
is the first eigenvalue
\(\lambda^{(1)}\).
Proof
First, we prove the necessity. Assume that y is a non-zero minimizer of \(R(y)\) and \(R(y)=\lambda\). Consider the one parameter family of curves
$$y=y+h\eta,\quad |h|\leq\epsilon, $$
where η and \({}^{\mathrm{c}}D_{q,0^{+}}^{\alpha} \) are \(C(A_{q,a}^{*})\) functions and \(\eta(0)=\eta(a)=0\) and \(\eta\neq0\). Define functions ϕ, ψ, ξ on \([-\epsilon,\epsilon]\) by
$$\phi(h):=I(y+ h\eta),\qquad \psi(h):=J(y+h\eta),\qquad \xi(h)=R(y+h\eta )= \frac{\psi(h)}{\phi(h)},\quad h\in[-\epsilon,\epsilon]. $$
Hence ξ is \(C^{1}\) function on \([-\epsilon,\epsilon]\). Since \(\xi (0)=R(y)\), 0 is a minimum value of ξ. Consequently, \(\xi_{i}'(0)=0\). But
$$\xi'(h)=\frac{1}{\phi(h)} \biggl(\psi'(h)- \frac{\psi(h)}{\phi (h)}\phi'(h) \biggr) $$
and
$$\begin{aligned}& \psi'(0) = 2 \int_{0}^{a} \bigl[p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}y {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}\eta+r(x)y\eta \bigr] \,d_{q}x, \\& \phi'(0) = 2 \int_{0}^{a}w_{\alpha}(x)y(x)\eta(x) \,d_{q}x, \\& \frac{\psi(0)}{\phi(0)} = R(y)=\lambda. \end{aligned}$$
Therefore,
$$\xi'(0)=\frac{2}{I(y)} \biggl( \int_{0}^{a} \bigl[p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}y {}^{\mathrm{c}}D_{q,0^{+}}^{\alpha}\eta+ \bigl(r(x)y-\lambda w_{\alpha } \bigr)\eta \bigr] \,d_{q}x \biggr). $$
Using (2.19), we obtain
$$\int_{0}^{a} \bigl[D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha}y(x)+ \bigl(r(x)-\lambda \bigr) w_{\alpha}(x)y(x) \bigr]\eta(x) \,d_{q}x=0. $$
Applying Lemma 4.3, we obtain
$$D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }y(x)+r(x)y(x)= \lambda w_{\alpha }(x)y(x),\quad x\in A_{q,qa}^{*}. $$
This proves the necessity. Now we prove the sufficiency. Assume that y is an eigenfunction of (5.1)-(5.2) associated with an eigenvalue λ. Then
$$ D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm{c}}D_{q,0^{+}}^{\alpha }y(x)+r(x)y(x)= \lambda w_{\alpha }(x)y(x),\quad x\in A_{q,qa}^{*}. $$
(5.25)
Multiplying (5.25) by y and calculating the q-integration from 0 to a, we obtain
$$\int_{0}^{a} \bigl[y(x)D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }y(x)+r(x)y^{2}(x) \bigr] \,d_{q}x=\lambda \int_{0}^{a}w_{\alpha}(x)y^{2}(x) \,d_{q}x. $$
Since \(y\neq0\), \(\int_{0}^{a}w_{\alpha}(x)y^{2}(x) \,d_{q}x>0\) and
$$\frac{\int_{0}^{a} [y(x)D_{q,a^{-}}^{\alpha}p(x){}^{\mathrm {c}}D_{q,0^{+}}^{\alpha }y(x)+r(x)y^{2}(x) ] \,d_{q}x}{\int_{0}^{a}w_{\alpha}(x)y^{2}(x) \,d_{q}x}=\lambda. $$
That is, \(R(y)=\lambda\). Therefore, any minimum value of J is an eigenvalue and it is attained at the associated eigenfunction. Therefore the minimum value of J is the smallest eigenvalue. □