Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 649 https://internationalpubls.com Solution of a Boundary Value Problem Involving General Class of Polynomial, Struve’s Function and PSI function of One Variable B. Satyanarayana 1 , * D. K. Pavan Kumar 2 ,Y. Pragathi Kumar 3 , Frederic Ayant 4 and N. Srimannarayana 5 1 Department of Mathematics, Acharya Nagarjuna University, Guntur, India. * 2 Department of Mathematics, Seshadri Rao Gudlavalleru Engineering College, Gudlavalleru, India. 3 Department of General Studies, University of the People (Remote), Pasadena, USA. 4 College Jean L’Herminier, Allee Des Nympheas, 83500, La Seyne-Sur-Mer, France. 5 Department of Mathematics, Koneru Lakshmaiah Education Foundation, Vaddeswaram, India. Article History: Received: 04-06-2024 Revised: 03-07-2024 Accepted: 30-07-2024 Abstract The authors of this paper have established some integrals that involves the PSI-function of one variable and Struve's function with the general class of polynomials. Additionally, solved a boundary value problem involving the steady state temperature distribution of a rectangular plate using PSI-function, Struve's function and general class of polynomials. This can be considered another novel technique to solve the boundary value problem. Finally, some special cases were incorporated. Keywords: PSI-function, General class of polynomial, Struve’s function and a Boundary value problem. 1. Introduction When tackling complex problems in physics and engineering, one essential concept often arises is the steady state solution. In many cases, these steady state solutions are derived from boundary value problems (BVPs), which involve solving differential equations with specific conditions at the boundaries. Since the steady-state heat equation is important in many domains, various techniques exist to obtain the solution analytically and numerically. There has been a lot of literature on the solutions of steady state heat equations. Many special functions such as Hypergeometric function [2], G-function [1], H-function [3, 10] and I-functions [8] involved in the solutions of different boundary value problems [4, 6]. In this paper, we use PSI-function (Pragathi - Satyanarayana’s I-function) [5] to solve a boundary value problem. This function was proved to be the generalization of I – functions defined by Saxena [8] and Arjun K Rathie [7]. The integral representation of PSI-function in terms of Mellin-Barnes type integration is as follows:           1, 1,, , ; 1, 1, , ; ; , ; 1 2, ; ; , ; i i i i j j j ji ji jin n pm n s p q r Lj j j ji ji jim m q a A a A z s z ds ib B b B                   (1.1) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 650 https://internationalpubls.com Where           1 1 1 1 1 1 1 j j i i ji ji m n B A j j j j j j r q p B A ji ji ji ji j m j n i b s a s s b s a s                                (1.2) Useful functions and Integral Formulae: 1. The general class of polynomials [11] is defined as follows:       ]/[ 0 , ! )(mn k k kn mkm n xA k n xS (1.3) where n = 0,1, 2,…; m is arbitrary positive integer and the coefficients An,k (n, k ≥ 0) are arbitrary constants. 2. The Struve’s function [9] is defined as follows:                 0 12 , ,, 2/1 t tvt k uyv utvykt z zH   (1.4) where Re (k)>0, Re ( )>0, Re (y)>0 and Re (v + u) >0. 3. The orthogonal property for cosine functions [2] is :                  c c mnforc mnfor dx c xm c xn 0 coscos  (1.5) 4. The integral formula due to Kumar [2] is :                               2/ 0 1 1 2 1 2 2 12 coscos a n mn m n m n na dx a x a x  (1.6) Boundary Value Problem: We consider a boundary value problem for a rectangular plate as shown in figure 1.1 with boundary conditions. Figure 1.1 Rectangular plate with insulated ends Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 651 https://internationalpubls.com In this article, we are going to find the steady state temperature  (x, y) for the rectangular plate whose edges x = 0 and x =a/2 are insulated. Here the Laplace’s equation is defined as 2 0, 2 0,0 2 2 2 2 b y a x yx        (1.7) Considering the boundary conditions: 0 2 0        a xx xx  2 0, b y  , (1.8)   00, x and                                             22 coscoscos)( 2 , a x z a x hS a x xf b x t l n 2 0, a x  (1.9) for section 3, and                                              22 , ,, coscoscos)( 2 , a x z a x hH a x xf b x k ulv n 2 0, a x  (1.10) for section 4. In the next section, it is to derive integral formulae that are useful in solving the given boundary value problem. 2. Main integral Theorem 1. Prove that dx a x z a x hS a px a xa t l n                                             2 2/ 0 2 coscos 2 coscos  [ / ] ,1 02 ! 4 kl t tk l kn k la h A k            *                                        1;, 2 ,1;, 2 ,;,,;, ;,,;,,1;2,2 4 ,1,1 ,1,1 1,. ;2,1    pk n pk n BbBb AaAakn z i i ii qmjijijimjjj pnjijijinjjj nm rqp (2.1) Proof: Substituitng corresponding formulae mentioned in (1.1) and (1.3) in the left hand side of the theorem 1, we will get dx a x z a x hS a px a xa t l n                                             2 2/ 0 2 coscos 2 coscos =                       2/ 0 ]/[ 0 2 , cos ! 2 coscos a tl k k k kl tk n a x hA k l a px a x           L s s dxds a x zs i     2 cos 2 1 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 652 https://internationalpubls.com Simplifying, =                            ]/[ 0 2/ 0 22 , 2 coscos 2 1 ! tl k L s a skn k kl tk dszdx a px a x s i hA k l     Applying (1.6), we get                              ]/[ 0 122 , 1 2 22 .1 2 22 2 122 2 1 ! tl k L s skn k kl tk dsz p skn p skn skna s i hA k l      Using the notation of (1.1) and (1.2), we can write the above integration as - =  [ / ] ,1 0 * 2 ! 4 kl t tk l kn k la h A k                                                  1;, 2 ,1;, 2 ,;,,;, ;,,;,,1;2,2 4 ,1,1 ,1,1 1,. ;2,1    pk n pk n BbBb AaAakn z i i ii qmjijijimjjj pnjijijinjjj nm rqp Theorem 2. Prove that dx a x z a x hH a px a xa k ulv n                                              2 2/ 0 2 , ,, coscos 2 coscos =   * 2 1 2 0 )( 21            t ty t n ha                1, 1, ,. 1 1, 4; 1, 1, 2 ( ),2 ;1 , , ; , , ; 4 , ; , , ; , ( ) , ;1 , 2 ( ) , ;1 , 1 ,0;1 , 1 ,0;1 2 i i i i j j j ji ji jin n p m n p q r j j j ji ji jim m q n y t a A a A z n b B b B y t p n y t p kt l v t u                                                  (2.2) Proof: Substituitng corresponding formulae mentioned in (1.1) and (1.4) in the left hand side, we will get dx a x z a x hH a px a xa k ulv n                                              2 2/ 0 2 , ,, coscos 2 coscos Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 653 https://internationalpubls.com =              0 12 2/1 t tvt utvlkt h                          L s a stvn dszdx a px a x s i 2/ 0 2)12(2 2 coscos 2 1     Applying (1.6), we get =              0 12 2/1 t tvt utvlkt h   L s i  2 1   dsz p stvn p stvn stvna s stvn                    1 2 12)12(2 .1 2 12)12(2 2 12)12(2 12)12(2    Let us assume y(t) = v + 2t + 1. Then         * 22 2/1 1 0 2       n t r tyt a ty h                        L ss dsz p styn p styn styna s i      22 1 2 12)(2 .1 2 12)(2 12)(2 2 1 Using the notation of (1.1) and (1.2) , we can write the above integral in PSI-function representation as - =             0 )( 21 * 2 1 2 t ty t n ha                1, 1, ,. 1 1, 4; 1, 1, 2 ( ),2 ;1 , , ; , , ; 4 , ; , , ; , ( ) , ;1 , 2 ( ) , ;1 , 1 ,0;1 , 1 ,0;1 2 i i i i j j j ji ji jin n p m n p q r j j j ji ji jim m q n y t a A a A z n b B b B y t p n y t p kt l v t u                                                  3. Solution of steady state temperature in rectangular plate involving General Class of Polynomial and PSI-function of one variable: We wish to find the steady-state temperature  (x, y) in a rectangular plate described in Section.1 whose vertical edges x=0, x=a/2 are insulated. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 654 https://internationalpubls.com Using General Class of polynomial [11] and PSI function [5], one of the boundary condition can be defined as                                             22 coscoscos)( 2 , a x z a x hS a x xf b x t l n 2 0, a x  (3.1) From the general solution of (1.7) given by Zill et al,[12], the superposition principle is -                   1 2 0, 2 0, 2 cos 2 ., r ro b y a x a xr a yr SinhAyAyx   (3.2) Substituting y= b/2 in (3.2) gives                      1 2 0, 2 0, 2 cos. 22 , r ro b y a x a xr a br SinhA b A b x   (3.3) Which is a half-range expansion in cosine series. We integrate (3.3) both sides w.r.t. x between the limits 0 and a/2.                          2/ 0 2/ 0 0 2/ 0 0 2 22 , a a r a r dx a xr Cos a br SinhAdx b Adx b x           2/ 0 0 2 . 22 , a ab Adx b x From (3.1), we have 2 . 2 .coscoscos 0 2/ 0 22 ab Adx a x z a x hS a xa t l n                                                                             2/ 0 22 coscoscos 4 a t l n o dx a x z a x hS a x ab A   Applying Theorem 1, and taking p = 0 in (2.1), we arrive to             1, 1,[ / ] ,. 1 0 , 1, 2;1 0 1, 1, 2 ,2 ;1 , , ; , , ; 1 .2 ! 4 4 , ; , , ; , , ;1 , , ;1 2 2 i i i i j j j ji ji jik n n pl t m ntk l k p q rn k j j j ji ji jim m q n k a A a A l h z A A n nb k b B b B k k                                                   (3.4) To find Ar, multiplying (3.3) both sides with       a xr Cos 2 and integrating w.r.t. x between the limits 0 and a/2, we get Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 655 https://internationalpubls.com                                    2/ 0 2/ 0 0 2/ 0 0 2 cos 2 cossinh 2 cos 2 2 cos 2 , a a r a r dx a bx a rx a br Adx a pxb Adx a rxb x   From (1.5),                                                  a br Sinh a Adx a x z a x hS a rx a x r a t l n   4 coscos 2 coscos 2 2/ 0 2                                                        2/ 0 22 cos 2 coscos . 4 a t l n r dx a x z a x CoshS a rx a x a br Sinha A    Applying Theorem.1 and assuming p = r in (2.1), we arrive to rA =                 ]/[ 0 , 1 . 4! 2sinh 1 tl k k kl tk n h A k l a br                                         1;, 2 ,1;, 2 ,;,,;, ;,,;,,1;2,2 4 ,1,1 ,1,1 1,. ;2,1    rk n rk n BbBb AaAakn z i i ii qmjijijimjjj pnjijijinjjj nm rqp (3.5) Hence the general solution from (3.4) and (3.5) is given by                                                  ]/[ 0 2 1 1 11 sinh2 2 cos 2 sinh 2.4 . ! )( , tl k r n n k k a br a xr a yr b yh k l yx      (3.6) Where  1                                        1;, 2 ,1;, 2 ,;,,;, ;,,;,,1;2,2 4 ,1,1 ,1,1 1,. ;2,1    k n k n BbBb AaAakn z i i ii qmjijijimjjj pnjijijinjjj nm rqp (3.7) and  2                                        1;, 2 ,1;, 2 ,;,,;, ;,,;,,1;2,2 4 ,1,1 ,1,1 1,. ;2,1    rk n rk n BbBb AaAakn z i i ii qmjijijimjjj pnjijijinjjj nm rqp (3.8) 4. Solution of steady state temperature in rectangular plate involving Struve’s function and Pragathi-Satyanarayana’s I-function of one variable: Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 656 https://internationalpubls.com we wish to find the steady-state temperature  (x, y) in a rectangular plate described in Section 1 whose vertical edges x=0, x=a/2 are insulated. Using Struve’s function [9] and PSI function [5], one of the boundary condition can be defined as                                              22 , ,, coscoscos)( 2 , a x z a x hH a x xf b x k ulv n 2 0, a x  (4.1) From the general solution of (1.7) given by Zill et al,[12], the superposition principle is –                   1 2 0, 2 0, 2 cos 2 ., r ro b y a x a xr a yr SinhAyAyx   (4.2) Substituting y= b/2 in (4.2) gives                      1 2 0, 2 0, 2 cos. 22 , r ro b y a x a xr a br SinhA b A b x   (4.3) which is a half-range expansion in cosine series. We integrate (4.3) both sides w.r.t. x between the limits 0 and a/2.                          2/ 0 2/ 0 0 2/ 0 0 2 22 , a a r a r dx a xr Cos a br SinhAdx b Adx b x           2/ 0 0 2 . 22 , a ab Adx b x From (4.1), we have 2 . 2 .coscoscos 0 2/ 0 22 , ,, ab Adx a x z a x hH a xa k ulv n                                         Applying Theorem.2 and taking p = 0 in (2.2), we arrive to     /2 1 2 10 0 4 1 , 1 * 2 .2 2 y t a t o n t b h A x dx ab b                                    , 1;0,1,1;0,1,1;),( 2 ,;,,;, ;,,;,,1;2),(2 4 ,1,1 ,1,1 1,. ;4,1                        utvlktty n BbBb AaAatyn z i i ii qmjijijimjjj pnjijijinjjj nm rqp    (4.4) To find Ar, multiplying (4.2) both sides with       a xr Cos 2 and integrating w.r.t. x between the limits 0 and a/2, we get, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 657 https://internationalpubls.com                   2/ 0 sinh 4 2 2 , a r a bra Adx a rx Cos b x   Using (4.1) and simplifying, we will get                     0 )( 12 1 . 2 1 sinh2 1 t ty t n r h a br A            1, 1, ,. 1 1, 4; 1, 1, 2 ( ), 2 ;1 , , ; , , ; 4 , ; , , ; , ( ) , ;1 , 2 i i i i j j j ji ji jin n p m n p q r j j j ji ji jim m q n y t a A a A z n b B b B y t r                                  ( ) , ;1 , 1 ,0;1 , 1 ,0;1 2 n y t r kt l v t u                   (4.5) Hence the general solution from (4.4) and (4.5) is given by         ( ) 1 22 1 1 10 1 2 2 sinh cos , 1 . 2 .2 2 sinh y t t n nk r r y r x h y a a x y r bb a                                               (4.6) Where  1                                     1;0,1,1;0,1,1;),( 2 ,;,,;, ;,,;,,1;2),(2 4 ,1,1 ,1,1 1,. ;4,1 utvlktty n BbBb AaAatyn z i i ii qmjijijimjjj pnjijijinjjj nm rqp    (4.7)  2           1, 1, ,. 1 1, 4; 1, 1, 2 ( ), 2 ;1 , , ; , , ; 4 , ; , , ; , ( ) , ;1 , 2 i i i i j j j ji ji jin n p m n p q r j j j ji ji jim m q n y t a A a A z n b B b B y t r                                  ( ) , ;1 , 1 ,0;1 , 1 ,0;1 2 n y t r kt l v t u                   (4.8) 5. Expansion Formulae (i) Substituting y = b/2 in the solution (3,6), we arrive to                                             22 coscoscos)( 2 , a x z a x hS a x xf b x t l n Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 658 https://internationalpubls.com     [ / ] 1 21 0 1 2 cos ( ) 1 ! 4 2 2 kl t tk n n k r rx l h a k                                 (5.1) Where  1 and  2 are assumed as in (3.7) and (3.8). (ii) Substituting y = b/2 in the solution (4.6), we arrive to                                              22 , ,, coscoscos)( 2 , a x z a x hH a x xf b x k ulv n     ( ) 1 22 1 1 0 1 2 cos 1 ( 1) 2 2 2 y t t n n k r rx h a                                   (5.2) Where  1 and  2 are assumed as in (4.7) and (4.8). 6. Special Cases  Taking Aji = Aj = Bji = Bj= 1 in (4.6), we get f(x) in terms of Struve’s function and I function of one variable given by Saxena [8] and we obtain modified (4.7) and (4.8) as -  1                                    1;0,1,1;0,1,1;),( 2 ,,,, ,,,,1;2),(2 4 ,1,1 ,1,1 1,. ;4,1 utvlktty n bb aatyn z I i i ii qmjijimjj pnjijinjj nm rqp    and                                             1;0,1,1;0,1,1;,)( 2 ,1;,)( 2 ,,,, ,,,,1;2),(2 4 ,1,1 ,1,1 1,. ;4,12 utvlktrty n rty n bb aatyn z I i i ii qmjijimjj pnjijinjj nm rqp      Taking r = 1, ji j ji j ji j ji j, ,A A ,B Ba = a b = b = = in (4.6), we get f(x) in terms of Struve’s function and I-function of one variable given by Arjun K Rathie [6], we obtain modified (4.7) and (4.8) as  1                               1;0,1,1;0,1,1;),( 2 ,;, ;,,1;2),(2 4 ,1 ,1 1,. 4,1 utvlktty n Bb Aatyn z i i ii qjjj pjjj nm qp    and  2                                     1;0,1,1;0,1,1;,1)( 2 ,1;,1)( 2 ,;, ;,,1;2),(2 4 ,1 ,1 1,. 4,1 utvlktty n ty n Bb Aatyn z i i ii qjjj pjjj nm qp    Similarly we can impose above said conditions to (3.6) and we can arrive the solution in terms of general class of polynomials and I-function of one variable by Saxena [8] and Arjun K Rathie [7] respectively. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 7s (2024) 659 https://internationalpubls.com 7. Conclusion In this paper, we presented a novel technique to obtain the solution of a boundary value problem using PSI-function. Initially, we have introduced two theorems that pertain to the multiplication of general class of polynomials and Struve's function in conjunction with the PSI-function. Subsequently, by using the outcomes of the established theorems, the solutions to the boundary value problems are derived in relation to the PSI-function of a single variable. In this study by specializing several parameters as well as variables lead to wide variety of useful special functions (or product of such special functions) expressible in terms of I-function defined by Saxena [8], defined by Rathie [7]. One can further specialize the results to H-function [3, 10], Meijer’s G-function [1], E-function [10] and hypergeometric function [2]. References [1] Agarwal, R.P. (1965). An expansion of Meijer’s G-function, Proc. Nat. Inst. Sci. India, 31, 536-546. [2] Kumar, H. (1993). Special functions and their applications in modern science and technology, PhD thesis Barkatullah University, Bhopal, M.P., India [3] Manilal Shah. (1973). On some applications related to Fox’s H-function of two variables, publications De Institute Mathematique, Novelle series tome 16(30), 123-133. [4] Neelam Pandey & Jyothi Mishra. (2014). I-function and boundary value problem in a rectangular plate, Research Journal of Mathematical and Statistical Science, 2(10), 5-7. [5] Pragathi Kumar, Y. & Satyanarayana, B . (2020). A study of Psi-function. Journal of Informatics and mathematical Sciences, Vol. 12(2), 159-171. [6] Pragathi Kumar, Y., Satyanarayana, B . , Srimannarayana, N. and Purnima, B.V. (2019). Solution of a Boundary Value Problem Involving I-Function and Struve’s Function, International Journal of Recent Technology and Engineering (IJRTE), Vol. 8(3), 411-415. [7] Rathie, A.K. (1997). A new generalization of generalized hypergeometric functions. Le Mathematiche, 52(2), 297-310. [8] Saxena, P. (2008). The I-function. Anamaya Publishers, New Delhi. [9] Sinha, L. J., (1993). Some relations between generalized Struve’s function and H-function, The Mathematics Education XXVII (3), 2000-2006. [10] Srivastava, H.M., Gupta, K.C., & Goyal, S.P. (1982). The H-functions of one and two variables with applications, South Asian Publishers, New Delhi. [11] Yashwant Singh & Nanda Kulkarni. (2015). A boundary value problem and expansion formula of I-function and general class of polynomials with applications, Int. J. of Engineering Research and Applications, 5(1), 105-108. [12] Zill, D.G., Cullen, M. R., (2008). Differential equations with boundary value problems, 7 th edition, Brooks/Cole Cengage Learning Publishers, USA.