Proc. Pakistan Acad. Sci. 42(3): 187-193.2005 ON A FOURTH ORDER PSEUDOPARABOLIC EQUATION Ye. A. Utkina and A. Maher Department of Mathematics,Faculty of Science, Department of Differential Equations, Kazan State University,18 Kremlyovskaya St., Kazan, 420008, Russia, and Assiut University, 71516, Egypt Received February 2005, accepted March 2005 Communicated by Prof. Dr. Q. K. Ghori Abstract: The present paper is devoted to investigation of the Goursat problem of a fourth order psedudoparabolic equation, with the help of Riemann function. Keywords: Partial differential equations, Goursat problem, Riemann function AMS Subject Classification: 35B, 35J, 35Q. Introduction 2 uy x, y0 \1 x xq , \ ,\1 C q , In the domain D ^`x0 x x1, y0 y y1 y p >@y0, y1 , xq >@x0, x1 . (2) we consider the equation Here we assume the following condition as L u u xxyy a21u xxy a12u xyy satisfied: a u a u a u a u 11 xy 20 xx 02 yy 10 x c y x , y x , c x y . M 0 \1 0 M 0 \ 0 \ 0 M1 0 i j a01u y a00u f ,aij C D , f 22 Formulation of the problem C i 0,1,2, j 0,1,2 , We will be searching for the above- (1) mentioned function u(x‘ y) with the help of the where the classCk l means the existence and conti- Riemann function, which is defined as the solution of the integral equation nuity for all derivatives r 0,...k;s 0,...,l . y w i j v x, y a x,W y W a x,W We will call the solution of the class ³ >@21 20 wx iwy j K C D i 0,1,2, j 0,1 as regular. The equation (1) is x v x,W dW a t, y the generalization of the Bussineska-Lyava equation ³ > 12 that describes longitudinal waves in a thin [ elastic shaft with allowance for the effects of x y the crosswise inertia. In this paper we study x t a t, y v t, y dt 02 @ ³ ³ the Goursat problem which consists of finding [ K of the function 2 2 1 0 0 1 a t,W x t a t,W y W a t,W uɋ D ɋ D p ɋ Dq , which in D > 11 01 10 ̓is a solution (1) and satisfies the following x t y W a t,W v t,W dWdt 1. (3) boundary conditions: 00 @ By virtue of [1], it exists and is unique. In u x0, y M y ,ux x0, y M1 y , y p , cases where it is necessary to emphasize the 2 dependence of v ̓not only on (x‘ y), but also on M,M1 C p ;u x, y0 \ x , [,K , we will indicate it, as usually, R x, y;[,K . Fourth order pseudoparabolic equation 188 Here we should note that v ̓does not coincide Q Py a20R ; with R as in ҏ[2-5]. Thus, here R x, y; x, y 1, T R a R a R a R ; while in the mentioned papers R x, y; x, y 0 . xy 21 x 12 y 11 However, v x, y remains the solution of the P Ry a21R ; equation adjoint to (1), namely, K N a R ; * x 02 L v vxxyy a21v xxy a12v xyy a11v xy F1 Ty a20R x a10R ; a20v xx a02v yy a10v x a01v y a00v 0. F2 Tx a02R y a01R . Let us now solve the Goursat problem (G). First, we will prove the auxiliary result. This enables us to write (4) in the form: Let u x, y belong to the class of the uR xxyy RL u >@uP xxy unknown functions of the Goursat problem. Then the following identity is true: >@uN xyy >@uQ >@uK yy >@uT xy xx uR { RL u u R a R xxyy > y 21 @xxy >@>@uF1 x uF2 y >@u x R y xy >@>u R a R u R a R x 12 xyy yy 21 y u R u a R u a R >@yy x y 21 x 20 x x a20 R @xx >@u Rxx a12 R x a02 R yy uR u a R u a R . (5) >@xxy x 12 y 02 y y u R a R a R a R > xy 21 x 12 y 11 @xy Let us collect the auxiliary identities which will be needed during the integration of (5). >u Rxyy a21R xy a12 R yy a11R y N x, y, x, y { P x, y, x, y { T x, y, x, y { 0 , a20 R x a10 R @ >u Rxxy a21R xx x Q x, y, x,K { K x, y,[, y { a R a R a R a R + 12 xy 11 x 002 y 01 @y F1 x, y, x,K { F2 x, y,[, y { 0 . For example, let us verify that N x,y,x,y {0. u x Ry ^u yy Rx ` ^uRxxy ` ^u y a21R x ` + xy x y x In reality, we put in (3) K y to obtain u a R u a R u a R , (4) x ^ x 12 y `y ^` 20 x x ^ 02 y `y v x, y >@a12 t, y x t a02 t, y v t, y dt 1 ³ [ where arguments of a i 0,1,2; j 0,1,2 are ij Differentiating this relation with respect to x x, y , and at R ̓and its derivatives- x, y;[,K . we obtain The relation (4) is checked directly. Let us put x . vx x, y a12 x, y v x, y ³ a02 t, y v t, y dt 0 [ N Rx a12 R ; Now let[ x . Hence 189 Ye. A. Utkina and A. Maher y vx x, y a12 x, y v x, y 0 . u R u a R u a R x ,K dK ³ > KK x K 21 x 20 x @ 0 Let us now start finding u x, y . We y0 change in (5) the roles of the variables x with[ u a12 R y x0 , y and y with K and calculate the integral in the right x uR u a R u a R [, y d[. and left parts in the bounds x0 d [ d x , ³ > [[y [ 12 y 02 y @ 0 x0 y0 dK d y : In view of (2), the last equality can be written in the form u xy x, y u xy R x, y0 u x Ry x, y0 u R x, y uR x, y y x 0 xy 0 c uxy x, y \1 x R x, y0 \ c x Ry x, y0 u xy R x0 , y u x Ry x0 , y u y Rx x0 , y \ x R x, y \ x R x, y 1 x 0 xy 0 uRxy x0 , y u xy R x0 , y0 c c u x Ry x0 , y0 u y Rx x0 , y0 uRxy x0 , y0 M1 y R x0 , y M1 y Ry x0 , y M y Rx x0 , y x y c M y R x , y M y R x , y ³³R [,K; x, y L u dKd[ xy 0 1 0 0 0 x00y M1 y0 Ry x0 , y0 \1 x0 Rx x0 , y0 x y >@u x P x0 , y >@uPx x0 , y >@u x P x, y0 M y0 Rxy x0 , y0 R [,K; x, y L u dKd[ >@uPx x, y0 >@u x P x0 , y0 ³³ x00y M y P x , y M y P x , y \ c x P x, y >@uPx x0 , y0 >u y N @ x0 , y >uN y @ x0 , y 1 0 x 0 0 \ x Px x, y0 M1 y0 P x0 , y0 >@u y N x, y0 >@uN y x, y0 M y0 Px x0 , y0 Mc y N x0 , y >u y N @ x 0 , y 0 >uN y @ x 0 , y 0 M y N y x0 , y \ 1 x N x, y0 \ x N y x, y0 y u Q uQ x ,K dK x N x , y y N x , y ³ >@x x 0 \ 1 0 0 0 M 0 y 0 0 y 0 y x Q x , Q x , d u K uK [ , y d[ uT x , y >@M1 K 0 K M K x 0 K K ³ >@y y 0 >@ 0 ³ y0 x0 y x uT x, y uT x , y uF x ,K dK \ 1 [ K [, y0 \ [ K y [, y0 d[ >@ 0 >@ 0 0 ³ >@1 0 ³ >@ y0 x0 x M y T x0 , y \ x T x, y0 >@uF2 [ , y 0 d[ u x R y x0 , y ³ >@ y x0 M y0 T x0 , y0 M K F1 x0 ,K dK >@u x R y x, y 0 >@u x R y x0 , y 0 ³ y0 >@u y R x x, y 0 >@uR xy x, y 0 x \ [ F [, y d[ u a 21 R x x, y 0 ³ 2 0 x0 M1 y Ry x0 , y \ c x Ry x, y0 M1 y0 Ry x0 , y0 Fourth order pseudoparabolic equation 190 \ 1 x Rx x, y0 \ x Rxy x, y0 y ª x º \ x a R x, y M K R x ,K u3 x, y exp« D1 t, y dt» 21 x 0 ³ > KK x 0 ³ « x0 » y0 ¬ ¼ x t Mc K a21R x x0 ,K M K a20 R x x0 ,K @dK ª ª º º «C y f t, y exp« D t , y dt»dt»; y a R x , y 3 ³ ³ 1 1 M 12 y 0 « x x » ¬ 0 ¬« 0 ¼» ¼ x ª y º \ [ R [, y \ c [ a R u x, y exp« E x,W dW » ³ > [[y 0 12 y 2 ³ 1 (6) y x0 ¬« 0 ¼» y [, y0 \ [ a02 R y [, y0 @d[.