To prove the existence of a strong solution \(V = \overline{L}^{-1}W =\overline{L ^{-1}}W \) of the problem (2.1)–(2.4) \(\forall W=(F,f,g) \in Y \), it suffices to prove that \(\overline{R(L)}=Y \), the density of the range \(R(L) \) in Y is equivalent to the orthogonality of a vector \(W=(F,f,g)\in Y \) to the set \(R(L) \). For this purpose, we begin by the following theorem (the proof of the density in a special case).
Theorem 4.1
For some function
\(G \in L_{p}^{2}(D) \), and for all
\(U \in D_{0}(L)= \lbrace U /U \in D(L),\ell _{1}U = 0, \ell _{2}U = 0 \rbrace \), we have
$$ ( \mathcal{L} U,G )_{L_{p}^{2}(D)} = 0. $$
(4.1)
Then
\(G=0 \)almost everywhere in the domainD.
Proof
Equation (4.1) can be written
$$ \biggl( \partial _{0t}^{\beta +1}U- \frac{1}{x}\frac{\partial }{ \partial x} \biggl( x\frac{\partial U}{\partial x} \biggr) - \frac{1}{x}\frac{\partial ^{2} U}{\partial y^{2}}, G \biggr)_{L _{p}^{2}(D)} = 0. $$
(4.2)
Let \(h(x,y,t) \) be a function that satisfies the boundary conditions (2.2)–(2.4) and
$$ h,h_{x},h_{y}, \int _{0}^{t}h(x,y,s)\,ds,x \int _{0}^{t}h_{x}(x,y,s)\,ds,x \int _{0}^{t}h_{y}(x,y,s)\,ds, \partial _{0t}^{\beta +1}h \in L^{2}(D). $$
(4.3)
Then we suppose
$$ U(x,y,t)= \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds. $$
(4.4)
By replacing \(U(x,y,t) \) in (4.2) we have
$$\begin{aligned} &\biggl( \partial _{0t}^{\beta +1} \biggl( x \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr) - \frac{\partial }{ \partial x} \biggl( x \int _{0}^{t} \int _{0}^{s}h_{x}(x,y,z)\,dz\,ds \biggr) \\ &\quad{}-\frac{\partial }{\partial y} \biggl( \int _{0}^{t} \int _{0} ^{s}h_{y}(x,y,z)\,dz\,ds \biggr), G \biggr) _{L^{2}(D)}=0. \end{aligned}$$
(4.5)
Now, we assume the function
$$ G(x,y,t)=- \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x,\eta,s)\,ds\,d \eta \,d\xi. $$
(4.6)
Then Eq. (4.5) becomes
$$\begin{aligned} &{-} \biggl( \partial _{0t}^{\beta +1} \biggl( \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x,\eta,s)\,ds\,d\eta \,d\xi \biggr) _{L_{p}^{2}(D)} \\ &\quad{}+ \biggl( \frac{\partial }{\partial x} \biggl( x \int _{0}^{t} \int _{0} ^{s}h_{x}(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x, \eta,s)\,ds\,d\eta \,d\xi \biggr) _{L^{2}(D)} \\ &\quad{}+ \biggl( \frac{\partial }{\partial y} \biggl( \int _{0}^{t} \int _{0} ^{s}h_{y}(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x, \eta,s)\,ds\,d\eta \,d\xi \biggr) _{L^{2}(D)}= 0. \end{aligned}$$
(4.7)
Taking into account that the function h verifies the conditions (2.2)–(2.4), then integrating by parts each term of (4.7) we have
$$\begin{aligned} &{-} \biggl( \partial _{0t}^{\beta +1} \biggl( \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x,\eta,s)\,ds\,d\eta \,d\xi \biggr) _{L_{p}^{2}(\varOmega )} \\ &\quad = \biggl( \partial _{0t}^{\beta } \biggl( \int _{0}^{y} \int _{0}^{t} h(x, \xi,s)\,ds\,d\xi \biggr), \int _{0}^{y} \int _{0}^{t}h(x,\xi,s)\,ds\,d\xi \biggr) _{L_{p}^{2}(\varOmega )}, \end{aligned}$$
(4.8)
$$\begin{aligned} & \biggl(\frac{\partial }{\partial x} \biggl( x \int _{0}^{t} \int _{0} ^{s}h_{x}(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x, \eta,s)\,ds\,d\eta \,d\xi \biggr) _{L^{2}(\varOmega )} \\ &\quad= \biggl( \int _{0}^{y} \int _{0}^{t} \int _{0}^{s}h_{x}(x,\xi,z)\,dz \,ds\,d \xi, \int _{0}^{y} \int _{0}^{t}h_{x}(x,\xi,s)\,ds\,d \xi \biggr) _{L_{p} ^{2}(\varOmega )} \\ &\quad=\frac{1}{2}\frac{\partial }{\partial t} \biggl\Vert \int _{0}^{y} \int _{0}^{t} \int _{0}^{s}h_{x}(x,\xi,z)\,dz \,ds\,d\xi \biggr\Vert _{L_{p}^{2}( \varOmega )}^{2}, \end{aligned}$$
(4.9)
$$\begin{aligned} & \biggl( \frac{\partial }{\partial y} \biggl( \int _{0}^{t} \int _{0}^{s}h _{y}(x,y,z)\,dz\,ds \biggr), \int _{0}^{y} \int _{0}^{\xi } \int _{0}^{t}h(x, \eta,s)\,ds\,d\eta \,d\xi \biggr) _{L^{2}(\varOmega )} \\ &\quad= \biggl( \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds, \int _{0}^{t}h(x,y,s)\,ds \biggr) _{L^{2}(\varOmega )} \\ &\quad=\frac{1}{2}\frac{\partial }{\partial t} \biggl\Vert \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr\Vert _{L^{2}(\varOmega )}^{2}. \end{aligned}$$
(4.10)
Substituting (4.8), (4.9), and (4.10) into (4.7) we get
$$\begin{aligned} &2 \biggl( \partial _{0t}^{\beta } \biggl( \int _{0}^{y} \int _{0}^{t} h(x, \xi,s)\,ds\,d\xi \biggr), \int _{0}^{y} \int _{0}^{t}h(x,\xi,s)\,ds\,d\xi \biggr) _{L_{p}^{2}(\varOmega )} \\ &\quad{}+\frac{\partial }{\partial t} \biggl\Vert \int _{0}^{y} \int _{0}^{t} \int _{0}^{s}h_{x}(x,\xi,z)\,dz \,ds\,d\xi \biggr\Vert _{L_{p}^{2}(\varOmega )}^{2} \\ &\quad{}+\frac{\partial }{\partial t} \biggl\Vert \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr\Vert _{L^{2}(\varOmega )}^{2} =0. \end{aligned}$$
(4.11)
According to Lemma (2.1) the first term on the LHS of (4.11) can be estimated as follows:
$$\begin{aligned} &2 \biggl( \partial _{0t}^{\beta } \biggl( \int _{0}^{y} \int _{0}^{t} h(x, \xi,s)\,ds\,d\xi \biggr), \int _{0}^{y} \int _{0}^{t}h(x,\xi,s)\,ds\,d\xi \biggr) _{L_{p}^{2}(\varOmega )} \\ &\quad \geq \partial _{0t}^{\beta } \biggl\Vert \int _{0} ^{y} \int _{0}^{t} h(x,\xi,s)\,ds\,d\xi \biggr\Vert _{L_{p}^{2}(\varOmega )} ^{2}. \end{aligned}$$
(4.12)
Equation (4.11) can be written
$$\begin{aligned} & \partial _{0t}^{\beta } \biggl\Vert \int _{0}^{y} \int _{0}^{t} h(x,\xi,s)\,ds\,d \xi \biggr\Vert _{L_{p}^{2}(\varOmega )}^{2} + \frac{\partial }{\partial t} \biggl\Vert \int _{0}^{y} \int _{0}^{t} \int _{0}^{s}h_{x}(x,\xi,z)\,dz \,ds\,d\xi \biggr\Vert _{L_{p}^{2}(\varOmega )}^{2} \\ &\quad{}+\frac{\partial }{\partial t} \biggl\Vert \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr\Vert _{L^{2}(\varOmega )}^{2} \leq 0. \end{aligned}$$
(4.13)
By replacing t by τ and integrating of (4.13) with respect to τ over \([0;t] \) gives
$$\begin{aligned} &D_{0t}^{\beta -1} \biggl\Vert \int _{0}^{y} \int _{0}^{t} h(x,\xi,s)\,ds\,d \xi \biggr\Vert _{L_{p}^{2}(\varOmega )}^{2} \\ &\quad{}+ \biggl\Vert \int _{0}^{y} \int _{0}^{t} \int _{0}^{s}h_{x}(x,\xi,z)\,dz \,ds\,d\xi \biggr\Vert _{L_{p}^{2}( \varOmega )}^{2} \\ &\quad{}+ \biggl\Vert \int _{0}^{t} \int _{0}^{s}h(x,y,z)\,dz\,ds \biggr\Vert _{L^{2}( \varOmega )}^{2} \leq 0. \end{aligned}$$
(4.14)
We find from inequality (4.14) that \(G\equiv 0 \) almost everywhere in D. □
Theorem 4.2
The range
\(R(L) \)of the operatorL, coincides with the whole spaceY.
Proof
Let \(W=(\varphi,g_{1},g_{2})\in R(L)^{\perp } \) such that
$$ ( \mathcal{L} u,\varphi )_{L_{p}^{2}(D)}+ (\ell _{1}u,g _{1} )_{V^{1}(\varOmega )} + (\ell _{2} u,g_{2} )_{L _{p}^{2}(\varOmega )} = 0. $$
(4.15)
If we put \(u \in D_{0}(L) \) into (4.15) we get
$$ ( \mathcal{L} u,\varphi )_{L_{p}^{2}(D)} = 0, \quad\forall u \in D_{0}(L). $$
(4.16)
By virtue of Theorem 4.1 we deduce that \(\varphi \equiv 0 \), thus (4.15) becomes
$$ (\ell _{1}u,g_{1} )_{V^{1}(\varOmega )} + (\ell _{2} u,g _{2} )_{L_{p}^{2}(\varOmega )} = 0, \quad \forall u \in D(L). $$
(4.17)
The trace operators \(\ell _{1} \) and \(\ell _{2} \) are independent, and \(R(\ell _{1}) \) and \(R(\ell _{2}) \) are everywhere dense in the spaces \(V^{1}(\varOmega ) \) and \(L_{p}^{2}(\varOmega ) \), respectively. Then \(g_{1} = 0 \), \(g_{2} = 0 \). Consequently \(W = 0 \). Hence \(R(L)^{\perp }={0} \) i.e. \(\overline{R(L)}=Y \). □