International Mathematical Forum, 1, 2006, no. 21, 1043 - 1050
Multi-Dimensional Laplace Transforms and Systems of Partial Differential Equations A. Aghili and B. Salkhordeh Moghaddam Department of Mathematics, Faculty of Science Guilan University, P. O. Box 1841, Rasht, Iran
[email protected] [email protected] Abstract The object of this paper is to establish new theorem and corollary involving systems of two-dimensional Laplace transforms containing several equations. This system can be used to calculate new Laplace transform pairs. In the second part, system of partial differential equations related to telegraph equation is solved by using the double Laplace transformation.
Mathematics Subject Classification: Primary 44A30, Secondary 35L05 Keywords: Multidimensional Laplace transforms, telegraph equation
1
Introduction
The two-dimensional Laplace transform of function f (x, y) is defined by Ditkin and Prudnikov [5]as follows: F (p, q) = pq
∞ ∞ 0
0
e−px−qy f (x, y) dx dy
and symbolically is denoted byF (p, q)=f ¨ (x, y)where the symbol=is ¨ called ”operational”. The correspondence between f (x, y) and F (p, q) may be interpreted as transformation which transforms the function f (x, y) into the function F (p, q). Thus, we call F (p, q) the image of f (x, y) and f (x, y) is the original of F (p, q) . In this article we consider a theorem in general case (con 1 1 i/2 j/2 sidered by Dahiya [7]) as a result we calculate the image of x y G nx , my ,x−i/2 y j/2 H
1 , 1 nx my
and x−i/2 y j/2T
i/2 j/2
1 , 1 nx my
1 , 1 nx my
when i = 1, 2, 3 and j = 3, 5, 7.
Also the image of x y G when i = ±3, ±5 and j = ±1, ±3 (see[3]) and in the case of i = 3, j = ±1, 3 (see[7]).
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2
A. Aghili and B. Salkhordeh Moghaddam
THE MAIN THEOREM
Theorem √2.1√ Let f ( x, y ) G(p, q)= ¨ √xy then
j
i
¨ x k +1 y − h +1 G (nx )−1 , (my)−1 = −(i−k)
j+h
j−h
i+k
16 p 2k q 2h n− 2k −1 m 2h√ −1 √ s p t q j−h √ √ × 0∞ 0∞ Besselk i+k , 2 Besselk , 2 f (s, t) ds dt k n h m − ki +1
x
y
j +1 h
G (nx ) , (my)
h−j 2h
i+k 2k
−1
−1
(2.1)
= ¨
− j+3h −1 2h
i−3k −1 2k
16 p q n− m √ √ s p t q j+h √ √ × 0∞ 0∞ Besselk k−i , 2 Besselk , 2 f (s, t) ds dt k n h m where i, j, k, n andmare positive. Proof: Let √ k−i i+k i+k i s2 i+k s p − − x k e nx ÷ 2 p 2k s k n 2k Besselk ,2 √ k n √ i−k s p ÷ 2 p s n Besselk x e ,2 √ k n √ t2 j h−j j+h j+h t q j + h − − ,2 √ y h e my ÷ 2 q 2h t h m 2h Besselk h m √ t2 j+h h−j j−h j−h t q − hj − my ,2 √ y e ÷ 2 q 2h t h m 2h Besselk h m 2
i+k 2k
s − ki − nx
k−i k
(2.2)
(2.3)
k−i 2k
(2.4)
(2.5) (2.6)
we start from (2.3) and (2.6) operational relations and multiply together (2.3) and(2.6) equations to get: i
t2
j
s2
x k y − h e− my − nx = ¨
4p
k−i 2k
q
j+h 2h
s
i+k k
t
h−j h
− i+k 2k
n
m
j−h 2h
Besselk
√ √ i+k s p j−h t q ,2 √ ,2 √ Besselk k n h m
now, we multiply both sides by f (s, t) and integrate with respect to s and t over the positive quarter plane to obtain i
j
x k y− h
∞ ∞ − t2 − s2 my nx 0
0
e
f (s, t) ds dt = k−i
j+h 2h
i+k
j−h
n− 2k m 2h √ √ ∞ ∞ i+k h−j i+k s p j−h t q ,2 √ ,2 √ × s k t h Besselk Besselk f (s, t) ds dt k n h m 0 0 4 p 2k q
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Multi-Dimensional Laplace Transforms
√ √ we make the change of variabless = u and t = v on the right hand side to v u √ √ j i obtain x k y − h 0∞ 0∞ e− my − nx f ( u, v) √1uv du dv = k−i
×
∞ ∞ 0
0
s
i+k k
t
h−j h
Besselk
16 p 2k q
j+h 2h
i+k
n− 2k m
j−h 2h
√ √ i+k s p j−h t q Besselk f (s, t) ds dt ,2 √ ,2 √ k n h m
√ √ f ( x, y ) finally by using G(p, q)= ¨ √xy j i 1 1 , my = ¨ x k +1 y − h +1 G nx
relation on the left hand side we get
k−i
j+h 2h
i+3 k
j−3 h
n− 2k m 2h √ √ ∞ ∞ k−i h−j i+k s p j−h t q ,2 √ ,2 √ × s k t h Besselk Besselk f (s, t) ds dt k n h m 0 0 16 p 2k q
from(2.4) and (2.5) operational relations, we have 1 1 − ki +1 hj +1 x y G nx , my = ¨ k+i
h−j 2h
j+3h
3k−i
n− 2k m− 2h √ √ ∞ ∞ k−i h+j k−i s p j+h t q ,2 √ ,2 √ × s k t h Besselk Besselk f (s, t) ds dt k n h m 0 0 16 p 2k q
Corollary2.1: Let i) F (p, q)=f ¨ (x, y) ∂ ii) φ(p, q) = −pq ∂q 2
∂ iii) σ(p, q) = pq ∂q 2
3
∂ iv) η(p, q) = −pq ∂q 3
F (p,q) pq
F (p,q) pq
F (p,q) pq
=yf ¨ (x, y)
=y ¨ 2f (x, y)
=y ¨ 3f (x, y)
√ √ −1/2 v) G(p, q)=(xy) ¨ f ( x, y) then 1 1 , my )= ¨ mπ√n q −1/2 φ(2 np , 2 mq )+ 2√πmn q −1 F (2 np , 2 mq ) (2.7) x1/2 y 3/2 G( nx
√π q −1 σ(2 np , 2 mq ) + 2m3π√n q −3/2 φ(2 np , 2 mq ) m mn q √3π q −2 F (2 p , 2 ) (2.8) 4 mn n m 1 1 x1/2 y 7/2 G( nx , my )= ¨ m2π√n q −3/2 η(2 np , 2 mq ) + m√3πmn q −2 σ(2 np , 2 mq ) q p q 15π 15π −3 √ q −5/2 φ(2 p , 2 √ )+ q F (2 , 2 ) (2.9) 4m n n m 8 mn n m
1 1 , my )= ¨ x1/2 y 5/2 G( nx
Moreover if we set
+ +
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A. Aghili and B. Salkhordeh Moghaddam
√ √ vi) H(p, q)=y ¨ −1/2 f ( x, y) then 1 1 , my )= ¨ x−1/2 y 5/2 H( nx
π 3π 1/2 −3/2 √ p1/2 q −1 σ(2 np , 2 mq )+ 2m p q φ(2 np , 2 mq )+ m m q √3π p1/2 q −2 F (2 p , 2 ) (2.10) 4 mn n m 1 1 π 1/2 −1/2 x−1/2 y 3/2 H( nx , my )= ¨ m p q φ(2 np , 2 mq ) + 2√πm p1/2 q −1 F (2 np , 2 mq ) (2.11) q 1 1 π 1/2 −3/2 −1/2 7/2 √ p1/2 q −2 σ(2 p , 2 x y H( nx , my )= ¨ m2 p q η(2 np , 2 mq )+ m3π )+ m n m p q p q 15π 1/2 −5/2 15π 1/2 −3 p q φ(2 n , 2 m )+ 8√m p q F (2 n , 2 m ) (2.12) 4m
and if we set √ √ vii) T (p, q)=xy ¨ −1/2 f ( x, y) μ(p, q)=xy ¨ 3 f (x, y) ψ(p, q)=xy ¨ 2 f (x, y) β(p, q)=xyf ¨ (x, y) γ(p, q)=xf ¨ (x, y) then 1 1 x−3/2 y 3/2 T ( nx , my )= ¨ √ √ π n π√ n p q p q p q nπ 1/2 −1/2 −1/2 −1 pq β(2 , 2 )+ pq γ(2 , 2 )+ p q ψ(2 , 2 )+ m 2 m n m 2m n m n m p q nπ 1/2 −1 √ p q F (2 n , 2 m ) (2.13) 4 m 1 1 −3/2 5/2 x y T ( nx , my )= ¨ √ √ √ π √n q √ n pq −2 γ(2 p , 2 pq −1 φ(2 np , 2 mq )+ 3π2mn pq −3/2 β(2 np , 2 mq )+ 3π )+ m m 4 m n m q nπ √ p1/2 q −1 σ(2 p , 2 ) 2m m n m q p q 3nπ 1/2 −3/2 √ p1/2 q −2 F (2 p , 2 + 83nπ )+ p q ψ(2 , 2 ) (2.14) m n m 4m n m 1 1 −3/2 7/2 x y T ( nx , my )= ¨ √ √ √ π n 3π√ n 15π n p q p q p q −3/2 −2 −5/2 pq μ(2 , 2 )+ pq φ(2 , 2 )+ pq β(2 , 2 )+ 2 n m m m 4m n m n m √m 15π n p q p q p q nπ 3nπ √ pq −3 γ(2 n , 2 m )+ 2m2 p1/2 q −3/2 η(2 n , 2 m )+ 2m√m p1/2 q −2 σ(2 n , 2 m )+ 8 m p q q 15nπ 1/2 −5/2 15nπ √ p1/2 q −3 F (2 p , 2 p q ψ(2 , 2 )+ ) (2.15) 8m n m 16 m n m
√ Example2.1. Letf (x, y) = y −1 sin(2 xy) , then i) F (p, q) = q(Ci(pq/2)sin(pq/2) − Ssi(pq/2)cos(pq/2)) ii) ψ(p, q) = (−1/2)pq(Ci(pq/2)cos(pq/2) + Ssi(pq/2)sin(pq/2)) iii) σ(p, q) = (1/4)p(2 − pqCi(pq/2)sin(pq/2) + pqSsi(pq/2)cos(pq/2))
1047
Multi-Dimensional Laplace Transforms
iv) η(p, q) = (1/8)pq −1 (4+(pq)2Ci(pq/2)cos(pq/2)+(pq)2Ssi(pq/2)sin(pq/2)) √ √ v) G(p, q) = q pπ(π − 2arctan( pq)) using(2.7)and(2.9),and simplifying , yields √ √ √ √ √ −2 πp 2 pq 2 pq 2 pq 2 pq 1 1/2 ¨ √mn (Ci( √mn )cos( √mn ) + Ssi( √mn )sin( √mn )) + y (π − 2arctan( √mnxy ))=
√ 2 pq
√ 2 pq
√ 2 pq
√ 2 pq
π/p(Ci( √mn )sin( √mn ) − Ssi( √mn )cos( √mn )) 1 y 3/2 (π − 2arctan( √mnxy ))= ¨ √ √ √ √ √ √ 2 pq 2 pq 2 pq 2 pq π −3/2 (4pqCi( √mn )sin( √mn ) − 2 mnpq − 4pqSsi( √mn )sin( √mn ) − 2mn q √ √ √ √ √ √ 2 pq 2 pq 2 pq 2 pq + 6 mnpqCi( √mn )cos( √mn ) + 6qmn mnpSsi( √mn )sin( √mn ) √ 2 pq
√ 2 pq
√ 2 pq
√ 2 pq
− 3mnCi( √mn )sin( √mn ) + 3mnSsi( √mn )cos( √mn )) 1 y 5/2 (π − 2arctan( √mnxy ))= ¨ √ √ −2 −5 √ 2 2 pq √ 2 πm n ((3mnpq − 15(mn) )Ssi( ) + ( mn(pq)3/2 8 mn √ √ √ √ √ 2 pq 2 pq 2 pq 2 pq 3/2 √ √ √ √ − 15 pq(mn) )(Ci( )cos( ) + Ssi( )sin( )) 4 mn √ mn √ mn mn √ 2 pq 2 pq 15(mn)2 + (−3mnpq + 8 )Ci( √mn )sin( √mn ) + 74 pq(mn)3/2 √ Example2.2. Let f (x, y) = xy J (2 xy), then 2 2 pq 3p2 q ψ(p, q) = (pq+1) F (p, q) = (pq+1) 3 4 σ(p, q) = φ(p, q) =
12p3 q (pq+1)5 12p2 q(3pq−2) (pq+1)6 3pq 2 (pq+1)4
180p3 q(pq−1) (pq+1)7 3pq(3pq−1) β(p, q) = (pq+1)5 √ √ T (p, q) = 32π (8 pq −
μ(p, q) =
−1 √ pq
γ(p, q) = 1)p−5/2 q −1 e 2 using(2.13)and(2.14),and simplifying, one has
√ √ √ √ − mnxy √ √ 32 πn2 m3 p(256 npq+20n mpq+mn3/2 ) √ √ x1/2 y 3/2 (m nx1/2 y−8 my 1/2 )e 2 = ¨− q(4 mnpq+mn)5
−
3
√ − mnxy √ √ x√1/2 y 5/2 (m √nx1/2 y −√8 my 1/2 )e 2 = ¨
√ 48 πn3 m3 p(272m npq+2048 m(pq)3/2 +24nm3/2 pq+m2 n3/2 ) √ 3/2 6 q (4 mnpq+mn)
Telegraph equation
The equation for the current I(x, t) and potential V (x, t) at a point x and time t of a transmission line containing resistance R, inductance L, Capacitance C and Leakage inductance G are: LIt + RI = −Vx
CVt + GV = −Ix
(3.1)
with the boundary data I(x, 0) = V (x, 0),
V (0, t) = v0 h(t)
(3.2)
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A. Aghili and B. Salkhordeh Moghaddam
v0 and is constant,h(t) ,is arbitrary function of t . We will use the following for the rest of this section [1] .If V (x, 0) = f (x), V (0, t) = g(t) and if their one-dimensional Laplace transformations are f (u, 0), f (0, v), respectively, then V (x, t)=uv ¨
∞ ∞ 0
0
e−uv−vt V (x, t) dx dt = U(u, v)
¨ 2U(u, v) − uvf (u, 0) Vt (x, t)=uv
(3.4)
¨ 2 vU(u, v) − uvf (0, v) Vx (x, t)=u
(3.5)
(3.3)
and I(x, 0) = β(x), I(0, t) = η(t) and if their one-dimensional Laplace transformations are l(u, 0), l(0, v), respectively, then I(x, t)=uv ¨
∞ ∞ 0
0
e−uv−vt I (x, t) dx dt = Q(u, v)
¨ 2 Q(u, v) − uvl(u, 0) It (x, t)=uv
(3.7)
¨ 2 vQ(u, v) − uvl(0, v) Ix (x, t)=u
(3.8)
(3.6)
by applying two Dimensional Laplace Transformation on system of P.D.E (3.1) we obtain V (x, t)=U(u, ¨ v) = uv
{(G + Cv) (l (0, v) + Lf (u, 0)) − u (Cl (u, 0) + f (0, v))} {(G + Cv) (R + Lv) − u2 }
(3.9)
I(x, t)=Q(u, ¨ v) = uv
{(R + Lv) (Cl (u, 0) + f (0, v)) − u (l (0, v) + Lf (u, 0))} {(G + Cv) (R + Lv) − u2 }
by using boundary data (3.2),we get V (x, t)=U(u, ¨ v) = uv
(G + Cv) l (0, v) − uf (0, v) (G + Cv) (R + Lv) − u2
(3.11)
I(x, t)=Q(u, ¨ v) = uv
(R + Lv) f (0, v) − ul (0, v) (G + Cv) (R + Lv) − u2
(3.12)
denominator has two different roots as follows
u = ± (G + Cv)(R + Lv)
(3.10)
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Multi-Dimensional Laplace Transforms
we consider only u = (G + Cv)(R + Lv), and this root must satisfy the numerator of U(u, v) and Q(u, v) . Therefore, by substitution of u u = (G + Cv)(R + Lv) in l(0, v) = G+Cv f (0, v) , we get :
l(0, v) =
R + Lv f (0, v) G + Cv
(3.13)
next, we substitute relation (3.13) in the (3.11) and (3.12) to obtain √ √ LCv 2 + GLv + CRv − u G + Cv R + Lv + GR f (0, v) V (x, t)=U(u, ¨ v) = (G + Cv) (R + Lv) (GR + GLv + CRv + LCv 2 − u2 ) at this point, in order to calculate original of V (x, t), I(x, t),we need to calculate U(u, v), Q(u, v), respectively. For the following cases with R = 0 and G = 0 : √ LCu − LCv f (0, v) √ V (x, t)=U(u, ¨ v) = LC (u2 − LCv 2 ) √ √ L LCv − u f (0, v) √ I(x, t)=Q(u, ¨ v) = C (LCv 2 − u2 ) (a) h(t) = Heaviside(t − 1)we have V (x, t) = v0 Heaviside(t − 1 −
I(x, t) =
√
LCx)
√ L v0 Heaviside(t − 1 − LCx) C
(b) h(t) = cos ωt we have √ √ V (x, t) = v0 cos(ω( LCx − t))Heaviside(t − 1 − LCx)
I(x, t) =
√ √ L v0 cos(ω( LCx − t))Heaviside(t − 1 − LCx) C
References 1. A.Aghili and B.Salkhordeh, Laplace transform pairs of n-dimensions and a Wave equation. Intern. Math. Journal, Vol. 5, 2004, no. 4,377382. 2. A.Aghili and B.Salkhordeh, Theorems on Two-Dimensional Laplace Transform and Boundary value problem.(to apear in FarEast J.of
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A. Aghili and B. Salkhordeh Moghaddam
Math,2006.) 3.A.Babakhani and .R.s.Dahya systems of Multi-dimensional lapace transforms and a heat equation ,Electronic Journal of Differential Equation ,Conf.07,2001,pp.25-36. 4. Debnath,L. Nonlinear Partial differential Equations for scientists and Engineers, Birkhauser Boston ,1997. 5.Ditkin,V.A. and Prudnikov,A.P. ,Operational Calculus In Two Variables and Its Application ,Pergamon Press, New York,1962. 6. Robrts,G.E. and Kaufman, Tables of Laplace Transforms, W.B.Saunders Company, Phladelphia and London ,1966. 7.R.s.Dahya systems of two-dimensional lapace transforms and their applications, Simon Stevin, A quarterly journal of Pure and Applied Mathematics 59 (1985) n0. 4, 373 384. 8. Voelker, D. Und Doestsch, G., Die Zweidimensionale Laplace Transformation, Verlag Birkuauser Basel, 1950.
Received: September 6, 2005
