**3. Main Result**

Consider the problem

$$\mathbf{P}\mathbf{x}(t) = h(t), \quad \mathbf{t} \in [-r, 0], \quad (\mathbf{P}\mathbf{x})^{(k)}(0) = \mathbf{x}\_k, \quad k = 0, 1, \ldots, m - 1,\tag{7}$$

for the degenerate fractional evolution equation with delay

$$LD\_t^\mathbf{x} \mathbf{x}(t) = Mx(t) + \int\_{-r}^0 \mathcal{K}(\mathbf{s}) \mathbf{x}(t+s) ds + \mathbf{g}(t), \quad t \in [0, T], \tag{8}$$

where *h*(0) = *x*0, K : [−*r*, 0] → L(X ; Y), *g* : [0, *T*] → Y.

A function *x* ∈ *C*([0, *T*); *DM*) ∩ *C*([−*r*, *T*); X ) is called a solution of problems (7) and (8), if *D<sup>α</sup> <sup>t</sup> x* ∈ *C*([0, *T*); X ), it satisfies Equalities (7) and (8).

+

**Theorem 3.** *Let <sup>p</sup>* <sup>∈</sup> <sup>N</sup>0*, an operator <sup>M</sup> be* (*L*, *<sup>p</sup>*)*-bounded, <sup>h</sup>* <sup>∈</sup> *<sup>C</sup>*([−*r*, 0]; <sup>X</sup> <sup>1</sup>)*, xk* ∈ X <sup>1</sup>*, <sup>k</sup>* <sup>=</sup> 0, 1, ... , *<sup>m</sup>* <sup>−</sup> <sup>1</sup>*, <sup>h</sup>*(0) = *<sup>x</sup>*0*, <sup>T</sup>*<sup>0</sup> <sup>&</sup>gt; <sup>0</sup>*, <sup>g</sup>* <sup>∈</sup> *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*0]; <sup>Y</sup>)*,* K ∈ *<sup>C</sup>m*(*p*+1)([−*r*, 0];L(<sup>X</sup> ; <sup>Y</sup>))*,* <sup>K</sup>(*n*)(−*r*) = <sup>K</sup>(*n*)(0) = <sup>0</sup> *at n* = 0, 1, ... , *m*(*p* + 1) − 1*. Then there exists T* ∈ (0, *T*0)*, such that problems* (7) *and* (8) *have a unique solution.*

**Proof.** Fix *<sup>T</sup>* <sup>&</sup>gt; 0 and consider on the segment [0, *<sup>T</sup>*] Equation (4) with some *<sup>f</sup>* <sup>∈</sup> *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*]; <sup>Y</sup>). Due to Theorem 2 and Remark 1 we have the solution *xf* of problems (4) and (5) with the given *xk*, *<sup>k</sup>* <sup>=</sup> 0, 1, ... , *<sup>m</sup>* <sup>−</sup> 1. For brevity denote *<sup>X</sup>β*(*t*) :<sup>=</sup> *<sup>E</sup>α*,*β*(*L*−<sup>1</sup> <sup>1</sup> *M*1*t <sup>α</sup>*)*P*, *<sup>β</sup>* <sup>&</sup>gt; 0, put at *<sup>t</sup>* <sup>∈</sup> [−*r*, 0) *xf*(*t*) = *<sup>h</sup>*(*t*) + *<sup>h</sup>*0(*t*) with some *<sup>h</sup>*<sup>0</sup> <sup>∈</sup> *<sup>C</sup>*([−*r*, 0]; <sup>X</sup> <sup>0</sup>) and define the operator

[Φ*f* ](*t*) := 0 −*r* <sup>K</sup>(*s*)*xf*(*<sup>t</sup>* <sup>+</sup> *<sup>s</sup>*)*ds* <sup>+</sup> *<sup>g</sup>*(*t*) = <sup>0</sup> −*t* K(*s*) *m*−1 ∑ *k*=0 (*t* + *s*)*kXk*<sup>+</sup>1(*t* + *s*)*xkds*+ + 0 −*t* K(*s*) *t*+*s* 0 *<sup>X</sup>α*(*<sup>t</sup>* <sup>+</sup> *<sup>s</sup>* <sup>−</sup> *<sup>τ</sup>*)*L*−<sup>1</sup> <sup>1</sup> *Q f*(*τ*)*dτds* − 0 −*t* K(*s*) *p* ∑ *l*=0 (*GD<sup>α</sup> t* )*l M*−<sup>1</sup> <sup>0</sup> (*I* − *Q*)*f*(*t* + *s*)*ds*+ + 0 *t*−*r* K(*s* − *t*)(*h*(*s*) + *h*0(*s*))*ds* + *g*(*t*), *t* ∈ [0,*r*), [Φ*f* ](*t*) := 0 −*r* <sup>K</sup>(*s*)*xf*(*<sup>t</sup>* <sup>+</sup> *<sup>s</sup>*)*ds* <sup>+</sup> *<sup>g</sup>*(*t*) = <sup>0</sup> −*r* K(*s*) *m*−1 ∑ *k*=0 (*t* + *s*)*kXk*<sup>+</sup>1(*t* + *s*)*xkds*+ 0 −*r* K(*s*) *t*+*s* 0 *<sup>X</sup>α*(*<sup>t</sup>* <sup>+</sup> *<sup>s</sup>* <sup>−</sup> *<sup>τ</sup>*)*L*−<sup>1</sup> <sup>1</sup> *Q f*(*τ*)*dτds* − 0 −*r* K(*s*) *p* ∑ *l*=0 (*D<sup>α</sup> <sup>t</sup> G*)*<sup>l</sup> M*−<sup>1</sup> <sup>0</sup> (*I* − *Q*)*f*(*t* + *s*)*ds* + *g*(*t*), *t* ∈ [*r*, *T*],

By induction, we can prove that at *t* ∈ [0, *T*], *n* = 0, 1, . . . , *m*(*p* + 1)

$$[\Phi f]^{(n)}(t) = \frac{d^n}{dt^n} \int\_{t-\tau}^t \mathcal{K}(\tau - t) \mathbf{x}\_f(\tau) d\tau + \mathcal{g}^{(n)}(t) = (-1)^n \int\_{-\tau}^0 \mathcal{K}^{(n)}(s) \mathbf{x}\_f(t + s) ds + \mathcal{g}^{(n)}(t), \tag{9}$$

since <sup>K</sup>(*n*)(−*r*) = <sup>K</sup>(*n*)(0) = 0 at *<sup>n</sup>* <sup>=</sup> 0, 1, ... , *<sup>m</sup>*(*<sup>p</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> 1. Therefore, for every *<sup>f</sup>* from the Banach space *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*]; <sup>Y</sup>) with the standard norm ·*m*(*p*+1) we have <sup>Φ</sup>*<sup>f</sup>* <sup>∈</sup> *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*]; <sup>Y</sup>).

$$\text{Let } t\_r := \min\{t, r\}. \text{ For } f\_1, f\_2 \in C^{m(p+1)}([0, T]; \mathcal{Y}), t \in [0, T], n = 0, 1, \dots, m(p+1), \text{ due to (9)}$$

$$\frac{d^n}{dt^n}([\Phi f\_1](t) - [\Phi f\_2](t)) = (-1)^n \int\_{-t\_r}^0 \mathcal{K}^{(n)}(s) \int\_0^{t+s} \mathcal{X}\_a(t+s-\tau) L\_1^{-1} \mathcal{Q}(f\_1(\tau) - f\_2(\tau)) d\tau ds - \int\_{-t\_r}^0 \mathcal{K}^{(n)}(s) \int\_0^{t+s} \mathcal{X}\_a(s) \mathcal{Q}(f\_1(\tau) - f\_2(\tau)) ds,$$

$$-(-1)^n \int\_{-t\_r}^0 \mathcal{K}^{(n)}(s) \sum\_{l=0}^p (GD\_t^n)^l \mathcal{M}\_0^{-1} (I-Q) (f\_1(t+s) - f\_2(t+s)) ds,$$

therefore, using Theorem 1 and Lemma 1, we obtain

$$\begin{aligned} \|\Phi f\_1 - \Phi f\_1\|\_{m(p+1)} &\le \mathcal{C}\_1 \int\_{-t\_r}^0 (t+s) \sum\_{n=0}^{m(p+1)} \|\mathcal{K}^{(n)}(s)\|\_{\mathcal{L}(\mathcal{X};\mathcal{Y})} ds \|f\_1 - f\_2\|\_0 + \\ &+ \mathcal{C}\_2 \int\_{-t\_r}^0 \sum\_{n=0}^{m(p+1)} \|\mathcal{K}^{(n)}(s)\|\_{\mathcal{L}(\mathcal{X};\mathcal{Y})} ds \|f\_1 - f\_2\|\_{mp} \le \mathcal{C} F(T) \|f\_1 - f\_2\|\_{m(p+1)}. \end{aligned}$$

where, for the monotonously non-decreasing non-negative function

$$F(t) := \int\_{-t\_r}^{0} \sum\_{n=0}^{m(p+1)} \|\mathcal{K}^{(n)}(s)\|\_{\mathcal{L}(\mathcal{X};\mathcal{Y})} ds$$

we have *F*(*t*) → 0 as *t* → 0+. So, the inequality Φ*f*<sup>1</sup> − Φ*f*1*m*(*p*+1) ≤ *q f*<sup>1</sup> − *f*2*m*(*p*+1) with some *q* ∈ (0, 1) is valid for sufficiently small *T* > 0 and there exists a unique fixed point *f*<sup>0</sup> of the operator Φ in *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*]; <sup>Y</sup>). Therefore,

$$LD\_t^a \mathbf{x}\_{f\_0}(t) - M\mathbf{x}\_{f\_0}(t) = f\_0(t) = [\Phi f\_0](t) = \int\_{-r}^0 \mathcal{K}(\mathbf{s}) \mathbf{x}\_{f\_0}(t+s) ds + \mathbf{g}(t),$$

and the function *xf*<sup>0</sup> , which is defined as in the beginning of this proof, is a solution of problems (7) and (8).

Note that the choice of function *h*<sup>0</sup> does not affect the proof, hence, we can choose *h*0(*t*) ≡ (*I* − *P*)*xf*<sup>0</sup> (0), then the obtained *xf*<sup>0</sup> is continuous on [−*r*, *T*].

Let there exist two solutions *<sup>x</sup>*1, *<sup>x</sup>*<sup>2</sup> of the problem, denoted as *fi*(*t*) = -0 K(*s*)*xi*(*t* + *s*)*ds*, *i* = 1, 2.

−*r* As before, we have *fi* <sup>∈</sup> *<sup>C</sup>m*(*p*+1)([0, *<sup>T</sup>*]; <sup>Y</sup>) and *LD<sup>α</sup> <sup>t</sup> xi* − *Mxi* = *fi*, hence, by the construction Φ*fi* = *fi*, *i* = 1, 2. Thus, Φ has two fixed points, it is a contradiction. Consequently, *f*<sup>1</sup> ≡ *f*2, for *y* := *x*<sup>1</sup> − *x*<sup>2</sup> we have *LD<sup>α</sup> <sup>t</sup> <sup>y</sup>* <sup>−</sup> *My* <sup>=</sup> 0, *<sup>y</sup>*(*k*)(0) = 0, *<sup>k</sup>* <sup>=</sup> 0, 1, ... , *<sup>m</sup>* <sup>−</sup> 1, therefore, *<sup>y</sup>* <sup>≡</sup> 0 due to Theorem 2. So, the solution of problems (7) and (8) is unique.

## **4. A Scott–Blair Type System**

Consider the problem

$$\frac{\partial^k \upsilon}{\partial t^k}(\mathbf{x}, 0) = z\_k(\mathbf{x}), \ \mathbf{x} \in \Omega, \ k = 0, 1, \ldots, m - 1,\tag{10}$$

$$w(\mathbf{x},t) = h(\mathbf{x},t), \ \mathbf{x} \in \Omega, \ t \in [-r,0], \tag{11}$$

$$w(\mathbf{x}, t) = 0, \ (\mathbf{x}, t) \in \partial \Omega \times [0, T]. \tag{12}$$

$$(1 - \chi \Delta) D\_t^\mathbb{R} v(\mathbf{x}, t) = -(\vec{v} \cdot \nabla) v(\mathbf{x}, t) - (v \cdot \nabla) \vec{v}(\mathbf{x}, t) - r(\mathbf{x}, t) + $$

$$\begin{split} \int\_{-r}^{0} (K\_1(\mathbf{s}) v(t + \mathbf{s}) + K\_2(\mathbf{s}) r(t + \mathbf{s})) d\mathbf{s}, \ (\mathbf{x}, t) \in \Omega \times [0, T], \end{split} \tag{13}$$

$$\nabla \cdot \boldsymbol{v}(\mathbf{x}, t) = 0, \ (\mathbf{x}, t) \in \Omega \times [0, T], \tag{14}$$

where <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup>*<sup>n</sup>* is a bounded region with a smooth boundary *<sup>∂</sup>*Ω, *<sup>χ</sup>* <sup>∈</sup> <sup>R</sup>, *<sup>v</sup>*˜ is a given function. Function of the fluid velocity *v* = (*v*1, *v*2, ... , *vn*) and of the pressure gradient *r* = (*r*1,*r*2, ... ,*rn*) = ∇*p* are unknown.

This system without delay can be obtained, if the dynamics of a Scott–Blair medium [22] are described by using a fractional derivative of the same order as in the rheological relation for this medium, with subsequent linearization.

Let L<sup>2</sup> := (*L*2(Ω))*n*, H<sup>1</sup> := (*W*<sup>1</sup> <sup>2</sup> (Ω))*n*, <sup>H</sup><sup>2</sup> := (*W*<sup>2</sup> <sup>2</sup> (Ω))*n*. The closure of {*<sup>v</sup>* <sup>∈</sup> (*C*<sup>∞</sup> <sup>0</sup> (Ω))*<sup>n</sup>* : ∇ · *<sup>v</sup>* <sup>=</sup> <sup>0</sup>} in the space <sup>L</sup><sup>2</sup> will be denoted by <sup>H</sup>*σ*, and in the space <sup>H</sup><sup>1</sup> it will be <sup>H</sup><sup>1</sup> *<sup>σ</sup>*. We have the decomposition <sup>L</sup><sup>2</sup> <sup>=</sup> <sup>H</sup>*<sup>σ</sup>* <sup>⊕</sup> <sup>H</sup>*π*, where <sup>H</sup>*<sup>π</sup>* is the orthogonal complement for <sup>H</sup>*σ*. Denote by <sup>Π</sup> : <sup>L</sup><sup>2</sup> <sup>→</sup> <sup>H</sup>*<sup>π</sup>* the corresponding to this decomposition orthoprojector, <sup>Σ</sup> <sup>=</sup> *<sup>I</sup>* <sup>−</sup> <sup>Π</sup>, <sup>H</sup><sup>2</sup> *<sup>σ</sup>* = H<sup>1</sup> *<sup>σ</sup>* <sup>∩</sup> <sup>H</sup>2.

The operator *A* := ΣΔ with the domain H<sup>2</sup> *<sup>σ</sup>* in the space H*<sup>σ</sup>* has a real, negative, discrete spectrum with a finite multiplicity, condensing at −∞ [23].

At *<sup>v</sup>*˜ <sup>∈</sup> <sup>H</sup><sup>1</sup> by the formula *Dw* <sup>=</sup> <sup>−</sup>(*v*˜ · ∇)*<sup>w</sup>* <sup>−</sup> (*<sup>w</sup>* · ∇)*v*˜ operator *<sup>D</sup>* ∈ L(H<sup>2</sup> *<sup>σ</sup>*;L2) is defined. Put

$$\mathcal{X} = \mathbb{H}\_{\sigma}^{2} \times \mathbb{H}\_{\pi\prime} \quad \mathcal{Y} = \mathbb{L}\_{2} = \mathbb{H}\_{\sigma} \times \mathbb{H}\_{\pi\prime} \tag{15}$$

$$L = \begin{pmatrix} I - \chi A & \mathbb{O} \\ -\chi \Pi \Delta & \mathbb{O} \end{pmatrix} \in \mathcal{L}(\mathcal{X}; \mathcal{Y}), \quad M = \begin{pmatrix} \Sigma D & \mathbb{O} \\ \Pi D & -I \end{pmatrix} \in \mathcal{L}(\mathcal{X}; \mathcal{Y}).\tag{16}$$

By the choice of the space X we take into account Equation (14) and condition (12). The function *<sup>r</sup>*(·, *<sup>t</sup>*) is a gradient, since it belongs to the space <sup>H</sup>*<sup>π</sup>* at *<sup>t</sup>* <sup>≥</sup> 0.

**Lemma 2.** *Ref. [24]. Let <sup>χ</sup>* <sup>=</sup> <sup>0</sup>*, <sup>χ</sup>*−<sup>1</sup> <sup>∈</sup>/ *<sup>σ</sup>*(*A*)*, the spaces* <sup>X</sup> *and* <sup>Y</sup> *and the operators <sup>L</sup> and <sup>M</sup> be defined by* (15) *and* (16) *respectively. Then the operator M is* (*L*, 0)*-bounded and the projectors have the form*

$$P = \left( \begin{array}{cccc} I & & \mathbb{O} \\ & \chi \Pi \Lambda (I - \chi A)^{-1} \Sigma D + \Pi \Pi & \mathbb{O} \end{array} \right), Q = \left( \begin{array}{cccc} I & & \mathbb{O} \\ & \begin{array}{ccc} \mathbb{I} & & \mathbb{O} \\ \end{array} \right).$$

The form of the projectors *<sup>P</sup>* and *<sup>Q</sup>* implies that <sup>X</sup> <sup>0</sup> <sup>=</sup> {0} × <sup>H</sup>*π*, <sup>X</sup> <sup>1</sup> <sup>=</sup> {(*w*1, *<sup>w</sup>*2) <sup>∈</sup> <sup>H</sup><sup>2</sup> *σ* × <sup>H</sup>*<sup>π</sup>* : *<sup>w</sup>*<sup>2</sup> = (*χ*ΠΔ(*<sup>I</sup>* <sup>−</sup> *<sup>χ</sup>A*)−1Σ*<sup>D</sup>* <sup>+</sup> <sup>Π</sup>*D*)*w*1}, <sup>Y</sup><sup>0</sup> <sup>=</sup> {0} × <sup>H</sup>*π*, <sup>Y</sup><sup>1</sup> <sup>=</sup> {(*w*1, *<sup>w</sup>*2) <sup>∈</sup> <sup>H</sup>*<sup>σ</sup>* <sup>×</sup> <sup>H</sup>*<sup>π</sup>* : *<sup>w</sup>*<sup>2</sup> <sup>=</sup> <sup>−</sup>*χ*ΠΔ(*<sup>I</sup>* <sup>−</sup> *<sup>χ</sup>A*)−1*w*1}.

**Theorem 4.** *Let <sup>h</sup>* <sup>∈</sup> *<sup>C</sup>*([−*r*, 0]; <sup>H</sup><sup>2</sup> *<sup>σ</sup>*)*, zk* <sup>∈</sup> <sup>H</sup><sup>2</sup> *<sup>σ</sup>, <sup>k</sup>* <sup>=</sup> 0, 1, ... , *<sup>m</sup>* <sup>−</sup> <sup>1</sup>*, <sup>h</sup>*(·, 0) = *<sup>z</sup>*0(·)*,* <sup>K</sup>*<sup>i</sup>* <sup>∈</sup> *<sup>C</sup>m*([−*r*, 0]; <sup>R</sup>)*,* K(*n*) *<sup>i</sup>* (−*r*) = <sup>K</sup>(*n*) *<sup>i</sup>* (0) = 0 *at n* = 0, 1, ... , *m* − 1*, i* = 1, 2*. Then there exists T* > 0*, such that problems* (10)*–*(14) *have a unique solution.*
