Applied Mahemaical Sciences, Vol. 8, 24, no. 82, 463-467 HIKARI Ld, www.m-hikari.com hp://dx.doi.org/.2988/ams.24.45355 On he Fourier Transform for Hea Equaion P. Haarsa and S. Poha 2 Deparmen of Mahemaics, Srinakharinwiro Bangkok, Thailand 2 Wad Ban-Koh School, Bandara, Amphoe Pichai Uaradi 5322, Thailand Copyrigh c 24 P. Haarsa and S. Poha. This is an open access aricle disribued under he Creaive Commons Aribuion License, which permis unresriced use, disribuion, and reproducion in any medium, provided he original work is properly cied. Absrac In his paper, we show ha he soluion of hea equaion can be obained by using he Fourier ransform, he convoluion, and he Fourier inversion. The soluion we obained is unique. Mahemaics Subjec Classificaion: 34B5, 35J5, 35J25, 35L5 Keywords: Hea equaion, Fourier ransform, Fourier inversion Inroducion The hea equaion [2] is of fundamenal imporance in diverse scienific fields. In mahemaics, i is he prooypical parabolic parial differenial equaion. In probabiliy heory, he hea equaion is conneced wih he sudy of Brownian moion via he Fokker-Planck equaion. In financial mahemaics i is used o solve he Black-Scholes parial differenial equaion. The diffusion equaion, a more general version of he hea equaion, arises in connecion wih he sudy of chemical diffusion and oher relaed processes. The hea equaion is used in probabiliy and describes random walks. Lunnaree and Nonlaopon [] presen he fundamenal soluion of operaor + m 2 ) k and his fundamenal soluion is called he diamond Klein-Gordon kernel. They also sudy he Fourier ransform of he diamond Klein-Gordon kernel and he Fourier ransform of heir
464 P. Haarsa and S. Poha convoluion. Romero and Cerui [3] inroduce a new definiion of he Fracional Fourier ransform of order α, <α, of a funcion which belongs o he Lizorkin space of funcions. In his paper, we show ha he soluion of he hea equaion can be obained by using he Fourier ransform, he convoluion and Fourier inversion heorem. The soluion we obained is unique. 2 Preliminaries We consider he hea equaion wih he iniial condiion in our problem. The Fourier ransform and he convoluion are used o solve he problem. Definiion 2.. Fourier ransform [4]. Consider a funcion fx) defined on he inerval, ). The Fourier ransform of f is denoed by he inegral F [f]x) = for x R and he inegral exiss. f)e ix) d. ) Definiion 2.2. Fourier ransform of a convoluion [4]. The convoluion of f and g is he funcion f g) defined by f g)x) = fy x)gy)dy. 2) I also can be wrien as f g)x) = variable. fy)gx y)dy, by he change of Lemma 2.3. Le f = fx) be an absoluely inegrable funcion on, ), and le ν be a real consan, and define he ranslae of f by ν according o he formula f ν x) =fx ν) for all <x<. The Fourier ransform of f ν saisfies he relaion F f ν )γ) =e iγν F f)γ) for all <x<. Proof. By definiion 2.., we have derived he Fourier ransfer o f ν x) as F f ν )γ) = = f ν x)e iγx dx fx ν)e iγx dx 3)
On he Fourier ransform 465 Le y = x ν. Then, dy = dx. The equaion 3) becomes F f ν )γ) = fy)e iγy+ν) dy = e iγν fy)e iγy dy = e iγν F f)γ). 4) 3 Main Resul We nex will show ha e x y)2 χy)dy is a soluion of u u xx +u x =in <x<, <<, ha can be derived by using he Fourier ransform mehod subjeced o he iniial condiion ux, ) = χx) for all <x<. Proof. By aking a Fourier ransform o boh sides of our given equaion and Lemma 2.3., hen we obained F u u xx + u x )γ) = F )γ) F u)γ)+iγ + γ2 )F u)γ) = ) F u)γ)e iγ+γ2 ) = F u)γ) = νγ)e γ2 iγ. 5) Therefore, F χ)γ) =F u., ))γ) =νγ) and F u)γ) =F χ)γ)e γ2 iγ. By able of Fourier ransform [4], i implies ha F e a )2 )γ) = γ 2 e 4a 2a. Take a = ) 2 e, and he above equaion can be rearranged as F 2 )γ) =e γ2. Therefore, ) 2 F u)γ) = F χ)γ)f e )γ)e iγ. 6) 2 where F f g)γ) = Ff)γ)F g)γ) is a Fourier ransform of he convo-
466 P. Haarsa and S. Poha luion [4] f g)x). So, he equaion 6) becomes ) ) 2 F u)γ) = F χ e γ)e iγ 2 = F χ ) e )2 [ = F χ ) e )2 γ)e iγ ] γ). 7) Equaion 7) can be derived from applying Lemma 2.3. by leing ν =. Consequenly, he inversion heorem [4] implies ux, ) = χ ) e )2 x) for all <x< and <<. As a resul, he soluion of he hea equaion for any iniial daa χ is ux, ) = χ ) e )2 x ) = e x y)2 χy)dy for all <x< and <<. Nex, we show ha our soluion is he only one soluion by he energy mehod [4]. Le Λ = u u 2 Then, i can be rewrien as. =.Λ =Λ cλ xx )Λ) = 2 Λ2 ) + cλ x Λ) x + cλ 2 x. 8) By inegraing over he inerval <x<l, we obain l = l 2 Λ2 ) dx cλ x Λ x=l x= + c Λ 2 xdx. 9) Because of he boundary condiions, Λ = a x =,l. d l d l 2 [Λx, )]2 dx = c Since Λ 2 dx is non-increasing, he equaion ) becomes l [Λx, )] 2 dx l [Λ x x, )] 2 dx. ) [Λx, )] 2 dx,. ) By imposing he iniial condiion of u, he righ side of equaion ) vanishes. I implies ha [Λx, )] 2 dx = for all >. Consequenly, Λ. I follows ha u = u 2 for all. Therefore, he soluion we obained is unique.
On he Fourier ransform 467 References [] A. Lunnaree and K. Nonlaopon, On he Fourier ransform of he diamond klein-gordon kernal, In. J. Pure and Appl. Mah., 68)2), 85-97. [2] Wikipedia: Hea equaion in mahemaics. [3] L.G. Romero and R.A. Cerui, A new fracional fourier ransform and convoluion producions, In. J. Pure. and Appl. Mah., 664)2), 397-48. [4] W.A. Srauss, Parial differenial equaions, John Wiley and Sons, Inc, Canada, 992. Received: May, 24