Generalization of Separation of Variables n-Harmonic Equation m Dimension and Unbounded Boundary Value Problem
aa r X i v : . [ m a t h . G M ] O c t Generalization Of Separation Of Variables n-Harmonic Equation m Dimensionand Unbounded Boundary Value Problem
Ibraheem OtufDepartment of mathematicsNorthern Border UniveristySaudi Arabiae-mail:[email protected]
ABSTRACT
The method of separation of variables is significant, it has been applied tophysics, engineering , chemistry and other fields. It allows to reduce the diffculity of problemsby separating the variables from partial differential equation system into ordinary differentialequations system. However, this method has complexity in higher order partial differentialequations. In this reserach, we generalize this method by using multinomial theorem of n-harmonicequation to solve n-harmonic equation with m dimension and then solving an important classof partial differential equations with unbounded boundary conditions. Additionaly, applicationof convolution. KEYWORDS : Fourier transform, boundary value problem, n-harmonic equation, applicationof convolution theorem.
The n-harmonic 2 dimesion has been studied by Doscharis in [1], and neumann Problem1-harmonic m dimesion have been solved by Otuf in [3]. We are intersted in n-harmonic equationm dimension which is in the form of △ nm u ( x ) = 0 , (1)where x = ( x , x , ........, x m ) . This equations has high complexity since it is a partial differentialequation of order 2n. However, separation of variables ia an intersting approch to solve thiskind of problem. At begining, we need to put our attention in driving new concept whichdecrease the complexity of solving this problem. Multinomial theorem of n-harmonic equationwhich is coming directly from the original multinomial theorem allows to wrtie the n-harmonicequation as summuation of ordinary differential equations. This lead to separate these equationsand write them as sequnce of equations. Solving these equations will desgin an importantgereralization. Moreover, we are intersted in solving unbounded boundary value problem thatsatisfying △ nm u ( x ) = 0 and then application of its convolutions theorem. We formalize multinomial theorem to desgin short approch to separate the variables of △ nm u ( x ) , and write them as summutions of ordinary differential equations. Supose for y i = X (2) i X i , we have y ( n ) i = X (2 n ) i X i . We let u ( x ) = Q m − i =1 X i ( x i ) X m ( x m ) , then apply △ nm to u ( x ) andthen divide by Q m − i =1 X i ( x i ) X m ( x m ) leads to the multinomial theorem of n-harmonic equationwhich is as following 1 m X r =1 y i ! ( n ) = X h + h + ...... + h m = n (cid:18) nh , h , ......h m (cid:19) m Y i =1 y h i i . (2)Where (cid:18) nh , h , ......, h m (cid:19) = n ! h ! h ! .........h m ! , (3)( n ) is order of dervative and h i = 0 . Furthermore, mutiplication of any two or more ordinarydifferential equations, yields sum of their orders. For example X (2)2 X (4)2 = X (6)2 We list someexamples For △ u ( x ) = 0 , we have( y + y ) (2) = y (2)1 + 2 y y + y (2)2 = X (4)1 X + 2 X (2)1 X X (2)2 X + X (4)2 X . (4)For △ u ( x ) = 0 , we have( y + y ) (3) = y (3)1 + 3 y (2)1 y + 3 y y (2)2 + y (3)2 = X (6)1 X + 3 X (4)1 X X (2)2 X + 3 X (2)1 X X (4)2 X + X (6)1 X . (5)For △ u ( x ) = 0 , we have( y + y + y ) (3) = X (6)1 X + X (6)2 X + X (6)3 X + 3 X (4)1 X X (2)2 X + 3 X (2)1 X X (4)2 X (6)+ 3 X (4)1 X X (2)3 X + 3 X (2)1 X X (4)3 X + 3 X (4)3 X X (2)2 X + 3 X (2)3 X X (4)2 X . (7) Let x = ( x , x , ........x m ) be vector in R m . The n-harmonic equation is △ nm u ( x ) = 0 . (8)Let λ i j to be an constants in R , where i = 1 , , , .......m, and j = 1 , , ......, n. We seek solutionof the form u ( x ) = m − Y i =1 X i ( x i ) X m ( x m ) . (9)Apply method of sepration of varibles to get multinomial theorem of n-harmonic equationthat is m X r =1 y i ! ( n ) = X h + h + ...... + h m = n (cid:18) nh , h , ......h m (cid:19) m Y i =1 y h i i . (10)Where y i = X (2) i X i , (11)and y ( n ) i = X (2 n ) i X i . (12)Let λ i j = X (2 j ) i X i , (13)2e get ordinary differential equations those are X (2 j ) i − λ i j X i = 0 , (14)and ( n ) X k =0 (cid:18) nk (cid:19) X (2 n − k ) m K (2 k ) m = 0 . (15)Where X (0) m = X, (2 j ) is the order of dervative, and k m = P m − i =1 λ i . Let solve (14) if j = 1 , we get X (2) i − λ i X i = 0 . (16)If λ i < , we have X i ( x i ) = a i cos p λ i x i + b i sin p λ i x i . (17)for j > . we will let λ i j = ( λ i ) j in (14), then the solutions of this forms have two terms.First one is (17), and second terms in which their constant forced to be zeros. The final resultis (17). Similary if λ i > , we have X i ( x i ) = a i cosh p λ i x i + b i sinh p λ i x i . (18)If λ i = 0 , we have X i ( x i ) = a i + b i x. (19)The solution of (15), elementary differetial equation yields n X k =0 ( X m + k m ) ( n ) = 0 . (20)Where X (0) = X. This gives n X k =0 ( s + k m ) n = 0 . (21)If k m < , we have X m ( x m ) = n X r =1 x r − m [ c r cosh p − k m x m + d r sinh p − k m x m ] . (22)If k m > , we have X m ( x m ) = n X r =1 x r − m [ c r cos p k m x m ) + d r sin p k m x m ] . (23)Finally if k m = 0 , we have X m ( x m ) = m X r =1 c r x r − . (24)The final solutions depends on choice of λ i for each i = 1 , , .........m − . We have u ( x ) = m − Y i =1 X i ( x i ) X m ( x m ) . (25)Remark: We may have different signs for λ i . Boundary Value Problem Unbounded Domain
We consider the Boundry value problem (BVP) replace x = ( x , x , ........, x m − , x m ) by( x, x m ) = ( x , x , ........, x m − , x m ) . △ nm u ( x, x m ) = 0 , −∞ < x i < ∞ ; L < x m < ∞ , (26) u ( x,
0) = f ( x ) , | u ( x, x m ) | < M. (27)Write u ( x, x m ) = Q m − i =1 X i ( x i ) X m ( x m ) . Apply gererlazation of sepration of varibles to get X (2 j ) i − λ i j X i = 0 , (28)and n X k =0 ( X m + k m ) ( n ) = 0 . (29)Where X (0) = X. From the gereraliztion we need only j = 1 i.e. X (2) i − λ i X i = 0 , (30)and by letting λ i = − w i . We get X (2) i + w i X i = 0 . (31)The solutions is X i ( x i ) = a i cos w i x i + b i sin w i x i . (32)The solutions for the mth variable is in the form X m ( x m ) = n X r =1 x r − m [ q r e −√− k m x m + f r e √− k m x m ] . (33)The final solutions u ( x, x m ) = m − Y i =1 [ a i cos w i x i + b i sin w i x i ] n X r =1 x r − m [ q r e −√− k m x m + f r e √− k m x m ] . (34)Where k m = − P m − i =1 w i . Since w i > , and x m → ∞ , we must have f r = 0 . It follows that u ( x, x m ) = Z (0 , ∞ ) m − n X r =1 m − Y i =1 [ A i r x r − m cos w i x i + x r − m B i r sin w i x i ][ e −√− k m x m ] dw dw · · · dw m − . (35)Since u ( x, L ) = f ( x ) , then f ( x ) = Z (0 , ∞ ) m − n X r =1 m − Y i =1 L r − e −√− k m L [ A i r cos w i x i + A i r sin w i x i ] dw dw · · · dw m − . (36)Where w = ( w , w , ........, w m − ) . Thus, we have A i r ( w ) = P nr =1 L − r e √− k m L π m − Z ( −∞ , ∞ ) m − f ( x ) m − Y i =1 cos w i x i dx dx · · · dx m − . (37)4 i r ( w ) = P nr =1 L − r e √− k m L π m − Z ( −∞ , ∞ ) m − f ( x ) m − Y i =1 sin w i x i dx dx · · · dx m − . (38)Where r = 1 , , ..........., n. The case 1-harmonic 2 dimenstion has been done in [2] , we extended to n-harmonic 2dimenstion. Let us the result of previous problem when m=2, u ( x , x ) = Z ∞ n X r =1 [ A r x r − cos w x + x r − B i r sin w x ][ e − w x ] dw . (39)where A r ( w ) = P nr =1 L − r e w L π Z ∞−∞ f ( x ) cos w x dx . (40) B r ( w ) = P nr =1 L − r e w L π Z ∞−∞ f ( x ) sin w x dx . (41)After we plug the coefficients into the general formula, we get u ( x , x ) = P nr =1 L − r e w Lπ Z ∞−∞ Z ∞ n X r =1 [ x r − cos w x cos w z f ( z ) . (42)+ x r − sin w x sin w z f ( z )][ e − w x ] dw dz (43)= P nr =1 L − r e w L π Z ∞−∞ Z ∞ n X r =1 [ x r − e − w x cos w ( z − x ) f ( z )] dw dz . (44)Let g ( x , x ) = Z ∞ n X r =1 cos w ( x ) x r − e − w x dw (45)= n X r =1 Z ∞ e iw x + e − iw x e − w x x r − dw (46)= n X r =1 x r − e − w x (cid:18) x − ix + 1 x + ix (cid:19) . (47)= n X r =1 x r − e − w x x x + x . (48)Define the convolution theorem in the form of u ( x , x ) = P nr =1 L − r e w L π Z ∞−∞ f ( z ) g ( x − z , x ) dz . (49)Let f ( z ) = (cid:26) z> , z< . (cid:27) (50)Now apply the theorem to get 5 ( x , x ) = P nr =1 L − r e w L π Z ∞−∞ n X r =1 x r − e − w x x x + ( x − z ) dz . (51)= P nr =1 L − r e w L π n X r =1 x r − e − w x m Z ∞−∞ e −√ k m x m x m x m + ( x i − z i ) dz i . (52)= P nr =1 L − r e w L π n X r =1 x r − e − w x (cid:20) π − (cid:18) x x (cid:19)(cid:21) . (53) The parabolic version of n-harmonic equation is α △ nm u ( x, t ) = u t ( x, t ) . (54)Where ( x, t ) = ( x , x , ........x m , t ) , t > , α ∈ R , and ( x, t ) ∈ R m × (0 , ∞ ) . The betweendifference between the eillptic version of n harmonic equation and prarbolic one is similar todifference between heat equation and 1-harmonic equation. The solution can presented as u ( x ) = Q mi =1 X i ( x i ) T ( t ) . X i ( x i ) has same solutions as the one in n harmonic equation. For T ( t ) we have T ′ ( t ) − αk m +1 T ( t ) = 0 . The solutions is T ( t ) = Ae αtk m +1 . The hyperbolic version of n-harmonic equation is β △ nm u ( x, t ) = u t ( x, t ) . (55)Where ( x, t ) = ( x , x , ........x m , t ) , t > , α ∈ R , and ( x, t ) ∈ R m × (0 , ∞ ) . The differencebetween the eillptic version of n harmonic equation and hyperbolic one is similar to differencebetween wave equation and 1-harmonic equation. The solution can be written as u ( x ) = Q mi =1 X i ( x i ) T ( t ) . X i ( x i ) has same solutions as the one in n harmonic equation. For T ( t ) wehave T ′′ ( t ) − β k m +1 T ( t ) = 0 . The solutions is T ( t ) = Ce β √ k m +1 t + De − βt √ k m +1 t . Sepations of variables is powerful method that assist to reduce the diffcuilty of partialdifferetion equation problems and solve them. The n-harmonic m dimension has been generlaized.Moreover, unbounded boundary value problem has been solved and applications of its convolution.Hyperbolic version and parabolic version are dicussed.
References [1] Michael Doschoris,
Towards a generaliztaion of seprations of variables,
Method AndApplication Of Analysis Vol. 19, No. 4, pp. 381-402, December 201262] Richard Haberman,
Applied Partial Differential Equation with Fourier Series, andBoundary Value Problems (5th edition), Pearson Education, 2013.[3] Ibraheem Otuf, Xingping Sun,