Let us look for a solution to (16) in the form of the sum of a transient solution and a steady-state solution,
$$ p ( x,t ) =U ( x,t ) +V(x,t). $$
(25)
By substitution into (16) we find
$$ \frac{\partial U}{\partial t}+\frac{\partial V}{\partial t}=K_{1}D^{\alpha } \biggl( \frac{{\partial}^{2}U}{\partial x^{2}}+\frac{{\partial}^{2}V}{ \partial x^{2}} \biggr) +K_{2}D^{\beta} \biggl( \frac{{\partial}^{2}U}{\partial x^{2}}+\frac{{\partial}^{2}V}{\partial x^{2}} \biggr) +f(x,t), $$
and the boundary conditions give
$$\begin{aligned}& U ( -L,t ) +V ( -L,t ) =l(t), \\ & {\biggl.\frac{\partial}{\partial x} \bigl[ K_{1}D^{\alpha} \bigl( U ( x,t ) +V ( x,t ) \bigr) +K_{2}D^{\beta} \bigl( U ( x,t ) +V ( x,t ) \bigr) \bigr] \biggr|_{x=0}}={\Psi } ( t ) . \end{aligned}$$
The problem will be partitioned into
$$ \left \{ \begin{array}{l} \frac{{\partial}^{2}V}{\partial x^{2}}=0\quad \mbox{in }{\Omega}_{0 } , \\ V ( -L,t ) =l ( t ) ,\quad 0< t\leq T, \\ \frac{\partial}{\partial x} [ K_{1}D^{\alpha}V ( x,t ) +K_{2}D^{\beta}V ( x,t ) ] |_{x=0}=\Psi ( t ) ,\quad 0<t\leq T \end{array} \right . $$
(26)
and
$$ \left \{ \begin{array}{l} \frac{\partial U}{\partial t}=K_{1}D^{\alpha}\frac{{\partial}^{2}U}{\partial x^{2}}+K_{2}D^{\beta}\frac{{\partial}^{2}U}{\partial x^{2}}-\frac{\partial V}{\partial t}+f ( x,t )\quad \mbox{in }{\Omega}_{0}, \\ U ( -L,t ) =0, \quad 0< t\leq T , \\ \frac{\partial}{\partial x} [ K_{1}D^{\alpha}U ( x,t ) +K_{2}D^{\beta}U ( x,t ) ] |_{x=0}=0,\quad 0<t\leq T, \\ U ( x,0 ) =g ( x ) -V ( x,0 ) ,\quad -L\leq x\leq 0. \end{array} \right . $$
(27)
We start by solving problem (26). We may look for a solution in the form
$$ V ( x,t ) =v_{1} ( t ) x+v_{2} ( t ) . $$
(28)
The boundary conditions imply
$$\begin{aligned}& v_{2} ( t ) -Lv_{1} ( t ) =l ( t ) , \end{aligned}$$
(29)
$$\begin{aligned}& K_{1}D^{\alpha}v_{1} ( t ) +K_{2}D^{\beta}v_{1} ( t ) =\Psi ( t ) . \end{aligned}$$
(30)
Therefore, from (28) and (29) we see that
$$ V ( x,t ) =(L+x)v_{1} ( t ) +l ( t ) . $$
(31)
The function \(v_{1} ( t ) \) may be found by the Laplace transform method. Indeed, assuming continuity of \(v_{1} ( t ) \) at zero, we have
$$ K_{1}s^{\alpha}{\overline{v}}_{1} ( s ) +K_{2}s^{\beta }{\overline{v}}_{1} ( s ) =\overline{ \Psi} ( s ) $$
or
$$ {\overline{v}}_{1} ( s ) =\frac{\overline{\Psi} ( s ) }{K_{1}s^{\alpha}+K_{2}s^{\beta}}= \frac{\overline{\Psi} ( s ) }{K_{1}} \frac{s^{-\beta}}{s^{\alpha-\beta}+\frac{K_{2}}{K_{1}}}. $$
Using (4), i.e.
$$ {{{\mathcal{L}}}^{-1} \biggl[ \frac{s^{\alpha-\beta}}{s^{\alpha}\mp a} \biggr] =t^{\beta-1}}E_{\alpha,\beta} \bigl( \pm at^{\alpha} \bigr), $$
we see that
$$ {{{\mathcal{L}}}^{-1} \biggl[ \frac{s^{-\beta}}{s^{\alpha-\beta }+K_{2}{K_{1}}^{-1}} \biggr] =t^{\alpha-1}}E_{\alpha-\beta,\alpha} \biggl( -\frac {K_{2}}{K_{1}}t^{\alpha-\beta} \biggr) . $$
Therefore
$$ v_{1} ( t ) = \biggl( \Psi\ast t^{\alpha-1}E_{\alpha-\beta ,\alpha } \biggl( -\frac{K_{2}}{K_{1}}t^{\alpha-\beta} \biggr) \biggr) ( t ) . $$
(32)
We pass now to problem (27). We write the equation there as
$$ \frac{\partial U}{\partial t}=K_{1}D^{\alpha}\frac{{\partial}^{2}U}{\partial x^{2}}+K_{2}D^{\beta} \frac{{\partial}^{2}U}{\partial x^{2}}+\tilde{f} ( x,t ) $$
(33)
with \(\tilde{f} ( x,t ) =f ( x,t ) -\frac{\partial V}{ \partial t}\). We look for a solution in the form
$$ U ( x,t ) =\sum_{n=0}^{\infty}{X_{n}}(x)T_{n}(t). $$
(34)
We assume that
$$ \tilde{f} ( x,t ) =\sum_{n=0}^{\infty}{A_{n}(t)} { X}_{n}(x), $$
(35)
where the coefficients \(A_{n}(t)\) may be computed by
$$ A_{n} ( t ) = \biggl( \int_{-L}^{0}{ \tilde{f} ( x,t ) {X}_{n}(x)\,dx} \biggr) { \biggl( \int _{-L}^{0}{{ X}_{n}(x){X}_{n}(x) \,dx} \biggr) }^{-1}. $$
The substitution of the expressions (34) and (35) in (33) yields
$$\begin{aligned} \frac{\partial}{\partial t} \Biggl( \sum_{n=0}^{\infty}{T_{n} ( t ) } {X}_{n} ( x ) \Biggr) =&K_{1}D^{\alpha} \Biggl( \sum_{n=0}^{\infty}{ T_{n} ( t ) X}^{\prime\prime}{_{n} ( x ) } \Biggr) \\ &{}+K_{2}D^{\beta} \Biggl( \sum_{n=0}^{\infty}{T_{n}(t)X}^{\prime\prime}{ _{n}(x)} \Biggr) +\sum_{n=0}^{\infty}{A_{n} ( t ) {X}_{n} ( x ) }. \end{aligned}$$
(36)
Before continuing with this identity we need to solve the homogeneous problem
$$ \left \{ \begin{array}{l} \frac{\partial W}{\partial t}=K_{1}D^{\alpha}\frac{{\partial}^{2}W}{ \partial x^{2}}+K_{2}D^{\beta}\frac{{\partial}^{2}W}{\partial x^{2}}\quad \mbox{in }{\Omega}_{e} , \\ W ( -L,t ) =0, \quad 0< t\leq T, \\ \frac{\partial}{\partial x} [ K_{1}D^{\alpha}W ( x,t ) +K_{2}D^{\beta}W ( x,t ) ] |_{x=0}=0,\quad 0<t\leq T. \end{array} \right . $$
(37)
Setting \(W ( x,t ) =X ( x ) T(t)\), we obtain by differentiation and substitution in (37)
$$\begin{aligned} X ( x ) T^{\prime} ( t ) &=K_{1}{X}^{\prime\prime }{(x)T}^{ ( \alpha ) } ( t ) +K_{2}{X}^{\prime\prime }{(x)T}^{ ( \beta ) } ( t ) \\ &= \bigl[ K_{1}T^{ ( \alpha ) } ( t ) +K_{2}T^{ ( \beta ) } ( t ) \bigr] X^{\prime\prime}(x) \end{aligned}$$
and separating variables it appears that
$$ \frac{T^{\prime} ( t ) }{K_{1}T^{ ( \alpha ) } ( t ) +K_{2}T^{ ( \beta ) } ( t ) }=\frac {X^{\prime \prime}(x)}{X ( x ) }=\mu=-{\lambda}^{2}. $$
The first boundary condition \(W ( -L,t ) =0\) implies \(X ( -L ) T ( t ) =0\), \(0< t\leq T\) and consequently \(X ( -L ) =0\). The second boundary condition gives
$$ K_{1}{X}^{\prime}{(0)T}^{ ( \alpha ) } ( t ) +K_{2}{ X^{\prime}(0)T}^{ ( \beta ) } ( t ) =0,\quad 0< t\leq T $$
or
$$ \bigl[ K_{1}T^{ ( \alpha ) } ( t ) +K_{2}T^{ ( \beta ) } ( t ) \bigr] X^{\prime} ( 0 ) =0,\quad 0< t\leq T. $$
We conclude that \(X^{\prime} ( 0 ) =0\). Solving the problem
$$ \left \{ \begin{array}{l} X^{\prime\prime}(x)+{\lambda}^{2}X(x)=0, -\quad L\leq x\leq0, \\ X ( -L ) =0,\qquad X^{\prime} ( 0 ) =0, \end{array} \right . $$
we obtain \({X}_{n} ( x ) ={\cos{\lambda}_{n}x}\) with \({\lambda}_{n}=\frac{ ( 2n+1 ) \pi}{2L}\). These eigenfunctions form a basis.
Now, in view of (36) we deduce that
$$ \sum_{n=0}^{\infty}{T}^{\prime}{_{n} ( t ) {X}_{n} ( x ) =\sum_{n=0}^{\infty}{ \bigl[ K_{1}T^{ ( \alpha ) } ( t ) +K_{2}T^{ ( \beta ) } ( t ) \bigr] } \bigl( -{\lambda}_{n}^{2}X_{n}(x) \bigr) }+\sum_{n=0}^{\infty}{A_{n} ( t ) {X}_{n}(x)}. $$
We get the system of equations
$$ T_{n}^{\prime} ( t ) +{\lambda}_{n}^{2} \bigl[ K_{1}T_{n}^{(\alpha )} ( t ) +K_{2}{T}_{n}^{{ ( \beta ) }} ( t ) \bigr] =A_{n} ( t ) ,\quad 0< t\leq T. $$
(38)
Notice that
$$ U ( x,0 ) =\sum_{n=0}^{\infty}{X_{n}}(x)T_{n}(0)=g(x)-V(x,0)=g(x)-l(0)=g(x)-g(-L). $$
This means that \(T_{n}(0)\) are the coefficients of the expansion of \(g(x)-g(-L)\) according to the basis \(\{{X_{n}}(x)\}_{n}\).
Applying Laplace transform to (38) we end up with
$$ s{\overline{T}}_{n} ( s ) +{\lambda}_{n}^{2} \bigl[ K_{1}s^{\alpha}{\overline{T}}_{n} ( s ) +K_{2}s^{\beta}{\overline {T}}_{n} ( s ) \bigr] ={ \overline{A}}_{n} ( s ) +T_{n}(0) $$
or
$$ \bigl[ s+{\lambda}_{n}^{2} \bigl( K_{1}s^{\alpha}+K_{2}s^{\beta} \bigr) \bigr] {\overline{T}}_{n} ( s ) ={\overline{A}}_{n} ( s ) +T_{n}(0). $$
That is,
$$ {\overline{T}}_{n} ( s ) =\frac{{\overline{A}}_{n} ( s ) +T_{n}(0)}{s+{\lambda}_{n}^{2} ( K_{1}s^{\alpha}+K_{2}s^{\beta } ) }. $$
(39)
Note that
$$\begin{aligned}& \frac{1}{s+{\lambda}_{n}^{2} ( K_{1}s^{\alpha}+K_{2}s^{\beta } ) } \\& \quad =\frac{s^{-\beta}}{s^{1-\beta}+K_{1}{\lambda}_{n}^{2}s^{\alpha-\beta }+K_{2}{\lambda}_{n}^{2}}=\frac{s^{-\beta}}{ ( s^{1-\beta}+K_{2}{ \lambda}_{n}^{2} ) ( 1+\frac{K_{1}{\lambda}_{n}^{2}s^{\alpha -\beta}}{s^{1-\beta}+K_{2}{\lambda}_{n}^{2}} ) } \\& \quad =\frac{s^{-\beta}}{s^{1-\beta}+K_{2}{\lambda}_{n}^{2}} \sum_{k=0}^{\infty}{{ \biggl( \frac{{-K}_{1}{\lambda}_{n}^{2}s^{\alpha -\beta}}{s^{1-\beta}+K_{2}{\lambda}_{n}^{2}} \biggr) }^{k}}= \sum _{k=0}^{\infty}\frac{{ ( {-K}_{1}{\lambda}_{n}^{2} ) } ^{k}s^{-\beta+k(\alpha-\beta)}}{{ ( s^{1-\beta}+K_{2}{\lambda} _{n}^{2} ) }^{k+1}}. \end{aligned}$$
(40)
Using (3) we see that
$$ \mathcal{L}^{-1} \biggl\{ \frac{s^{-\beta+k ( \alpha-\beta ) }}{{ ( s^{1-\beta}+K_{2}{\lambda}_{n}^{2} ) }^{k+1}} \biggr\} =t^{k ( 1-\alpha ) }E_{1-\beta,k ( 1-\alpha ) +1}^{k+1} \bigl( -K_{2}{ \lambda}_{n}^{2}t^{1-\beta} \bigr). $$
(41)
Therefore (39)-(41) imply that
$$\begin{aligned} \begin{aligned}[b] T_{n} ( t ) ={}&\int_{0}^{t}{A_{n} ( t-s ) \sum_{k=0}^{\infty }{{ \bigl( {-K}_{1}{\lambda}_{n}^{2} \bigr) }^{k}s^{k ( 1-\alpha ) }E_{1-\beta,k ( 1-\alpha ) +1}^{k+1} \bigl( -K_{2}{\lambda}_{n}^{2}s^{1-\beta} \bigr) \,ds}} \\ &{}+T_{n}(0){\sum_{k=0}^{\infty}{{ \bigl( {-K}_{1}{\lambda}_{n}^{2} \bigr) }^{k}t^{k ( 1-\alpha ) }E_{1-\beta,k ( 1-\alpha ) +1}^{k+1} \bigl( -K_{2}{\lambda}_{n}^{2}t^{1-\beta} \bigr) }}\end{aligned} \end{aligned}$$
(42)
and by (34) we have
$$ U ( x,t ) =\sum_{n=0}^{\infty}{T_{n} ( t ) } {\cos \frac{ ( 2n+1 ) \pi}{2L}x } $$
(43)
with \(T_{n} ( t ) \) as in (42). Gathering (43), (42), (32), and (31) in (25) we obtain the solution \(p(x,t)\) to the problem in \({\Omega}_{0}\).
As a conclusion we have proved the following result.
Theorem 3
The solution of problem (16) is given by (25) where
\(U ( x,t ) \)
is as in (43). The functions
\(v_{1} ( t ) \)
in
\(V ( x,t ) \) (see (31)) is given in (32). That is,
$$\begin{aligned} p ( x,t ) =&\sum_{n=0}^{\infty}{\cos \frac{ ( 2n+1 ) \pi}{2L}x } \\ &{}\times\Biggl[ \int_{0}^{t}{A_{n} ( t-s ) \sum_{k=0}^{\infty}{{ \bigl( {-K}_{1}{\lambda}_{n}^{2} \bigr) }^{k}s^{k ( 1-\alpha ) }E_{1-\beta ,k ( 1-\alpha ) +1}^{k+1} \bigl( -K_{2}{\lambda }_{n}^{2}s^{1-\beta } \bigr) \,ds}} \\ &{}+T_{n}(0){\sum_{k=0}^{\infty}{{ \bigl( {-K}_{1}{\lambda}_{n}^{2} \bigr) }^{k}t^{k ( 1-\alpha ) }E_{1-\beta,k ( 1-\alpha ) +1}^{k+1} \bigl( -K_{2}{\lambda}_{n}^{2}t^{1-\beta } \bigr) }} \Biggr] \\ &{}+(L+x) \biggl( \Psi\ast t^{\alpha-1}E_{\alpha-\beta,\alpha} \biggl( - \frac{K_{2}}{K_{1}}t^{\alpha-\beta} \biggr) \biggr) ( t ) +l(t), \end{aligned}$$
where
$$ \Psi(t) =\frac{-1}{\sqrt{K_{1}}}\int_{0}^{t}s^{(\alpha-1)/2}E_{\alpha-\beta ,(\alpha+1)/2}^{1/2} \bigl( -K_{2}K_{1}^{-1}s^{\alpha-\beta} \bigr) \bigl( K_{1}D^{\alpha}+K_{2}D^{\beta} \bigr) h^{\prime}(t-s)\,ds,$$
\({A_{n} ( t ) }\)
and
\(T_{n}(0)\)
are the coefficients of the expansions of
\(\tilde{f} ( x,t ) =f ( x,t ) -\frac{\partial V}{\partial t}\)
and
\(g(x)-g(-L)\), respectively, according to the basis
\(\{{X_{n}}(x)\}_{n}\).
The graphs (see Figure 1) correspond to \(p(x,0)=g(x)=\sin(4\pi x)\), \(p(-1,t)=l(t)=h(t)=t\).