A Note on Integral Transforms and Partial Differential Equations

Applied Mathematical Sciences, Vol 4, 200, no 3, 09-8 A Note on Integral Transforms and Partial Differential Equations Adem Kılıçman and Hassan Eltayeb Department of Mathematics and Institute for Mathematical Research University Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia akilicman@putraupmedumy eltayeb@putraupmedumy Abstract In this study, we apply double integral transforms to solve partial differential equation namely double Laplace and Sumudu transforms, in particular the wave and poisson s equations were solved by double Sumudu transform and the same result can be obtained by double Laplace transform Mathematics Subject Classification: 44A35 Primary 35G5, 44A85; Secondary Keywords: Laplace transform, Sumudu transform, convolution introduction The topic of partial differential equations is very important subject yet there is no general method to solve all the PDEs The behavior of the solutions very much depend essentially on the classification of PDEs therefore the problem of classification for partial differential equations is very natural and well known since the classification governs the sufficient number and the type of the conditions in order to determine whether the problem is well posed and has a unique solution It is also well known that some of second-order linear partial differential equations can be classified as Parabolic, Hyperbolic or Elliptic however if a PDE has coefficients which are not constant, it is rather a mixed typein many applications of partial differential equations the coefficients are not constant in fact they are a function

0 A Kılıçman and H Eltayeb of two or more independent variables and possible dependent variables Therefore the analysis to describe the solution may not be hold globally for equations with variable coefficients that we have for the equations having constant coefficients On the other side there are some very useful physical problems where its type can be changed One of the best known example is for the transonic flow, where the equation is in the form of ( u2 c 2 φ xx 2uv c 2 φ xy + ( v2 c 2 φ yy + f(φ =0 where u and v are the velocity components and c is a constant, see [] Similarly, partial differential equations with variable coefficients are also used in finance, for example, the arbitrage-free value C of many derivatives C τ + σ2 (s, τ 2 C s2 + b(s, τ C r(s, τc =0 2 s 2 s with three variable coefficients σ(s, τ, b(s, τ and r(s, τ In fact this partial differential equation holds whenever C is twice differentiable with respect to s and once with respect to τ, see [8] However, in the literature there was no systematic way to generate a partial differential equations by using the equations with constant coefficients, the most of the partial differential equations with variable coefficients depend on nature of particular problems Recently, A Kılıcman and H Eltayeb in [5], introduced a new method producing a partial differential equation by using the PDEs with constant coefficients and classification of partial differential equations having polynomial coefficients Later the same authors extended this setting in [7] to the finite product of convolution of hyperbolic and elliptic PDEs where the authors considered the positive coefficients of polynomials In this study we are going to solve PDEs with variable boundary conditions by using double integral transform methods: double Laplace transform and double Sumudu transform First of all we give the following definition and during this study we use the following convolution notation: double convolution between two continuous functions F (x, y

and G(x, y given by ( F (x, y F 2 (x, y = Integral transforms and PDE y x 0 0 F (x θ,y θ 2 F 2 (θ,θ 2 dθ dθ 2 for further details and properties of the double convolutions and derivatives we refer to [3] And also the definition of double Laplace and Sumudu transforms given by (2 L x L t [f(x, s] = F (p, s = where x, t > 0 and p, s complex value 0 e px e st f(x, tdtdx 0 and F (v, u = S 2 [f(t, x; (v, u] = uv where x, t > 0 and u, v complex value 0 0 e ( t v + x u f(t, xdtdx respectively More details see [4] and [6] Now since we are going to apply the integral transform methods to PDE, first we must know the integral transform of partial derivatives as follows: double Laplace transform of the first order partial derivative with respect to x given by [ ] f(x, t L x L t = pf (p, s F (0,s x Also the double Laplace transform for second partial derivative with respect to x is L xx [ 2 f(x, t 2 x ] = p 2 F (p, s pf (0,s F(0,s, x the double Laplace transform for second partial derivative with respect to t similarly as above given by [ ] 2 f(x, t L tt = s 2 F(p, 0 F (p, s sf (p, 0 2 t t In a similar manner the double Laplace transform of a mixed partial derivative can be deduced from single Laplace transform as [ ] 2 f(x, t L x L t = psf (p, s pf (p, 0 sf (0,s F (0, 0 x t Similarly, the double Sumudu transform for second partial derivative with respect to x given by (3 S 2 [ 2 f(t, x x 2 ;(v, u ] = u F (v, u 2 u F (v, 0 2 u f(v, 0 x

2 A Kılıçman and H Eltayeb Finally, the double Sumudu transform of 2 f(t,x t 2 (4 S, given by [ ] 2 f(t, x ;(v, u = t 2 v F (v, u 2 v F (0,u 2 v f(0,u t It is well known that in order to obtain the solution of partial differential equations by integral transform methods we need the following two steps: Firstly, we transform the partial differential equations to algebraic equations by using the double integral transform methods Secondly, on using the double inverse transform to get the solution of PDEs Now, let us solve the linear second order partial differential equations by using the two transforms as follows: (5 au xx + bu xy + cu yy + du x + eu y + fu = g (x; y g 2 (x; y under boundary conditions u(x, 0 = f (x f 2 (x, u(0,y=w (y w 2 (y u y (x, 0 = d dx (f (x f 2 (x, u x (0,y= d dy (w (y w 2 (y and u(0, 0 = 0 where the symbol means double convolution [2] and a, b, c, d, e and f are constant coefficient, g (x, y is the exponential function where as g 2 (x, y is a polynomial In this study we consider that (5 has a solution on using the both transforms; further the inverse double Laplace and Sumudu transforms exists In the next we discuss the solution of above equation by using double Sumudu transform, but here we use the matrix forum, before we deal with this transform we give some concept as a k follows, let P (x, y = x + m b l be a function which can be represented by the k y l following matrix product: a a 2 a n ( M P (x, y = x x 2 x a 2 a 3 a n 0 a 3 x n 3 a n 0 0 a n 0 0 + ( y y 2 y 3 y m b b 2 b m b 2 b 3 b m 0 b 3 b m 0 0 b m 0 0

Integral transforms and PDE 3 then we see that M P (x, yu = M P (xu + M P (yu where M P (x, and M P (y define a linear mappings in an obvious way Now we shall write vectors u in C n C m as rows vectors or columns vectors interchangeably, whichever are convenient although, when M P (x, yu is to be computed and of course the vector u is written as a column vector (6 M P (x, yu = x i n i a i+k u k + m m j b y j j+l u l for any u k =(u 0 (y,u 2 (yu n (y and u l =(u 0 (x,u 2 (xu n (x C n C m In particular case, we apply the double Sumudu transform for a linear second order partial differential equation with constant coefficient Let g(x, y be a continuous on (0, (0,, zero on (, 0 (, 0 then it is locally integrable and Laplace transformable Let u(x, y be differentiable on (0, (0,, and satisfy j= (7 au xx + bu xy + cu yy + du x + eu y + fu = g(x, y under the boundary conditions u(x, 0 = f (x f 2 (x, u(0,y=w (y w 2 (y u y (x, 0 = d dx (f (x f 2 (x, u x (0,y= d dy (w (y w 2 (y and u(0, 0 = 0 then the transform of this equation lead us to the ( ( ( d a U(0,β A (α, β S x S y [u(x, y] (α, β = S x S y [g(x, y] (α, β+ α α 2 a 0 U x (0,β ( ( e c + β β 2 c 0 ( U(α, 0 U y (α, 0 +0 where A(p, q = a α + bαβ + c + dα + eβ + f and Sumudu transform of single convolution and derivative of single convolution are given 2 β2 by S x [f (x f 2 (x] = αf (αf 2 (α, S y [w (y w 2 (y] = βw (βw 2 (β and [ ] d S x dx (f (x f 2 (x [ ] d S y dy (w (y w 2 (y = α [F (α F (0] F 2 (α, = β [W (β W (0] W 2 (β,

4 A Kılıçman and H Eltayeb respectively Proposition Let f(x, t be n, m times differentiable on (0, (0, and f(x, t =0for t, x < 0 Then for any polynomial P of degree (m, n then double Sumudu transform S x S t [P (Df(x, y] (α, β =P (α, βs x S t [f(x, t] (α, β M P (βψ(f,n M P (αφ(f,m In particular [ ] m f S x S t x m = α m S xs t [f(x, t] (α, β ( α, m α,, m α and similarly we have [ ] n f S x S t = t n β S xs n t [f(x, t] (α, β ( β, n β,, n β for n =2and m =2we have a a 2 a n a 2 a 3 a n 0 a 3 a n 0 0 a n 0 0 b b 2 b n b 2 b 3 b n 0 b 3 b n 0 0 b n 0 0 f 0 (0,β x f(0,β m x m f(0,β f 0 (α, 0 (α, 0 y n y n (α, 0 [ ] 2 f S x S t (α, β = ( ( x 2 α S 0 xs 2 t [f(x, t] (α, β α α 2 0 and ( f 0 (0,β x f(0,β [ ] 2 f S x S t (α, β = ( ( t 2 β S 0 xs 2 t [f(x, t] (α, β β β 2 0 ( for n =and m =, we have [ ] f (8 S x S t (α, β = x α S xs t [f(x, t] (α, β α f(0+,β f 0 (α, 0 (α, 0 y

and Integral transforms and PDE 5 (9 [ ] f S x S t (α, β = t β S xs t [f(x, t] (α, β f(α, 0+ β Proof We use induction on n, m The result is trivially true if n =0,m =0, and the case n =,m = are equivalent to (8 and (9 respectively Suppose now that the result is true for some n, m > 0 and let P (x, y = n+ m+ a k x + k b l t l having degree (n +,m+ The first two statements follow by putting h = f x, z = f and using the induction hypothesis and (8 and (9 Now write t P (x, t =a 0 + x ζ(x+b 0 + t η(t, where ζ(x = n a k+ x k and η(t = m b l+ Then t l P ( Df(x, t =a 0 f + ζ( Dh + b 0 f + η( Dz and therefore the double Laplace transform given by S x S t [P (Df](p, s = a 0 S x S t [f]+s x S t [ζ(dh]+b 0 S x S t [f]+s x S t [η(dz] = a 0 S x S t [f]+ζ(ps x S t [h]+b 0 S x S t [f]+η(ss x S t [z] M ζ (pψ(f,n M η (sφ(f,m = a 0 S x S t [f](α, β+b 0 S x S t [f](α, β+ζ(α ( +η(β β S xs t [f](α, β f (α, 0+ β n i α i a i+k+ f (k+ (0+,β m j= ( α S xs t [f](α, β α f (0+,β m j β j b j+l+ f (l+ (α, 0+

6 A Kılıçman and H Eltayeb on using (6 and h = f (k+ and z = f (l+ The summation above can also be written = = n i+ a α i i+k f (k (0+,β+ k= n i+ a α i i+k f (k (0+,β+ n+ m+ + α i a if (0+,β n i+ α i j= m j+ β j m j+ m b β j j+l f (l (α, 0+ j= j= l= m j+ m b β j j+l f (l (α, 0+ β j b jf (α, 0+ a i+k f (k (0+,β α n+ a n+f (0+,β b j+l f (l (α, 0+ β m+ b m+f (α, 0+ α i a if (0+,β β j b jf (α, 0+ = M P (αψ(f,n + ζ(αf (0+,β+M p (βφ(f,m + η(βf (α, 0+ Thus ( S x S t [P (Df](α, β = a 0 S x S t [f](α, β+b 0 S x S t [f](α, β+ζ(α α S xs t [f](α, β α f (0+,β ( +η(β β S xs t [f](α, β f (α, 0+ β finally we have M P (αψ(f,n + ζ(αf (0+,β+M p (βφ(f,m + η(βf (α, 0+ S x S t [P (Df(x, y] (α, β =P (α, βs x S t [f(x, t] (α, β M P (βψ(f,n M P (αφ(f,m In particular, if we consider the wave equation in the form of (0 u tt u xx = 3e 2x+t (x, t R 2 + u(x, 0 = e 2x + e x, u t (x, 0 = e 2x + e x u(0,t = 2e t, u x (0,t=3e t then on using the above proposition we have

Integral transforms and PDE 7 3 A (α, β S x S y [u(x, y] (α, β = ( β( 2α ( ( 0 + α α 2 0 ( ( 0 + β β 2 0 where A (α, β = β by simplify and arrangement, we have 2 α2 2 ( β 3 ( β 2 3α ( 2α( α 2 3α ( 2α( α S x S y [u(x, y] (α, β = 3α 2 β 2 (α 2 β 2 ( 2α( β β 2 (2 + 3α (α 2 β 2 ( β ( + α2 (2 3α 3αβ +2β (α 2 β 2 ( 2α( α in the next step we replace the complex variables α, β in eq( by p, q respectively and multiply each terms by then by applying the inverse transform and we pq compute the poles by using MAPLE software, then we obtain the solution of eq(0 as follows u(x, t =e 2x+t + e x+t In the following example we apply the above proposition and double Sumudu transform to the Poisson s equation in the form of (2 u xx + u yy = 4e x+y sin(x + y u(x, 0 = e x cos(x, u(0,y=e y cos(y u x (0,y = e y cos(y e y sin(y u y (x, y = e x cos(x e x sin(x, then on using the above proposition and inverse double Sumudu transform we obtain the solution of eq(2 in the form of u(x, y =e x+y cos(x + y Thus, the same result can easily be obtained on using the double Laplace transform

8 A Kılıçman and H Eltayeb Acknowledgement: The authors gratefully acknowledge that this research partially supported by Ministry of Science, Technology and Innovations(MOSTI, Malaysia under the e-science Grant 06-0-04-SF05 References [] G-Q Chen and M Feldman, Multidimensional transonic shocks and free boundary problems for nonlinear equations of mixed type, J Amer Math Soc, 6 (2003, pp 46 494 [2] H Eltayeb and A Kılıçman A Note on Solutions of Wave, Laplace s and Heat Equations with Convolution Terms by Using Double Laplace Transform: Appl, Math, Lett 2 (2008 324 329 [3] H Eltayeb and A Kılıçman A note on double Sumudu transform and double Laplace transform, Int Pure and Appl Math, to appear [4] A Kılıçman and H Eltayeb A Note on Defining Singular Integral as Distribution And Partial Deferential Equation with Convolution term: Mathematical and Computer Modelling 49 (2009 327 336 [5] A Kılıçman and H Eltayeb A note on classification of hyperbolic and elliptic equations with polynomial coefficients, Applied Mathematics Letters 2( (2008, pp 24 28 [6] A Kılıçman and H E Gadain An Application of Double Laplace Transform and Double Sumudu Transform, Lobachevskii Journal of Mathematics, 2009, Vol 30, No 3, pp 24 223 [7] A Kılıçman and H Eltayeb On finite product of convolutions and classifications of hyperbolic and elliptic equations, arxiv: 0902422v [ mathap], 6 jan 2009 [8] R C Merton Theory of rational option pricing, Bell Journal of Economics and Management Science 4(Spring(973, pp 4 83 Received: May, 2009