Existence of solutions for a class of impulsive Burgers equation

The Burgers equation is a fundamental partial differential equation in fluid mechanics. It is also a very important model encountered in several areas of applied mathematics such as heat conduction, acoustic waves, gas dynamics and traffic flow. Analytical solutions of the partial differential equations modeling physical phenomena exist only in few of the cases. Therefore the need for the construction of efficient numerical methods for the approximate solution of these models always exists. Many of the analytical solutions to the Burgers equation involve Fourier series. There are several approximate solutions of the Burgers equation (see [1]). These solutions are always fixed to areas before and after the shock formation. For an area where the shock wave is forming no approximate solution has yet been found. It is therefore necessary to solve the Burgers equation numerically in this area (see [2, 3]). Numerical solutions themselves have difficulties with stability and accuracy.

(О 2024 Georgiev, S. G. and Hakem, A.

Here, in this paper, we focus our attention on a class of Burgers equations and we will investigate it for existence of classical solutions. More precisely, consider the problem ut + uux =0, t G Jq, x G R,

u (0 , x ) = u q ( x ) , x G R ,                                   (1 . 1)

u(tk+, x) = u(tk,x) + Ik(u(tk, x)), x G R, k G {1,... , m}, where

(H1) T> 0, 0 = t o 1 <...m < tm+1 = T, J = [0,T ], Jq = J\{tk}m₌₁, m G N.

(H2) I k G C ([0 ,T ] x R ), | I k ( u ) | C A | u | r k , u G R , r k >  0, k G { 1 ,... ,m } , A is a positive constant.

(H3) u q G C 1 ( R ), |u q | C B on R , B is a positive constant.

Additional conditions for the constants A and B will be given below. Here

u(t k ,x )= lim u ( t,x ) , u(t k + ,x )= lim u ( t,x ) ,x G R . t ^ t k -                         t ^ t k +

Whereas impulsive differential equations are well studied, the literature concerning impulsive partial differential equations does not see to be very rich. To the best of our knowledge, there are no any references devoted on investigations of the impulsive Burgers equation for existence and uniqueness of classical solutions.

The paper is organized as follows. In the next section, we give some preliminary results. In Section 3, we prove existence of at least one solution for the problem (1.1). In Section 4, we prove existence of at least two nonnegative solutions of the problem (1.1). In Section 5, we give an example that illustrates our main results.

2.    Preliminary Results

Below, assume that X is a real Banach space. Now, we will recall the definitions of compact and completely continuous mappings in Banach spaces.

Definition 2.1. Let K : M С X ^ X be a map. We say that K is compact if K(M ) is contained in a compact subset of X . The map K is called a completely continuous map if it is continuous and it maps any bounded set into a relatively compact set.

Proposition 2.1 (Leray–Schauder Nonlinear Alternative [4]) . Let C be a convex, closed subset of a Banach space E, 0 G U С C, where U is an open set. Let f : U ^ C be a continuous, compact map. Then

• (a)    either f has a fixed point in U ;

• (b)    or there exist x G dU, and A G (0 , 1) , such that x = Xf ( x ) .

To prove our existence result we will use the following fixed point theorem which is a consequence of Proposition 2.1.

Theorem 2.1. Let E be a Banach space, Y a closed, convex subset of E, U be any open subset of Y with 0 G U. Consider two operators T and S, where

Tx = e x G U, for e > 0 and S : U ^ E be such that

• (i)    I — S : U ^ Y continuous, compact and

• (ii)    { x G U : x = X(I — S^x, x G dU } = 0, for any X G (0 , 1 ) .

Then there exists x* G U such that

Tx * + Sx * = x *

• <1    We have that the operator 1 (I — S') : U ^ Y is continuous and compact. Suppose that there exist x o G dU and ^ 0 G (0 , 1) , such that

xo = ^o I (I - S)xo, that is xo = Ao (I — S)xo, where Ao = ^o | G (0, 1} . This contradicts the condition (ii). From Leray-Schauder nonlinear alternative, it follows that there exists x* G U, so that

x* = I (I — S)x*, or

Ex* + Sx* = x*, or

Tx * + Sx * = x * . >

Definition 2.2. Let X and Y be real Banach spaces. A map K : X ^ Y is called expansive if there exists a constant h> 1 for which one has the following inequality

\\Kx — Ky^Y > h\x — y^x, for any x, y G X .

Now, we will recall the definition for a cone in a Banach space.

Definition 2.3. A closed, convex set P in X is said to be cone if

• 1)    ax G P for any a ^ 0 and for any x G P ,

• 2)    x, — x G P implies x = 0.

Denote P * = P \{ 0 } . The next result is a fixed point theorem which we will use to prove existence of at least two nonnegative global classical solutions of the IVP (1.1). For its proof, we refer the reader to [5] and [6].

Theorem 2.2. Let P be a cone of a Banach space E; Q a subset of P and U i , U 2 and U 3 three open bounded subsets of P, such that U 1 C U 2 C U 3 and 0 G U i . Assume that T : Q ^ P is an expansive mapping, S : U 3 ^ E is a completely continuous map and S(U 3 ) C (I — T )(Q) . Suppose that (U 2 \ U 1 ) П Q = 0, (U 3 \ U 2 ) П Q = 0, and there exists u o G P * such that the following conditions hold:

• (i)    Sx = (I — T)(x — Au o ), for any A >  0 and x G dU 1 П (Q + Au o ),

• (ii)    there exists e ^ 0 , such that Sx = (I — T)(Ax), for any A ^ 1 + e, x G dU 2 and Ax G Q ,

• (iii)    Sx = (I — T)(x — Au o ), for any A >  0 and x G dU 3 П (Q + Au o) .

Then T + S has at least two non-zero fixed points xi,x2 G P, such that x1 G dU2 П Q and x2 G (U3 \ U2) П Q, or xi G (U2 \ Ui) П Q and x2 G (U3 \ U2) П Q.

Define the spaces PC (J ) , PC 1 (J ) and PC 1 (J, C ¹ ( R )) by

^{PC ( J} ) = {g ^: g ^{G C} ( J o ) , ( 3 g ⁽ t + ) , g^t j}) and g ⁽ j ) = g ⁽ t j ) , j ^{G{1 ,} ...,k ^} } ,

PC 1 ( J ) = { g : g G PC (J ) П C 1 ( J o ) , ( 3 g ’ ( t - ) , g ’ ( t j ⁺ )) and g ’ ( t - ) = g’(t j ) , j G { 1 ,... , k } } and

PC 1 (J, C 1 ( R ))

= { ^{u :}

Suppose that X =

J x R ^ R : u(^x) G PC 1 (J ) , x G R and u(t, • ) G C ¹ ( R ) , t G j}. (2.1)

PC 1 ( J, C 1 ( R )) is endowed with the norm

Hull = sup |

sup         | u ( t, x ) | ,

( t,x ) ^ \ t j ,t j +i ] x R

sup        | u x ( t,x ) | ,

( t,x ) e [ t j ,t j +x ] x R

sup       | u t ( t,x ) | , j G{ 1 ,...,k } к

(t,x)e[tj ,tj+i]xR                                       J provided it exists. Note that X is a Banach space. For u G X, define the operator

t

S 1 u ( t,x )= u ( t, x ) — u o ( x ) — ^ I k (u(t k ,x ))+ / u ( s,x ) u x ( s,x ) ds, (t,x) G J x R . o k                0

Lemma 2.1. Suppose that (H 1) - ( H 3) hold. If u G X satisfies the equation

S ₁ u(t,x') = 0 , (t,x) G J x R ,

(2 . 2)

then it is a solution of the problem (1 . 1) .

<1 Let u G X be a solution of the equation (2.2). We differentiate the equation (2.2) with respect to t and we find ut(t, x) + u(t, x)ux(t, x) = 0,

(t,x) G J x R, and u(0,x) = uo(x), x G R. Next, we put t = tj + and t = tj, j G {1,... ,m}, in the equation (2.2) and we obtain tj

0 = u(tj+,x) — uo(x) —   £ Ik (u(tk ,x))+ / u(s,x) Ux(s,x) ds, j G{1,...,m},x G R, o

0 = u(tj ,x) — uo(x) — £ Ik(u(tk ,x))+ / u(s,x) Ux(s,x) ds, j G{1,...,m},x G R, o

u(tj+, x) = u(tj, x) + Ij(u(tj, x)), j G {1,..., m}, x G R.

Consequently u satisfies (1.1). This completes the proof. ⊲

Lemma 2.2. Suppose that (H 1)-(H3) hold. If u G X, ||u|| C B, then

\Siu(t, x)| C 2B + A jm Br + TB2, (t, x) G J x R.

k=1

<1 We have

\Siu(t,x)\ =

C

u(t, x) - Uo(x)

-

t

^2 Ik(u(tk,x)) + /

0k₀

u(s, x)ux(s, x) ds

\u(t,x)\ + \uo(x)\ + 52 \Ik(u(tk,x))\ +

0_k

t

/ 0

\u(s, x)u_x(s, x)\ ds

C

\u(t,x)\ + \uo(x)\ + A 52 \u(tk,x)\^rk +

0_k

t

\u(s, x)u_x(s, x)\ ds

C 2B + A 52 B^rk + TB2, (t,x) G J x R.

k=1

This completes the proof. >

(H4) Suppose that g G C(J x R) is a nonnegative function, such that tx

216 (1 + t + t²+ t³) (1 + \x\ + \x\²+ \x\³+ \x\4 + \x\⁵+

\xv/ f

g(t1, x1) dx1

dt1 C D,

(t, x) G J x R, for some constant D > 0. In the last section, we will give an example for a function g and a constant D that satisfy (H4).

For u G X, define the operator tx

S2u(t,x) =11

(t - t1)²(x - x1)³g(t1 , x1)S1u(t1 , x1) dx1 dt1 ,

(t,x) G J x R.

Lemma 2.3. Suppose that (H 1)-(H4) hold. For u G X, ||u|| C B, we have

ISUI C D

m

2B + A²B^rk + TB² | ,

(t, x) G J x R.

< We have tx

\S₂u(t,x)\ = У y(t — t1)²(x — x1)3g(t1, x1)S1u(t1, x1) dx1 dt1

t

x

C

J J(t - ti)²\x

-

x1\3g(t1, x1)\S1u(t1, x1)\dx1 dt1 C

m

2B + A 2 B^rk + TB²|

t

x

x

У У(t - ti)²\x - xi\³g(ti,xi)(1+ ti)(1 + \xi\²+ \xi\³) dxi dti

and

and

m

2B + A ^ Brk + TB²J t²(1+ t)|x|³(1 + |x|²

t

x

+ |x|3) У У g(t_i,x_i) dx1 dt1

m

2B + A ^ B^rk + TB²,  (t,x) G J x R,

tx

—S2u(t,x) = 2^ ({t — t_i)(x — x1 )3g(t_i, x_i)S_iu(t_i, x_i) dx1 dt1

m

2B+A ^

k=1

t

x

< 2 У ^ (t — t1)|x — x1 |³g(t1, x1)|S1u(t1, x1 )| dx1 dt1

t

Brk + TB²  J

x

У^{(t —}ti^{)|x —} 0

Xi|³g(ti,xi)(1 + ti)(1 + |xi|²+ |xi|³)dxi dti

m

2B + A ^ Brk + TB² I t(1 + t) |x|³(1 + |x|²

t

x

+ |x|3) У У g(t_i,x_i) dx1 dti

m

2B + A ^ B ^rk + TB²\ ,  (t,x) G J x R,

tx

—S₂u(t,x) = 3 / ((t — t1)²(x — x1)²g(t1, x_i)S1u(t₁, x1) dx1 dti dx             J J

t

x

< 3y J(t — t1)²(x — x₁)²g(t₁,x_i)|S1u(t₁,x1)| dx1 dt1

m

2B+A ^

k=1

t

Brk +TB²] I

x

У(t — ti)²(x — xi)²g(ti,xi)(1 + ti)(1 + |xi|²+ |xi|³) dxi dti 0

m

2B + A ^ Brk + TB²\ t²(1+ t)|x|²(1 + |x|²

t

x

+ |x|3) У У g(t_i,x_i) dxi dt1

m

2B + A ^ B ^rk + TB²\ ,  (t,x) G J x R.

Consequently

m

2B + A ^ B^rk + TB².

This completes the proof. >

Lemma 2.4. Suppose that (H 1)-(H4) hold. If u G X satisfies the equation

S₂u(t,x) = C, (t,x) G J x R,

(2.3)

for some constant C, then u is a solution to the problem (1.1).

• <1 We differentiate three times with respect to t the equation (2.3) and we get

x

2 J(x - x1)3g(t, x1 )S1u(t, x1) dx1 = 0, (t, x) G J x R, or

x

У (x — x1)³g(t, x1)S1u(t, x1) = 0, (t, x) G J x R. 0

Now, we differentiate four times with respect to x the last equation and we find

6g(t, x)S1u(t, x) = 0,   (t,x) G J x R, or

g(t,x)S1 u(t,x) = 0,   (t,x) G J x R, whereupon

Siu(t, x) = 0, (t, x) G (0, T] x (R\{0}).

Since S1u(^, •) G C(J x R), we get

0 = lim S1u(t, x) = S1u(0, x) = lim S1u(t, x), (t, x) G J x R. t→0                       x→0

Therefore

S1u(t,x) = 0,   (t,x) G J x R.

Hence and Lemma 2.1, we conclude that u is a solution to the problem (1.1). This completes the proof. >

Now, suppose that

(H5)

m

2B + A £ B^r + TB²]

< B.

(H6)

For u

ǫ

(B+D(

2B + A у; B^r + TB k=1

G X, define the operators

))

< B.

Tu(t,x) = —eu(t,x),

Su(t, x) = (1 + e)u(t, x) + eS₂u(t, x)   (t, x) G J x R.

By Lemma 2.4, it follows that any fixed point of the operator T + S is a solution to the problem (1.1).

Lemma 3.1. Suppose that (H 1)-(H6) hold. For u G X, we have

\\(I — S)u|| ^ B and ||((1 + e)I — S)u^

< Applying Lemma 2.3, we get

\(I - S)u| = II

— eu — eS2u\< e|u| + e||S2u|| < e

(^B+D(

2B + A jm Brk + TB k=1

< B

and

||((1 + e)I - S)u|| = ||eS2u|| = €»M < eD

m

2B + A £ Brk + TB²I

< ǫB.

This completes the proof. >

Our main result in this section is as follows.

Theorem 3.1. Suppose that (H 1)-(H6) hold. Then the problem (1.1) has at least one solution.

• <1 Let Y denote the set of all equi-continuous families in X with respect to the norm || • |. Let also, Y = Y be the closure of Y,

U = {u e Y : |u| < B} .

For u E U and e > 0, define the operators

T (u)(t,x') = eu(t,x),

S(u)(t, x) = u(t, x) — eu(t, x) — eS2(u)(t, x), (t, x) E J x R. For u E U, we have

|(I — S)u| = ||eu + eS2u| < e|u| + e|S2u| < eBi + eABi.

Thus, S : U ^ X is continuous and (I — S](U) resides in a compact subset of Y. Now, suppose that there is a u E dU, so that u = X(I — S )u, or u = Xe (u + S2u),                                    (3.1)

for some X E (0, |). Then, using that S2u(0,x) = 0, we get

u(0, x) = Xe(u(0, x) + S2u(0, x)) = Xeu(0, x) x E R, whereupon Xe = 1, which is a contradiction. Consequently

{u E U : u = Xi (I — S)u, u E dU} = 0

for any Xi E (0,|). Then, from Theorem 2.1, it follows that the operator T + S has a fixed point u* E Y. Therefore

u*(t,x) = Tu*(t,x) + Su*(t,x') = eu*(t,x) + u*(t,x) — eu*(t,x) — eS2u*(t,x),   (t,x) E J x R, whereupon

S2u^*(t, x) = 0,   (t, x) E J x R.

From here, u∗ is a solution to the problem (1.1). From here and from Lemma 2.4, it follows that u is a solution to the IVP (1.1). This completes the proof. >

4. Existence of at Least Two Nonnegative Solutions

Suppose:

(H7) Let m > 0 be large enough and A, B, r, L, Ri be positive constants that satisfy the following conditions r 0, Ri > (m- + 1) L,

D ^2R1 + A ^VRik + TR^

L

< 5 ’

L e 10 ’

For x G X, define the operators

T₁u(t, x) = (1 + me)u(t, x) —

S3u(t, x) = —eS₂u(t, x) — meu(t, x) — e—,   (t, x) G J x R.

Our main result in this section is as follows.

Theorem 4.1. Suppose that (H 1)-(H4) and (H7) hold. Then the problem (1.1) has at least two nonnegative solutions.

<1 Define

U1 = P~ = {v G P : ||v|

U2 = Pl = {v G P : ||v|| <

U3 = Pri = {v G P : ||v||

D^m

+ A 5m

R2 = R1 + m ( 2R1 + A ^Rjk + TRl)

Q = PR2 = {v G P : ||v|| C R2}.

• 1)    For v1,v2 G Q, we have

|Trv1 — T1v21| = (1 + me) IА — v21, whereupon Ti : Q ^ E is an expansive operator with a constant 1 + me.

• 2)    For v G PR1, we get

||S3v|| ^{C e |S}2 ^vH + ^mell^v|l + E10 ^{C E}^^D ^^2R1 ^{+ A}V Rrk ^{+ TR}1j ^{+ mR}1⁺ 10^ *

Therefore S3(PR1) is uniformly bounded. Since S3 : PR1 ^ E is continuous, we have that S3(PR1) is equi-continuous. Consequently S3 : PR1 ^ E is a 0-set contraction.

• 3)    Let v1 G PR1. Set

V2 = V1 + —S2 V1 + m 5m

Note that S2V1 + LL ^ 0 on J x R. We have V2 ^ 0 on J x R and

1         L       D        ,m

I All C llvl II +||S2v1 II + 7— ^ R1 +I 2R1 + A V R1 k + TR1 I + 7— = m      5m     m

Therefore V2 G Q and

—emv₂= —emv₁— eS₂v₁

L e 10

L e 10 ’

or

(I — T1)v2 = —emv2 + e L = S3v1*

Consequently S3(PR1) С (I — T₁)(Q).

• 4)    Assume that for any uo G P* there exist A ^ 0 and u G dPr П (Q + Auo) or u G dPR1 П (Q + Auo), such that

• S3u = (I — Ti)(u — Auo).

  Then

  
  

  -eS2u

  
  

LL

• — meu — e — = —me(u — Auo) + e—,

or

• —S₂u = Amu₀+ —. 5

  Hence,

  
  

  L

  
  

  |S2u| = Amuo + 5 > L.

  
  
  
  

  This is a contradiction.

  5) Suppose that for any ǫ1 such that Aiui G PR1 and

  
  

  ^ 0 small enough there exist a ui G dPL and Ai ^ 1 + ei,

  
  

  S3u1 = (I - Ti^)(Ai^ui⁾.

  
  

  (4.1)

  
  

  In particular, for ǫ1 > ₅²_m, we

  
  

  Since ui G dPl and Aiui G PR1, it follows that

  
  

  have u_iG dPL, A_iu_iG PR1, A_i> 1 + e_iand (4.1) holds.

  
  

  (5m +1) L < ^AiL = ^Ai^|ui^|^ Ri.

  
  

  Moreover,

  
  

  —eS2ui — meui

  
  

  _-

  
  

  e— = —Aimeui + e—,

  10 ^{i i}10’

  
  

  or

  
  

  ^L

  S2u + 7

  
  

  = (Ai — 1)mui.

  
  

  From here,

  
  

  2 L

  
  

  >S2^ui + — = ^(Ai 5

  
  

  — 1)m|u_i| = (A_i— 1)mL

  
  

  and

  
  

  1 ^u(0,x)   1+ x^{2 ,}

  
  

  x ∈ R,

  
  

u(tk+, x) = u(tk, x) + (u(tk, x))⁴,   x G R, k G {1,2, 3}.

Here

Then

Take

Then and

A = 1, B = 1,  C = 1, r = 1.

2B + A ^fBr + TB²= 2 + 3 + 1 = 6.

k=1

1      _    3    _          «    1    _    1

D = e = —rx,  r = -,  Ri = 1, r = -,  L = -,  m = 1050.

1050 ’           2 ’       ¹      ’           8 ’           2 ’

m

2B + A^fB^rk + TB² | =

Л < 1 = b 1050

'(^B+ D ^B + A f ^Brk + TB²j ^)-^ (w²^)

< 1 = B.

Thus, (H5) and (H6) hold. Hence and Theorem 3.1, it follows that the considered problem has at least one solution. Moreover, r < L < r1,

R = 1 > (—2— + 1p = f— + Л L’

1        \5 • 1050    /2    \5m J ’ and

D (^2R1 + A f Rr^k+ T R' j =10650 < ^ = L■

Consequently (H7) holds. By Theorem 4.1, it follows that the considered problem has at least two nonnegative solutions.

Now, we will construct a function g for arbitrary n. Let

h(s) = log

1 + s¹¹ V2 + s²²

1 - s¹¹V2 + s²²’

l(s) = arctan

s11 v^

1 - s²²’

s ∈ R.

Then h‘(s) = — (1

22V2s¹⁰(1 - s²²)

- s¹¹\2 + s²²)(1 + s¹¹v^ + s²²) ’

l‘(s)

11V2s¹⁰(1 + s²⁰) i+s⁴⁰

s ∈ R.

Therefore

-w < lim (1 + |s| + ••• + s⁶)h(s) < to, s→±∞

-∞ < lim (1 + |s| + • • • + s⁶)l(s) < w.

s→±∞

Hence, there exists a positive constant C1 so that

/        2              1 , 1 + sⁿV2 + s²²    1        s¹¹^

(1 + |s| + s²+-----+ s⁶) ---7= log----—7=---— +--= arctan  ----— < C,

V                        44^2    1 - s11 V2 + s22   22^2       1 - s22        ’ s E R. Note that by [7, p. 707, Integral 79], we have dz

1 + z⁴

1      1 + z \2 + z²

= --7= log ------=----

4 \2    1 - zV2 + z²

1        z V2

+--t= arctan-----т.

2\/2        1 - z²

Let

s¹⁰

^Q(s) = (1 + s²)⁴(1 + s⁴⁴)(1 + s + s²)^{2 ,}

s E R.

Then there exists a positive constant C2 so that

216(1 + t + t² + t³) (1 + |x| + • •

t

•⁺x⁶⁾/

Q(s) ds

x

J ^Q(y) dy

< C2,

(t,x) E J x R.

Take g(t,x) = DQ(t)Q(x), (t, x) E J x R. Hence, tx

216(1 + t + t² + t³) (1 + |x| + • •

• + x⁶

g(s, y) dy

ds< D,

(t,x) E J x R.