Kolomiiets, T. Yu.

Pogorui, A. A.

Rodríguez-Dagnino, R. M.

Pogorui A. A.

Kolomiiets T. Yu.

Rodríguez-Dagnino R. M.

Zhytomyr State University Library

Analele Universităţii OradeaFasc. Matematica, Tom XXVI (2019), Issue No. 1, 153–160AN ALGEBRAIC APPROACH FOR SOLVING FOURTH-ORDERPARTIAL DIFFERENTIAL EQUATIONSA. POGORUI1, T. KOLOMIIETS2 AND R. M. RODRÍGUEZ-DAGNINO3Abstract. It is well-known that any solution of the Laplace equation is a real or imag-inary part of a complex holomorphic function. In this paper, in some sense, we extendthis property into four order hyperbolic and elliptic type PDEs. To be more specific,the extension is for a c-biwave PDE with constant coefficients, and we show that thecomponents of a differentiable function on the associated hypercomplex algebras providesolutions for the equation.1. IntroductionIn this paper we are interested in finding the solution of the following equation(∂4∂x4− 2c ∂4∂x2∂y2+∂4∂y4)u (x, y) = 0, c > 0. (1.1)Depending on the value of c we may consider three cases. Namely, the case where 0 < c < 1and we call it as the c-biwave equation of the elliptic type, the case where c > 1 and wecall it as the c-biwave equation of hyperbolic type, and in the case where c = 1 Eq.(1.1) isthe well-known biwave equation. The biwave equation has been used in modeling of d-wavesuperconductors (see for instance [1], and references therein) or in probability theory [2, 3].In [4] the author studied Eq.(1.1) in the case where c < −1 and considered its applicationto theory of plain orthotropy.It is easily verified that any equation of the form(A∂4∂x4+ 2B∂4∂x2∂y2+ C∂4∂y4)u (x, y) = 0,where AC > 0 and AB < 0 can be reduced to Eq.(1.1) by changing variables. To obtain allsolutions of Eq. (1.1) for 1 6= c > 0 we will use the method developed in [7]. According tosuch approach we need a commutative algebra with basis containing e1, e2 such thate41 − 2c e21e22 + e42 = 0. (1.2)Then, we study monogenic functions on the subspace of this algebra containing e1, e2and show that any solution of Eq. (1.1) can be obtained as a component of such monogenicfunctions.2010 Mathematics Subject Classification. Primary 32W50; Secondary 30C45, 35C99.Key words and phrases. Monogenic function; commutative algebra; partial differential equations; biwaveequation.153154 A. POGORUI, T. KOLOMIIETS AND R. M. RODRÍGUEZ-DAGNINO2. Hyperbolic caseFirstly we study Eq. (1.1) in the case where c > 1, which is said to be hyperbolic. Letus consider an associative commutative algebra over the real field RAc = {xu + yf + ze + vfe : x, y, z, v ∈ R}with a basis u, f , e, fe, where u is the identity element of Ac and the following Cayley tableholds fe = ef , f2 = u, e2 = u−mfe, where m =√2(c− 1).The basis elements u, e satisfy Eq. (1.2).It is easily verified that for c > 1 algebra Ac has the following idempotentsi1 =k1k1 + k2u− f√2k1 + k2e,i2 =k2k1 + k2u +f√2k1 + k2e, (2.1)where k1 =√c+ 1−√c− 1, k2 =√c+ 1 +√c− 1.Therefore, we havei1 + i2 = uandi1 i2 =k1k2(k1 + k2)2 u−√2k2(k1 + k2)2 fe +√2k1(k1 + k2)2 fe− 2(k1 + k2)2 u +2m(k1 + k2)2 fe = 0.It is easily seen thate = fk1√2i2 − fk2√2i1. (2.2)Consider a subspace Bc of algebra Ac of the following formBc = {xu + ye | x, y ∈ R} .Definition 2.1. A function g : Bc → Ac is called differentiable (or monogenic) onBc if for any Bc 3 w = xu + ye there exists a unique element g′ (w) such that for anyh ∈ BclimR3ε→0g (w + εh)− g (w)ε= hg′ (w) ,where hg′ (w) is the product of h and g′ (w) as elements of Ac.It follows from [7] that a function g (w) = uu1 (x, y)+ f u2 (x, y)+eu3 (x, y)+ f eu4 (x, y)is monogenic if and only if there exist continuous partial derivatives ∂ui(x,y)∂x ,∂ui(x,y)∂y , i =1, 2, 3, 4 and it satisfies the following Cauchy-Riemann type conditionse∂∂xg (w) = u∂∂yg (w) , ∀w ∈ Bc,or∂u1 (x, y)∂y=∂u3 (x, y)∂x,∂u2 (x, y)∂y=∂u4 (x, y)∂x,AN ALGEBRAIC APPROACH FOR SOLVING PARTIAL DIFFERENTIAL EQUATIONS 155∂u3 (x, y)∂y=∂u1 (x, y)∂x−m∂u4 (x, y)∂x,∂u4 (x, y)∂y=∂u2 (x, y)∂x−m∂u3 (x, y)∂x.It is also proved in [7] that if g is monogenic then its components ui (x, y) satisfies Eq.(1.1).By passing in Bc from the basis u, e to the basis i1, i2, we havew = xu + y e =(x− f k2√2y)i1 +(x+ fk1√2y)i2.Lemma 2.1. A function g : Bc → Ac, where c > 1, is differentiable if and only if it canbe represented as followsg (w) = α (w1) i1 + β (w2) i2, (2.3)where w1 = x−f k2√2y, w2 = x+fk1√2y and α (w1), β (w2) have continuous partial derivatives∂∂xα (w1) ,∂∂yα (w1) ,∂∂xβ (w2) ,∂∂yβ (w2) satisfying∂∂yα (w1) = −fk2√2∂∂xα (w1) ,∂∂yβ (w2) = fk1√2∂∂xβ (w2) .Proof. Sufficiency can be verified directly. Indeed,∂∂yg (w) =∂∂yα (w1) i1 +∂∂yβ (w2) i2= −f k2√2∂∂xα (w1) i1 + fk1√2∂∂xβ (w2) i2On the other hand, taking into account Eqs. (2.1), (2.2), we havee∂∂xg (w) =(fk1√2i2 − fk2√2i1)(∂∂xα (w1) i1 +∂∂xβ (w2) i2)= −f k2√2∂∂xα (w1) i1 + fk1√2∂∂xβ (w2) i2.Hence,e∂∂xg (w) = u∂∂yg (w) .Now let us prove necessity. Suppose that a functiong (w) = uu1 (x, y) + f u2 (x, y) + eu3 (x, y) + f eu4 (x, y)is monogenic on Bc. Let us defineα (w1) = u(u1 (x, y)−k2√2u4 (x, y))+ f(u2 (x, y)−k2√2u3 (x, y)),β (w2) = u(u1 (x, y) +k1√2u4 (x, y))+ f(u2 (x, y) +k1√2u3 (x, y)).Thus, we have∂∂yα (w1) = u(∂u3 (x, y)∂x− k2√2(∂u2 (x, y)∂x−m∂u3 (x, y)∂x))+f(∂u4 (x, y)∂x− k2√2(∂u1 (x, y)∂x−m∂u4 (x, y)∂x))156 A. POGORUI, T. KOLOMIIETS AND R. M. RODRÍGUEZ-DAGNINO= −f k2√2∂u1 (x, y)∂x− u k2√2∂u2 (x, y)∂x+ u(k2√2m+ 1)∂u3 (x, y)∂x+f(k2√2m+ 1)∂u4 (x, y)∂x.Taking into account thatk2√2m+ 1 =√c2 − 1 + c = k222,we have ∂∂yα (w1) = −fk2√2∂∂xα (w1).Much in the same manner, it can be shown that ∂∂yβ (w2) = fk1√2∂∂xβ (w2). Remark 2.1. Considering variables x, y1 = − k2√2y and x, y2 =k1√2y, we have∂∂y1α = f∂∂xα, (2.4)∂∂y2β = f∂∂xβ.Hence, it is easily verified that if the components α1, α2 of α (w1) = α1 (w1) + fα2 (w1)have continuous partial derivatives ∂2∂x2αk (w1) and∂2∂y21αk (w1), k = 1, 2 then they satisfythe wave equation (∂2∂x2− ∂2∂y21)u (x, y1) = 0.Similarly, the components β1, β2 of β (w2) = β1 (w2) + fβ2 (w2) satisfy the wave equation(∂2∂x2− ∂2∂y22)u (x, y2) = 0.Theorem 2.2. u (x, y) is a solution of Eq. (1.1) for c > 1 if and only if for somei, j ∈ {1, 2} it can be represented as followsu (x, y) = αi (ω1) + βj (ω2) ,where αi (ω1) , βj (ω2) are four times continuous differentiable components of α (ω1) andβ (ω2) of monogenic function g (ω) in the decomposition (2.3) i.e.,g (ω) = α (ω1) i1 + β (ω2) i2,where α (ω1) = α1 (ω1) + fα2 (ω1), β (ω2) = β1 (ω2) + fβ2 (ω2) satisfy Eq. (2.4).Proof. As it was mentioned above u (x, y) = αi (ω1) + βj (ω2) is a solution for c > 1 of Eq.(1.1).Now suppose that u (x, y) is a solution of Eq. (1.1). It is easily verified that(∂4∂x4− 2c ∂4∂x2∂y2+∂4∂y4)u (x, y) =(∂2∂x2− ∂2∂y21)(∂2∂x2− ∂2∂y22)u (x, y) = 0. (2.5)It is easily seen that Eq. (2.5) is equivalent to the set of the following systems(∂2∂x2 −∂2∂y22)u (x, y) = v1 (x, y) ,(∂2∂x2 −∂2∂y21)v1 (x, y) = 0orAN ALGEBRAIC APPROACH FOR SOLVING PARTIAL DIFFERENTIAL EQUATIONS 157(∂2∂x2 −∂2∂y21)u (x, y) = v2 (x, y) ,(∂2∂x2 −∂2∂y22)v2 (x, y) = 0.Let us consider the first system. Since any solution of(∂2∂x2 −∂2∂y2k)vk (x, y) = 0 is of theform vk (x, y) = f1 (x+ yk) +f2 (x− yk), where fi, i = 1, 2 are arbitrary twice differentiablefunctions it follows from the second equation of the system thatv1 (x, y) = f1 (x+ y1) + f2 (x− y1) .Thus, the first equation of the system is(∂2∂x2− ∂2∂y22)u (x, y) = f1 (x+ y1) + f2 (x− y1) . (2.6)It is easily seen that a partial solution of Eq. (2.6) isU (x, y) =k21k21 − k22(F1(x− k2√2y)+ F2(x+k2√2y))=k21k21 − k22(F1 (x+ y1) + F2 (x− y1)) ,where F′′k = fk, k = 1, 2.Thus, the general solution of the system is as followsu (x, y) = g1 (x+ y2) + g2 (x− y2) +k21k21 − k22(F1 (x+ y1) + F2 (x− y1))Let us put α1 (ω1) =k21k21−k22(F1 (x+ y1) + F2 (x− y1)) and β2 (ω2) = g1 (x+ y2)+g2 (x− y2).Taking into account that(∂2∂x2 −∂2∂y21) [k21k21−k22(F1 (x+ y1) + F2 (x− y1))]= 0 and(∂2∂x2 −∂2∂y22)[g1 (x+ y2) + g2 (x− y2)] = 0 we conclude the proof for the first system.The case of the second system can be proved similarly. 3. Elliptic caseNow we consider an associative commutative algebra Ac, where 0 < c < 1, over thecomplex field C with a basis u, e and the following Cayley table u e = e u = e, e2 = u+ iµe,where µ =√2(1− c). The matrix representations of u and e areu =(1 00 1), e =(0 11 iµ).Hence, we have the following traces of these representationstr (uu) = 2, tr (ue) = iµ, tr (ee) = 2− µ2.Sincedet(tr (uu) tr (ue)tr (ue) tr (ee))= 2(1 + c) 6= 0,then, Ac is a semi-simple algebra [8].158 A. POGORUI, T. KOLOMIIETS AND R. M. RODRÍGUEZ-DAGNINOBy following similar steps as in Eq. (2.1) we can show that for 0 < c < 1 algebra Ac hasthe following idempotentsI− =k1k1 + k2u +√2k1 + k2e, I+ =k2k1 + k2u−√2k1 + k2e, (3.1)where k1 =√c+ 1− i√1− c, k2 =√c+ 1 + i√1− c.It is also easily verified that these idempotents also satisfyI− + I+ = uandI− I+ = 0.It is straightforward to see thate =k2√2I− −k1√2I+. (3.2)Lemma 3.1. All non-zero elements of subspace Bc = {xu + y e | x, y ∈ R} of algebraAc are invertible, that is, if 0 6= w ∈ Bc then there exists w−1 ∈ Bc.Proof. Suppose w = su + t e ∈ Bc. Let us show that there exists w−1 = xu + y e, x, y ∈ Rsuch that ww−1 = 1. Indeed, the equation(su + t e) (xu + y e) = uhas a unique solution since the determinant of the systemsx+ ty = 1,tx+ (s+ iµt) y = 0,where x, y are unknown, is ∆ = s2− t2 + iµts and ∆ = 0 if and only if s = t = 0. A functionf (w), w ∈ Bc is said to be differentiable if it is differentiable in the common sense, i.e., forall w ∈ Bc there exists the following limitlimBc3∆w→0f (w + ∆w)− f (w)∆w= f ′ (w) .It is easily seen that if f is differentiable then it is monogenic and hence, it satisfies thefollowing Cauchy-Riemann type of conditions [7]e∂∂xf (w) = u∂∂yf (w)or in this case we have∂u1 (x, y)∂y=∂u3 (x, y)∂x,∂u2 (x, y)∂y=∂u4 (x, y)∂x,∂u3 (x, y)∂y=∂u1 (x, y)∂x− µ∂u4 (x, y)∂x,∂u4 (x, y)∂y=∂u2 (x, y)∂x+ µ∂u3 (x, y)∂x.In [7] it is also proved that if a function f (x, y) = uu1 (x, y) + iu2 (x, y) + eu3 (x, y) +i eu4 (x, y) is monogenic then ui (x, y) satisfies Eq. (1.1). We should mention that a con-structive description of monogenic functions in a three-dimensional harmonic algebra wasstudied in [5, 6].By passing from the basis u, e to the basis I−, I+ we havew = xu + y e =(x+k2√2y)I− +(x− k1√2y)I+.AN ALGEBRAIC APPROACH FOR SOLVING PARTIAL DIFFERENTIAL EQUATIONS 159Lemma 3.2. A function f : Bc → Ac, 0 < c < 1, is differentiable if and only if it canbe represented as followsf (w) = α (w1) I− + β (w2) I+,where w1 = x1 + i y1, x1 = x, y1 = −i k2√2y, w2 = x2 + i y2, x2 = x, y2 = ik1√2y and α (w1),β (w2) are analytical functions of variables w1, w2, respectively, as follows∂∂y1α (w1) = i∂∂xα (w1) ,∂∂y2β (w2) = i∂∂xβ (w2) .Proof. The sufficiency can be verified directly. Indeed,∂∂yf (w) =∂∂y1α (w1)∂y1∂yI− +∂∂y2β (w2)∂y2∂yI+= −i k2√2∂∂y1α (w1) I− + ik1√2∂∂y2β (w2) I+=k2√2∂∂xα (w1) I− −k1√2∂∂xβ (w2) I+.On the other hand, taking into account Eqs. (2.4), (3.1), we havee∂∂xf (w) =(k2√2I− −k1√2I+)(∂∂xα (w1) I− +∂∂xβ (w2) I+)=k2√2∂∂xα (w1) I− −k1√2∂∂xβ (w2) I+.Hence,e∂∂xf (w) =∂∂yf (w) .Now let us prove necessity. Suppose that a functionf (w) = u1 (x, y) + iu2 (x, y) + eu3 (x, y) + i eu4 (x, y)is monogenic on Bc, i.e., e∂∂xf (w) =∂∂yf (w).By using Eq. (3.1) we can represent f (w) in the following mannerf (w) = α (w1) I− + β (w2) I+,whereα (w1) = u1 (x, y) +k2√2u3 (x, y) + i(u2 (x, y) +k2√2u4 (x, y)),β (w2) = u1 (x, y)−k1√2u3 (x, y) + i(u2 (x, y)−k1√2u4 (x, y)).Consideru∂∂yf =∂∂y1α (w1)∂y1∂yI− +∂∂y2β (w2)∂y2∂yI+= − i k2√2∂α∂y1I− +i k1√2∂β∂y2I+.Then, taking into account Eq. (3.1), we havee∂∂xf (w) =(k2√2I− −k1√2I+)(∂α∂xI− +∂β∂xI+)=k2√2∂α∂xI− −k1√2∂β∂xI+.160 A. POGORUI, T. KOLOMIIETS AND R. M. RODRÍGUEZ-DAGNINOTherefore,∂∂y1α (w1) = i∂∂xα (w1) ,∂∂y2β (w2) = i∂∂xβ (w2) .Suppose α (ω1) = α1 (ω1) + iα2 (ω1) and β (ω2) = β1 (ω2) + iβ2 (ω2). It follows fromthe proof of Lemma 3.2 that αi (ω1) + βj (ω2), i, j ∈ {1, 2} are solutions of Eq. (1.1) for0 < c < 1. Theorem 3.3. u (x, y) is a solution of Eq. (1.1) for 0 < c < 1 if and only if for somei, j ∈ {1, 2} it can be represented as followsu (x, y) = αi (ω1) + βj (ω2) ,where αi (ω1) , βj (ω2) are components of α (ω1) and β (ω2) of monogenic function g (ω)in the decomposition (2.3) i.e.,f (ω) = α (ω1) I− + β (ω2) I+,where α (ω1), β (ω2) are complex analytical functions of respective variables.Proof. As mentioned above αi (ω1) + βj (ω2), i, j ∈ {1, 2} are solutions of Eq. (1.1) for0 < c < 1.If u (x, y) is a solution of Eq. (1.1) for 0 < c < 1 much in the same way as in provingTheorem 2.2 we can show that u (x, y) = αi (ω1) + βj (ω2). References[1] Feng X, Michael Neilan M, Discontinuous finite element methods for a bi-wave equation modelingd-wave superconductors, Math. Comp., 2011;80(275):1303–1333.[2] Pogorui AA, Rodŕıguez-Dagnino RM, One-dimensional semi-Markov evolution with general Erlangsojourn times, Random Oper. Stochastic Equations, 2005;13(4):399–405.[3] Kolomiiets T, Pogorui AA, Rodŕıguez-Dagnino RM, The distribution of random motion with 3-Erlangsojourn times, Random Oper. Stochastic Equations. 2015;23(2):69–79.[4] Gryshchuk SV, Commutative complex algedras of the second range with identity element and somecases of plain orthotropy, Ukrainian Math. J., 2018; vol.70, no. 8 , 1058–1071 (in Ukrainian).[5] Plaksa SA, Shpakovskii VS, Constructive description of monogenic functions in a harmonic algebra ofthe third rank, Ukrainian Math. J. 2010;62(08):1078–1091.[6] Plaksa SA, Pukhtaevich RP, Constructive description of monogenic functions in a three-dimensionalharmonic algebra with one-dimensional radical, Ukrainian Math. J. 2013;65(5):740–751.[7] Pogorui AA, Rodŕıguez-Dagnino RM, Shapiro M, Solutions for PDEs with constant coefficients andderivability of functions ranged in commutative algebras, Math. Methods Appl. Sci. 2014;37(17):2799–2810.[8] Van der Waerden BL, Algebra, 4th ed., Berlin Heidelberg (Springer Verlag); 1959.Received 22 December 20181,2 Department of Algebra and Geometry, Zhytomyr Ivan Franko State University, Zhytomyr10008, Ukraine3 School of Engineering and Sciences, Tecnologico de Monterrey, Av. Eugenio Garza Sada 2501Sur, C.P. 64849, Monterrey, N.L., MéxicoE-mail address: 1 pogor@zu.edu.ua, 2 tamila.kolomiiets@gmail.com, 3 rmrodrig@tec.mx

An algebraic approach for solving fourth-order partial differential equations

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An algebraic approach for solving fourth-order partial differential equations

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