Non-Local Problems for the Fractional Order Diffusion Equation and the Degenerate Hyperbolic Equation

Abstract

This research explores nonlocal problems associated with fractional diffusion equations and degenerate hyperbolic equations featuring singular coefficients in their lower-order terms. The uniqueness of the solution is established using the energy integral method, while the existence of the solution is equivalently reduced to solving Volterra integral equations of the second kind and a fractional differential equation. The study focuses on a mixed domain where the parabolic section aligns with the upper half-plane, and the hyperbolic section is bounded by two characteristics of the equation under consideration and a segment of the x-axis. By utilizing the solution representation of the fractional-order diffusion equation, a primary functional relationship is derived between the traces of the sought function on the x-axis segment from the parabolic part of the mixed domain. An explicit solution form for the modified Cauchy problem in the hyperbolic section of the mixed domain is presented. This solution, combined with the problem’s boundary condition, yields a fundamental functional relationship between the traces of the unknown function, mapped to the interval of the equation’s degeneration line. Through the conjugation condition of the problem, an equation with fractional derivatives is obtained by eliminating one unknown function from two functional relationships. The solution to this equation is explicitly formulated. For a specific solution of the proposed problem, visualizations are provided for various orders of the fractional derivative. The analysis demonstrates that the derivative order influences both the intensity of the diffusion (or subdiffusion) process and the shape of the wave front.

1. Introduction

Fractional partial differential equations can arise in the mathematical modeling of physical media with fractal geometry [1,2,3]. Boundary value problems for the fractional order diffusion equation were studied in [4,5,6,7,8,9,10,11]. In [12], a certain family of generalized derived Riemann–Liouville operators $D_{a+}^{\alpha ,\beta }$ of orders $\alpha$ and $\beta$ was studied. Applications of this operator are given in [13]. In [14], the unique solvability of the problem for the Riemann–Liouville partial fractional derivative equation with a boundary condition containing a generalized fractional integro-differentiation operator was investigated. The problem in which the boundary condition contains a linear combination of generalized fractional operators with a hyper-geometric Gaussian function for a mixed-type equation with the partial fractional Riemann–Liouville derivative was studied in [15]. The nonlocal boundary value problem for a mixed-type equation with the Riemann-Liouville fractional partial derivative was studied in [16]. The analysis of a non-local boundary value problem for the Gellerstedt equation featuring a singular coefficient in an unbounded domain was conducted in the study referenced in [17]. Reference [18] deals with the fractional generalization of the integro-differential diffusion-wave equation for the Heisenberg sub-Laplacian, with homogeneous Bitsadze–Samarskiy-type time-nonlocal conditions. In [19], important properties of the Riemann–Liouville derivative, one of the most used fractional derivatives, were studied.

The article’s research plan is structured as follows. Section 1 provides information about the problem under study and provides relevant links to articles. Section 2 provides the problem statement. Section 3 provides the main results: the issues of existence and uniqueness of the solution are investigated. Section 4 provides the statement of the proposed problem in a particular case, and Section 5 provides the methodology for solving this problem with its visualization.

We consider the equation

$$\left\{\begin{array}{c}{u}_{xx}-D_{0,y}^{\gamma }u=0,\gamma \in (0,1),y>0,\hfill \\ -{(-y)}^m{u}_{xx}+u_{yy}+{\displaystyle \frac{{\alpha }_0}{{(-y)}^{1-\frac{m}{2}}}}{u}_x+{\displaystyle \frac{{\beta }_0}{y}}{u}_y=0,y<0,\hfill \end{array}\right.$$

(1)

where $D_{0,y}^{\gamma }$ is the Riemann–Liouville fractional derivative of order $\gamma (0<\gamma <1)$ of $u(x,y)$ [20,21]

$$\left(D_{0,y}^{\gamma }u\right)(x,y)={\displaystyle \frac{\partial }{\partial y}}{\displaystyle \frac{1}{\Gamma (1-\gamma )}}\underset{0}{\overset{y}{\int }}{\displaystyle \frac{u(x,t)dt}{{(y-t)}^{\gamma }}},0<\gamma <1,y>0,$$

In (1) $m,{\alpha }_0,{\beta }_0$ are some real numbers satisfying conditions $m>0,|{\alpha }_0|<{\displaystyle \frac{m+2}{2}}$, $-{\displaystyle \frac{m}{2}}<{\beta }_0<1$.

Equation (1) for $y<0$ is an equation of a hyperbolic-type with a parabolic degeneration along the straight line $y=0$. As $m=2$, ${\beta }_0=0$, it passes to the moisture transfer equation [1], and for ${\alpha }_0=0$, ${\beta }_0=0$ from Equation (1) we arrive at the Gellerstedt equation, which finds application in the problem of determining the shape of the dam gate. A special case of Equation (1) for $y<0$ is also the Tricomi equation, which is the theoretical basis of transonic gas dynamics.

2. Formulation of a Problem

The Equation (1) is considered in domain $D=D^{+}\cup D^{-}\cup J$, where $D^{+}$ is the half-plane $y>0$, $D^{-}$ is the finite region of the fourth quadrant of the plane, limited by the characteristics

$$OC:x-{\displaystyle \frac{2}{m+2}}{(-y)}^{{\displaystyle \frac{m+2}{2}}}=0,BC:x+{\displaystyle \frac{2}{m+2}}{(-y)}^{{\displaystyle \frac{m+2}{2}}}=1$$

of Equation (1) coming from points $O(0,0)$, $B(1,0)$ and intersecting at point $C({\displaystyle \frac{1}{2}}$, $-{\left({\displaystyle \frac{m+2}{4}}\right)}^{{\displaystyle \frac{2}{m+2}}})$ and the segment $OB$ of the straight line $y=0$.

Let us introduce the notation: $J=(0,1)$, which is the unit interval of the line $y=0$, ${\Theta }_0\left(x\right)=\left({\displaystyle \frac{x}{2}},-{\left({\displaystyle \frac{m+2}{4}}x\right)}^{{\displaystyle \frac{2}{m+2}}}\right)$ is the intersection point of the characteristic of Equation (1) coming from the point $(x,0)\in J$ with the characteristic $OC$.

$\left(I_{0+}^{\mu ,\rho ,\eta }f\right)\left(x\right)$ is a generalized fractional integro-differentiation operator with a hypergeometric Gauss function $F(a,b,c;z)$ introduced by M. Saigo [22], with the form for real $\mu ,\rho ,\eta$ and $x>0$

$$\left(I_{0+}^{\mu ,\rho ,\eta }f\right)\left(x\right)=\left\{\begin{array}{c}{\displaystyle \frac{x^{-\mu -\rho }}{\Gamma \left(\mu \right)}}\underset{0}{\overset{x}{\int }}{(x-t)}^{\mu -1}F\left(\mu +\rho ,-\eta ;\mu ;1-{\displaystyle \frac{t}{x}}\right)f\left(t\right)dt,\mu >0,\hfill \\ {\left({\displaystyle \frac{d}{dx}}\right)}^n\left(I_{0+}^{\mu +n,\rho -n,\eta -n}f\right)\left(x\right),\mu \le 0,n=[-\mu ]+1.\hfill \end{array}\right.$$

(2)

in particular

$$\left(I_{0+}^{0,0,\eta }f\right)\left(x\right)=f\left(x\right),$$

(3)

Note that if $\mu >0$, then the formulas are valid

$$\left(I_{0+}^{\mu ,-\mu ,\eta }f\right)\left(x\right)=\left(I_{0+}^{\mu }f\right)\left(x\right),\left(I_{0+}^{-\mu ,\mu ,\eta }f\right)\left(x\right)=\left(D_{0+}^{\mu }f\right)\left(x\right),$$

(4)

where $\left(I_{0+}^{\mu }f\right)\left(x\right)$ and $\left(D_{0+}^{\mu }f\right)\left(x\right)$ are fractional Riemann–Liouville integration and differentiation operators of the order $\mu >0$ [20];

$$\left(I_{0+}^{\mu }f\right)\left(x\right)={\displaystyle \frac{1}{\Gamma \left(\mu \right)}}\underset{0}{\overset{x}{\int }}{(x-t)}^{\mu -1}f\left(t\right)dt,\mu >0,x>0,$$

$$\left(D_{0+}^{\mu }f\right)\left(x\right)={\left({\displaystyle \frac{d}{dx}}\right)}^n{\displaystyle \frac{1}{\Gamma (n-\mu )}}\underset{0}{\overset{x}{\int }}{(x-t)}^{n-\mu -1}f\left(t\right)dt,\mu >0,n=\left[\mu \right]+1.$$

(5)

Problem A. Find a solution $u(x,y)$ to Equation (1) in domain D satisfying the boundary conditions

$$y^{1-\gamma }{u|}_{y=0}=0,(-\infty <x\le 0,1\le x<\infty ),$$

(6)

$$A_1{x}^{1+b-\alpha -\beta }\left(I_{0+}^{a,b,-a-\alpha }{t}^{\alpha +\beta -1}u\left[{\Theta }_0\left(t\right)\right]\right)\left(x\right)+A_2\left(I_{0+}^{a+\alpha ,0,\beta -1-a-b}u(t,0)\right)=g\left(x\right),(x\in J)$$

(7)

and the transmission conditions

$$\underset{y\to 0-}{lim}u(x,y)=c\left(x\right)\underset{y\to 0+}{lim}{y}^{1-\gamma }u(x,y),\forall x\in \overline{J},$$

(8)

$$\underset{y\to 0-}{lim}{(-y)}^{{\beta }_0}{u}_y(x,y)=d\left(x\right)\underset{y\to 0+}{lim}{y}^{1-\gamma }{\left(y^{1-\gamma }u(x,y)\right)}_y,\forall x\in J,$$

(9)

Here, $\alpha ={\displaystyle \frac{m+2({\beta }_0+{\alpha }_0)}{2(m+2)}},\beta ={\displaystyle \frac{m+2({\beta }_0-{\alpha }_0)}{2(m+2)}},$ $0<\alpha <{\displaystyle \frac{1}{2}},0<\beta <{\displaystyle \frac{1}{2}},$$a>max\{-\alpha ,1-\beta \}$, $A_1,A_2$ are valid constants, $g\left(x\right),c\left(x\right),d\left(x\right)$ are given, functions such that

$$g\left(x\right)\in C^1\left(\overline{J}\right)\cap C^2\left(J\right),c\left(x\right),d\left(x\right)\in C^2\left(\overline{J}\right)\cap C^3\left(J\right),c\left(x\right)d\left(x\right)>0,{\displaystyle \frac{d^2}{d{x}^2}}\left[c\left(x\right)d\left(x\right)\right]\le 0.$$

(10)

Non-local boundary value problems for Equation (1) in unbounded and bounded domains are studied in [23,24].

We will look for a solution to the $u(x,y)$ problem in the class of doubly differentiable functions in domain D, such that

$$y^{1-\gamma }u(x,y)\in C\left({\overline{D}}^{+}\right),u(x,y)\in C\left({\overline{D}}^{-}\right),$$

$$y^{1-\gamma }{\left(y^{1-\gamma }u(x,y)\right)}_y\in C(D^{+}\cup \{(x,y):0<x<1,y=0\}),$$

$$u_{xx}\in C(D^{+}\cup D^{-}),u_{yy}\in C\left(D^{-}\right).$$

3. Main Results

3.1. Uniqueness of the Solution of the Problem

Theorem 1. 

Let us $A_1\le 0$, $A_2>0,$ $c\left(x\right)d\left(x\right)>0,{\displaystyle \frac{d^2}{d{x}^2}}\left(c\left(x\right)d\left(x\right)\right)\le 0.$ Then, problem A, has only a trivial solution.

Proof of Theorem 1. 

Let us introduce the following notation

$$\underset{y\to 0+}{lim}{y}^{1-\gamma }u(x,y)={\tau }_1\left(x\right),\underset{y\to 0-}{lim}u(x,y)={\tau }_2\left(x\right),$$

(11)

$$\underset{y\to 0+}{lim}{y}^{1-\gamma }{\left(y^{1-\gamma }u(x,y)\right)}_y={\nu }_1\left(x\right),\underset{y\to 0-}{lim}{(-y)}^{{\beta }_0}{u}_y(x,y)={\nu }_2\left(x\right).$$

(12)

It is known [25] that the solution of Equation (1) in domain $D^{+}$ satisfies the condition (6) and the condition

$$\underset{y\to 0+}{lim}{y}^{1-\gamma }u(x,y)={\tau }_1\left(x\right),\forall x\in \overline{J}$$

(13)

is given by the formula

$$u(x,y)=\underset{0}{\overset{1}{\int }}G(x,y,t){\tau }_1\left(t\right)dt,$$

(14)

where $G(x,y,t)={\displaystyle \frac{\Gamma \left(\gamma \right)}{2}}{y}^{{\displaystyle \frac{\gamma }{2}}-1}{e}_{1,{\displaystyle \frac{\gamma }{2}}}^{1,{\displaystyle \frac{\gamma }{2}}}\left(-|x-t|y^{-{\displaystyle \frac{\gamma }{2}}}\right),$ $e_{1,{\displaystyle \frac{\gamma }{2}}}^{1,{\displaystyle \frac{\gamma }{2}}}\left(z\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{z^k}{k!\Gamma \left(\frac{(1-k)\gamma }{2}\right)}}$ is a Wright-type function [25].

It is also known [26] that the functional relation between ${\tau }_1\left(x\right)$ and ${\nu }_1\left(x\right)$, brought from the parabolic part of $D^{+}$ to the line $y=0$ has the form

$${\nu }_1\left(x\right)={\displaystyle \frac{1}{\Gamma (1+\gamma )}}{\tau }_1^{\prime \prime }\left(x\right).$$

(15)

Let us find the ratio between ${\tau }_2\left(x\right)$ and ${\nu }_2\left(x\right)$ brought to the line $y=0$ from the hyperbolic part $D^{-}$ of domain D.

The solution of the modified Cauchy problem, in domain $D^{-}$, has the form [23]

$$u(x,y)={\gamma }_1\underset{0}{\overset{1}{\int }}{\tau }_2\left(x+{\displaystyle \frac{2}{m+2}}(2t-1){(-y)}^{{\displaystyle \frac{m+2}{2}}}\right)t^{\beta -1}{(1-t)}^{\alpha -1}dt+$$

$$+{\gamma }_2{(-y)}^{1-{\beta }_0}\underset{0}{\overset{1}{\int }}{\nu }_2\left(x+{\displaystyle \frac{2}{m+2}}(2t-1){(-y)}^{{\displaystyle \frac{m+2}{2}}}\right)t^{-\alpha }{(1-t)}^{-\beta }dt,$$

(16)

where ${\gamma }_1={\displaystyle \frac{\Gamma (\alpha +\beta )}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}},{\gamma }_2=-{\displaystyle \frac{2\Gamma (1-\alpha -\beta )}{(m+2)\Gamma (1-\alpha )\Gamma (1-\beta )}}.$

Using the Formula (16) and the ratio (2), we have

$$u\left[{\Theta }_0\left(x\right)\right]={\gamma }_1\Gamma \left(\alpha \right)\left(I_{0+}^{\alpha ,0,\beta -1}{\tau }_2\right)\left(x\right)+{\gamma }_2{\left({\displaystyle \frac{m+2}{4}}\right)}^{1-\alpha -\beta }\Gamma (1-\beta )\left(I_{0+}^{1-\beta ,\alpha +\beta -1,\beta -1}{\nu }_2\right)\left(x\right).$$

Substituting $u\left[{\Theta }_0\left(x\right)\right]$ into the boundary condition (7) and applying successively the relations [22]

$$x^{\mu +\rho +\eta }\left(I_{0+}^{\mu ,\rho ,\eta }\phi \right)\left(x\right)=\left(I_{0+}^{\mu ,-\mu -\eta ,-\mu -\rho }\phi \right)\left(x\right),\mu >0,$$

(17)

$$\left(I_{0+}^{\mu ,\rho ,\eta }{I}_{0+}^{\sigma ,\delta ,\mu +\eta }\phi \right)\left(x\right)=\left(I_{0+}^{\mu +\sigma ,\rho +\delta ,\eta }\phi \right)\left(x\right),\mu >0,$$

(18)

after simple calculations, we obtain

$$k_1\left(I_{0+}^{a+\alpha ,0,\beta -1-a-b}{\tau }_2\right)\left(x\right)+k_2\left(I_{0+}^{a+1-\beta ,\alpha +\beta -1,\beta -1-a-b}{\nu }_2\right)\left(x\right)=g\left(x\right)$$

(19)

where $k_1=A_1{\gamma }_1\Gamma \left(\alpha \right)-A_2,k_2=A_1{\gamma }_2\Gamma (1-\beta ){\left({\displaystyle \frac{m+2}{2}}\right)}^{1-\alpha -\beta }.$

Applying the operator ${\left(I_{0+}^{a+\alpha ,0,\beta -1-a-b}\right)}^{-1}=I_{0+}^{-a-\alpha ,0,\alpha +\beta -1-b}$ to both parts of the resulting equality, taking into account (18), (3), (4), and $\left(D_{0+}^{\alpha }\left(I_{0+}^{\alpha }f\right)\left(t\right)\right)\left(x\right)=f\left(x\right)$ we have

$$k_1{\tau }_2\left(x\right)+k_2\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\right)\left(x\right)=g_1\left(x\right)$$

(20)

where $g_1\left(x\right)=\left(I_{0+}^{-a-\alpha ,0,\alpha +\beta -1-b}g\right)\left(x\right).$

Consider the corresponding homogeneous problem: $g\left(x\right)=0.$

Consider the following two cases:

(a)

Let $k_1\ne 0$:

$$A_1\Gamma (\alpha +\beta )-A_2\Gamma \left(\beta \right)\ne 0.$$

(21)

Then, the ratio (20) takes the form

$${\tau }_2\left(x\right)=k_3\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\right)\left(x\right)+{\displaystyle \frac{1}{k_1}}{g}_1\left(x\right),$$

(22)

where $k_3=-{\displaystyle \frac{k_2}{k_1}}.$

Let us evaluate the following integral

$$I=\underset{0}{\overset{1}{\int }}{\tau }_2\left(x\right){\nu }_2\left(x\right)dx.$$

Due to the conjugation conditions (8), (9) and ratio (11), (12), we have

$${\tau }_2\left(x\right)=c\left(x\right){\tau }_1\left(x\right),{\nu }_2\left(x\right)=d\left(x\right){\nu }_1\left(x\right),$$

(23)

and, therefore, by virtue of ratio (15), we have

$$I={\displaystyle \frac{1}{\Gamma (1+\gamma )}}\underset{0}{\overset{1}{\int }}c\left(x\right){\tau }_1\left(x\right)d\left(x\right){\tau }_1^{″}\left(x\right)dx.$$

Integrating in parts and considering that according to (6) and (13) ${\tau }_1\left(0\right)={\tau }_1\left(1\right)=0$, we obtain

$$I={\displaystyle \frac{1}{\Gamma (1+\gamma )}}\left[{\displaystyle \frac{1}{2}}\underset{0}{\overset{1}{\int }}{\tau }_1^2\left(x\right){\displaystyle \frac{d^2}{d{x}^2}}\left(c\left(x\right)d\left(x\right)\right)dx-\underset{0}{\overset{1}{\int }}{\left({\tau }_1^{\prime }\left(x\right)\right)}^{2}c\left(x\right)d\left(x\right)dx\right].$$

(24)

Hence, due to the conditions (10), we obtain an estimate in the domain of $D^{+}$ for the integral:

$$I\le 0.$$

(25)

Now, let us find the estimate of the integral I in domain $D^{-}$.

For $g\left(x\right)=0$, the equality (22) takes the form

$${\tau }_2\left(x\right)=k_3\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\left(t\right)\right)\left(x\right)={\displaystyle \frac{k_3}{\Gamma (1-\alpha -\beta )}}\underset{0}{\overset{x}{\int }}{\nu }_2\left(t\right){(x-t)}^{-\alpha -\beta }dt$$

and, therefore,

$$I={\displaystyle \frac{k_3}{\Gamma (1-\alpha -\beta )}}\underset{0}{\overset{1}{\int }}{\nu }_2\left(x\right)dx\underset{0}{\overset{x}{\int }}{(x-t)}^{-\alpha -\beta }{\nu }_2\left(t\right)dt.$$

Next, we will employ the well-established formula for the gamma function $\Gamma \left(\sigma \right)$ [27]

$$\underset{0}{\overset{\infty }{\int }}{s}^{\sigma -1}cos\left(ks\right)ds={\displaystyle \frac{\Gamma \left(\sigma \right)}{k^{\sigma }}}cos\left({\displaystyle \frac{\sigma \pi }{2}}\right),(k>0,0<\sigma <1).$$

Assuming $k=\mid x-t\mid ,\sigma =\alpha +\beta$ to it, we obtain

$$\mid x-t{\mid }^{-\alpha -\beta }=$$

$$={\displaystyle \frac{1}{\Gamma (\alpha +\beta )cos\left(\pi \frac{\alpha +\beta }{2}\right)}}\underset{0}{\overset{\infty }{\int }}{s}^{\alpha +\beta -1}cos(s\mid x-t\mid )ds,(0<\alpha +\beta <1).$$

Applying this formula and the Dirichlet formula of the permutation of the order of integration in the repeated integral, we arrive at the relation

$$I={\displaystyle \frac{k_{3}sin\left(\pi \frac{\alpha +\beta }{2}\right)}{\pi }}\underset{0}{\overset{\infty }{\int }}{s}^{\alpha +\beta -1}\left[{\left(\underset{0}{\overset{1}{\int }}{\nu }_2\left(x\right)cos\left(sx\right)dx\right)}^2+{\left(\underset{0}{\overset{1}{\int }}{\nu }_2\left(x\right)sin\left(sx\right)dx\right)}^2\right]ds.$$

From the conditions of the theorem, we obtain

$$I\ge 0.$$

(26)

It follows from (25) and (26) that $I=0$, and, therefore, according to (24)

$${\displaystyle \frac{1}{2}}\underset{0}{\overset{1}{\int }}{\tau }_1^2\left(x\right){\displaystyle \frac{d^2}{d{x}^2}}\left(c\left(x\right)d\left(x\right)\right)dx-\underset{0}{\overset{1}{\int }}{\left({\tau }_1^{\prime }\left(x\right)\right)}^{2}c\left(x\right)d\left(x\right)dx=0.$$

Hence, by virtue of the conditions (10) and the equalities ${\tau }_1\left(0\right)={\tau }_1\left(1\right)=0$, we obtain

$${\tau }_1\left(x\right)=0,x\in \overline{J}.$$

(27)

(b)

Let $k_1=0,k_2\ne 0$:

$$A_1\Gamma (\alpha +\beta )-A_2\Gamma \left(\beta \right)=0,A_1\ne 0.$$

(28)

Then, (20) is a homogeneous Abel equation:

$$k_2\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\right)\left(x\right)=0$$

having only a trivial solution ${\nu }_2\left(x\right)=0$.

Then, by virtue of the second Formula (23) ${\nu }_1\left(x\right)=0$, the ratio (15) in the place with the conditions ${\tau }_1\left(0\right)={\tau }_1\left(1\right)=0$ leads to the equality (27). This, according to (14), means that $u(x,y)\equiv 0$ in domain of $\overline{D}$, which proves the uniqueness of the solution to the original problem under the conditions (21) and (28). □

3.2. The Existence of a Solution to Problem A

Theorem 2. 

Let (a) $k_1\ne 0,k_2=0;$ (b) $k_1=0,k_2\ne 0;$ (c) $c\left(x\right)=c=\mathrm{const};$ (d) $c\left(x\right)=c=\mathrm{const},$ $d\left(x\right)=d=\mathrm{const}.$ Then, there is a solution to problem A.

Proof of Theorem 2. 

(According to (14) and (15), to prove the existence of a solution to problem A, it is enough to find ${\nu }_1\left(x\right)$.

By virtue of (23), Equation (20) takes the form

$$k_{1}c\left(x\right){\tau }_1\left(x\right)+k_2\left(I_{0+}^{1-\alpha -\beta }d\left(t\right){\nu }_1\left(t\right)\right)\left(x\right)=g_1\left(x\right).$$

(29)

Consider the case $k_1\ne 0,k_2=0,$ then (29) provides an explicit expression for ${\tau }_1\left(x\right),$ that is, ${\tau }_1\left(x\right)={\displaystyle \frac{g_1\left(x\right)}{k_{1}c\left(x\right)}}$ and ${\nu }_1\left(x\right)$ are found by the Formula (15).

If $k_1=0,k_2\ne 0,$ then (29) is an Abel integral equation of the first kind.

$$k_2\left(I_{0+}^{1-\alpha -\beta }d\left(t\right){\nu }_1\left(t\right)\right)\left(x\right)=g_1\left(x\right).$$

(30)

with $0<\alpha +\beta <1.$ According to the condition (10) $g\left(x\right)\in C^1\left(\overline{J}\right)\cup C^2\left(J\right).$ Function $g_1\left(x\right)$ is also continuous [15] and the well-known solution of the Equation (30) [20] provides an explicit expression for ${\nu }_1\left(x\right)$ in the form

$${\nu }_1\left(x\right)={\displaystyle \frac{1}{k_{2}d\left(x\right)}}{\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{\Gamma (\alpha +\beta )}}\underset{0}{\overset{x}{\int }}{(x-t)}^{\alpha +\beta -1}{g}_1\left(t\right)dt.$$

If $c\left(x\right)=c\ne 0,$ then Equation (29) takes the form

$$c{k}_1{\tau }_1\left(x\right)+k_2\left(I_{0+}^{1-\alpha -\beta }d\left(t\right){\nu }_1\left(t\right)\right)\left(x\right)=g_1\left(x\right).$$

(31)

Differentiate both parts of the ratio (31) by x twice and taking into account (5), (15), we obtain

$$c{k}_1\Gamma (1+\gamma ){\nu }_1\left(x\right)+k_2\left(D_{0+}^{1+\alpha +\beta }d\left(t\right){\nu }_1\left(t\right)\right)\left(x\right)=g_1^{″}\left(x\right).$$

(32)

As you know, [20] if

$${\nu }_2\left(x\right)=d\left(x\right){\nu }_1\left(x\right)=\underset{y\to 0-}{lim}{(-y)}^{{\beta }_0}{u}_y(x,y)\in C\left([0,1]\right),$$

(33)

then the formula is correct

$$\left(I_{0+}^{1+\alpha +\beta }{D}_{0+}^{1+\alpha +\beta }{\nu }_2\left(t\right)\right)\left(x\right)={\nu }_2\left(x\right)-c_1{x}^{\alpha +\beta }-c_2{x}^{\alpha +\beta -1},$$

(34)

where

$$c_1={\displaystyle \frac{1}{\Gamma (\alpha +\beta +1)}}{\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\right)}^{\prime }(0+),c_1={\displaystyle \frac{1}{\Gamma (\alpha +\beta )}}\left(I_{0+}^{1-\alpha -\beta }{\nu }_2\right)(0+)$$

(35)

If the condition (33) is satisfied, then applying the operator $\left(I_{0+}^{1+\alpha +\beta }\right)$ to both parts of (32) and considering (34) we arrive at the integral equation:

$$k_{2}d\left(x\right){\nu }_1\left(x\right)+c{k}_1\Gamma (1+\gamma )\left(I_{0+}^{1+\alpha +\beta }{\nu }_1\left(t\right)\right)\left(x\right)=g_2\left(x\right),$$

(36)

where

$$g_2\left(x\right)=\left(I_{0+}^{1+\alpha +\beta }{g}_1^{″}\left(t\right)\right)\left(x\right)+k_2[c_1{x}^{\alpha +\beta }+c_2{x}^{\alpha +\beta -1}].$$

(37)

If $k_{2}d\left(x\right)\ne 0,$ then (36) Volterra integral equation of the second kind [28]

$${\nu }_1\left(x\right)+\underset{0}{\overset{x}{\int }}K(x,t){\nu }_1\left(t\right)dt=F\left(x\right)$$

(38)

with a continuous core $K(x,t)={\displaystyle \frac{c{k}_1\Gamma (1+\gamma )}{k_2\Gamma (1+\alpha +\beta )d\left(x\right)}}{(x-t)}^{\alpha +\beta }$ and the free term $F\left(x\right)={\displaystyle \frac{1}{k_{2}d\left(x\right)}}{g}_2\left(x\right),$ where function $g_2\left(x\right)$ is given by Formula (37), and the constants $c_1,c_2$ are given by Formula (35).

It is known [6] that Equation (38) has a unique solution ${\nu }_1\left(x\right)$.

If $c\left(x\right)=c\ne 0,$ $d\left(x\right)=d\ne 0$, then Equation (29) reduces to a fractional differential equation [6]:

$$\left(D_{0+}^{1+\alpha +\beta }{\nu }_1\left(t\right)\right)\left(x\right)+k_4{\nu }_1\left(x\right)=\mathsf{\Phi }\left(x\right),$$

(39)

where $k_4={\displaystyle \frac{c{k}_1\Gamma (1+\gamma )}{k_{2}d}},\mathsf{\Phi }\left(x\right)={\displaystyle \frac{1}{k_{2}d}}{g}_1^{″}\left(x\right).$

In [6], an explicit solution of the ${\nu }_1\left(x\right)$ Equation (39) is written out, which, according to (14), completes the proof of the existence of a solution to the original problem. □

4. Formulation of a Problem in Case ${\beta }_0=-{\displaystyle \frac{m}{2}}$

Let ${\alpha }_0=0$, ${\beta }_0=-{\displaystyle \frac{m}{2}}$, then the Equation (1) takes the form

$$\left\{\begin{array}{c}{u}_{xx}-D_{0,y}^{\gamma }u=0,\gamma \in (0,1),y>0,\hfill \\ -{(-y)}^m{u}_{xx}+u_{yy}-{\displaystyle \frac{m}{2y}}{u}_y=0,y<0.\hfill \end{array}\right.$$

(40)

In this case, let us study the following problem.

Problem B. Find a solution $u(x,y)$ of Equation (40) satisfying the boundary condition (6) and the condition

$${\displaystyle \frac{d}{dx}}u\left[{\Theta }_0\left(x\right)\right]={\displaystyle \frac{d}{dx}}u(x,0)+\delta \left(x\right),$$

(41)

and the transmission conditions

$$\underset{y\to 0-}{lim}u(x,y)=\underset{y\to 0+}{lim}{y}^{1-\gamma }u(x,y),\forall x\in \overline{J},$$

(42)

$$\underset{y\to 0-}{lim}{(-y)}^{-{\displaystyle \frac{m}{2}}}{u}_y(x,y)=\underset{y\to 0+}{lim}{y}^{1-\gamma }{\left(y^{1-\gamma }u(x,y)\right)}_y,\forall x\in J.$$

(43)

Here $\delta \left(x\right)$ is given function such that

$$\delta \left(x\right)\in C^1\left(\overline{J}\right)\cap C^2\left(J\right).$$

5. Solution Methodology

Solving a modified Cauchy problem with initial data

$$u(x,0)=\tau \left(x\right),x\in \overline{J},\underset{y\to 0-}{lim}{(-y)}^{-{\displaystyle \frac{m}{2}}}{u}_y(x,y)=\nu \left(x\right),x\in J,$$

in domain $D^{-}$ for equation $-{(-y)}^m{u}_{xx}+u_{yy}-{\displaystyle \frac{m}{2y}}{u}_y=0$ is provided by the d’Alembert’s formula

$$u(x,y)={\displaystyle \frac{1}{2}}\left[\tau \left(x-{\displaystyle \frac{2}{m+2}}{(-y)}^{{\displaystyle \frac{m+2}{2}}}\right)+\tau \left(x+{\displaystyle \frac{2}{m+2}}{(-y)}^{{\displaystyle \frac{m+2}{2}}}\right)\right]-$$

$$-{\displaystyle \frac{2}{m+2}}{(-y)}^{{\displaystyle \frac{m+2}{2}}}\underset{0}{\overset{1}{\int }}\nu \left(x+{\displaystyle \frac{2}{m+2}}(2t-1){(-y)}^{{\displaystyle \frac{m+2}{2}}}\right)dt.$$

(44)

From (44), we calculate

$$u\left[{\Theta }_0\left(x\right)\right]={\displaystyle \frac{1}{2}}\tau \left(x\right)-{\displaystyle \frac{1}{2}}\underset{0}{\overset{x}{\int }}\nu \left(z\right)dz.$$

Substituting $u\left[{\Theta }_0\left(x\right)\right]$ into the condition (41), we obtain the second functional relation between the unknown functions $\tau \left(x\right)$ and $\nu \left(x\right):$

$${\tau }^{\prime }\left(x\right)+\nu \left(x\right)=-2\delta \left(x\right).$$

(45)

Excluding $\nu \left(x\right)$ from Equations (15) and (45), we obtain

$${\tau }^{\prime }\left(x\right)+{\displaystyle \frac{1}{\Gamma (1+\gamma )}}{\tau }^{″}\left(x\right)=-2\delta \left(x\right).$$

(46)

Applying the method of variation of constants to Equation (46), we have

$$\tau \left(x\right)=C_1+C_2{e}^{-\lambda x}+{\displaystyle \frac{2}{\lambda }}\underset{0}{\overset{x}{\int }}\delta \left(s\right)\left(e^{\lambda (s-x)}-1\right)ds,$$

(47)

where $\lambda =\Gamma (1+\gamma ),C_1,C_2-$ are constant values.

From (47), taking into account (45), we obtain

$$\nu \left(x\right)=\left({\displaystyle \frac{2}{\lambda }}-2\right)\delta \left(x\right)+C_2\lambda e^{-\lambda x}+2\underset{0}{\overset{x}{\int }}{e}^{\lambda (s-x)}\delta \left(s\right)ds,$$

Given that $\tau \left(0\right)=\tau \left(1\right)=0$, from (47), it is easy to show that

$C_1={\displaystyle \frac{2}{1-e^{\lambda }}}\underset{0}{\overset{1}{\int }}\delta \left(s\right)\left(e^{\lambda (s-1)}-1\right)ds,$$C_2=-{\displaystyle \frac{2}{1-e^{\lambda }}}\underset{0}{\overset{1}{\int }}\delta \left(s\right)\left(e^{\lambda (s-1)}-1\right)ds.$

Using the found $\tau \left(x\right)$ and $\nu \left(x\right)$, it is easy to obtain a solution to problem B in domains $D^{+}$ and $D^{-}$, which means that the solution to the problem (40), (6), and (41) in a given class of functions in the domain D, satisfying the boundary conditions (6), (41) and the gluing conditions (42) and (43).

Let us consider an example of solving problem B for a specific type of function $\delta \left(x\right)$ with its visualization. Visualization was performed using PyCharm 2024.1 software in the Python language [29].

Example 1. 

Let $\delta \left(x\right)=x^{{\displaystyle \frac{3}{2}}}{(1-x)}^{{\displaystyle \frac{3}{2}}}$.

Then, Equalities (47) and () take the following form

$$\tau \left(x\right)=C_1+C_2{e}^{-\lambda x}+{\displaystyle \frac{2}{\lambda }}\underset{0}{\overset{x}{\int }}{s}^{{\displaystyle \frac{3}{2}}}{(1-s)}^{{\displaystyle \frac{3}{2}}}{e}^{\lambda (s-x)}ds,$$

$$\nu \left(x\right)=\left({\displaystyle \frac{2}{\lambda }}-2\right)x^{{\displaystyle \frac{3}{2}}}{(1-x)}^{{\displaystyle \frac{3}{2}}}+C_2\lambda e^{-\lambda x}+2\underset{0}{\overset{x}{\int }}{s}^{{\displaystyle \frac{3}{2}}}{(1-s)}^{{\displaystyle \frac{3}{2}}}{e}^{\lambda (s-x)}ds,$$

where

$$C_1={\displaystyle \frac{2\sqrt{\pi }\Gamma \left(2.5\right)e^{\frac{-\lambda }{2}}{I}_2(-0.5\lambda )}{{\lambda }^2(1-e^{\lambda })}}-{\displaystyle \frac{2B(2.5,2.5)}{1-e^{\lambda }}},C_2={\displaystyle \frac{2\sqrt{\pi }\Gamma \left(2.5\right)e^{\frac{-\lambda }{2}}{I}_2(-0.5\lambda )}{{\lambda }^2(1-e^{\lambda })}}+{\displaystyle \frac{2B(2.5,2.5)}{1-e^{\lambda }}}.$$

Here, $I_2\left(x\right)$ is the modified Bessel function, $\Gamma \left(x\right)$ is the gamma function, and $B(x,y)$ is the beta function.

Figure 1 shows the graphs of functions $\tau \left(x\right)$ and $\nu \left(x\right)$ for $\gamma =0.9$ and $\delta \left(x\right)=x^{{\displaystyle \frac{3}{2}}}{\left(1-x\right)}^{{\displaystyle \frac{3}{2}}}$.

Figure 1. Graphs of functions for $\delta \left(x\right)=x^{{\displaystyle \frac{3}{2}}}{\left(1-x\right)}^{{\displaystyle \frac{3}{2}}}$: (a) $\tau \left(x\right)$; (b) $\nu \left(x\right)$.

Using the obtained functions $\tau \left(x\right)$ and $\nu \left(x\right)$, we can obtain a solution to the problem in domain $D^{+}$ and $D^{-}$, respectively, using Formulas (14) and (42).

Let us present graphs of the solution $u\left(x,y\right)$ of Problem B using Formulas (14) and (42) depending on the values of parameter $\gamma$.

In Figure 2, we can see that when changing the parameter $\gamma$ in domain $D^{+}$, the subdiffusion mode is enhanced due to the fact that the diffusion process proceeds more slowly than normal diffusion. We see that the region of positive values of the solution function $u(x,y)$ expands, and the region of negative values, on the contrary, narrows. In domain $D^{-}$ the wave mode proceeds, the shape of which is also affected by the values of parameter $\gamma$.

Figure 2. Graphs of the solution of Problem B for $m=1.5$ and different values of $\gamma$: (a) $\gamma =1$; (b) $\gamma =0.8$; (c) $\gamma =0.6$; (d) $\gamma =0.4$.

6. Discussion

The properties of solution of Equation (1) at $y<0$ essentially depend on coefficients ${\alpha }_0$ and ${\beta }_0$, at the lowest terms of Equation (1). If ${\beta }_0=1$, then the solution to Equation (1) on the parabolic degeneracy line has a logarithmic singularity. In this case, boundary value problems for Equation (1) at $y<0$ are studied with different conditions.

7. Conclusions

In this work, we study a boundary value problems for a differential equation with partial fractional derivative and degenerate hyperbolic equation. The main results are new. Using these results, we can explore various boundary value problems for differential equations with a partial fractional derivative of the second and higher orders.

The research illustrates a particular case solution for a non-local problem, complete with graphical representations of the functions. It is demonstrated that the fractional derivative’s order affects the diffusion process, leading to a slowdown consistent with subdiffusion phenomena. Additionally, the study reveals that the fractional derivative’s order plays a role in determining the wave front’s configuration.

The theoretical results obtained in the article will be used in modeling various modes of transport of radioactive radon gas as a precursor to strong earthquakes [30].