THE METHOD OF SEPARATION OF VARIABLES AND THE EIGENVALUE PROBLEM FALL 2011 “ The universe has always been presented by mathematical problems “ The Minh Tran 1. Solution : Solve the eigenvalue problem : ( ) ( ) ( ) ( ) 2 2 0 2 0 2 d d d subject to and dx dx dx φ φ φ λφ φ φ π π = − = = We have equation form : ( ) ( ) ( ) ( ) ( ) " 2 2 2 2 2 2 2 0 : 0 (*) 1 : 0 (*) 0 0 0 2 0 2 sx sx s s s s We have auxiliary equations r Case r r r s x Ae Be A B Ae Be d d A B Ae Be dx dx There are not exist π π π π φ λφ λ λ λ λ λ φ φ φ π φ φ π λ − − − + = + = < ⇔ + = ⇔ = − > ⇔ = ± − = ± ⇒ = +  =  + = +   ⇔   = − = −     ( ) 2 1 2 2 : 0 (*) 0 0 Case r r r x A Bx λ φ = ⇔ = ⇔ = = ⇒ = + ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 0 A A B d d B B dx dx B x A A φ φ π π φ φ π φ  = = +   ⇔   = =    =  ⇔ ⇒ =  ∀  We have not had eigenvalue ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 : 0 (*) 0 0 cos sin 0 2 cos 2 sin 2 0 2 sin 2 cos 2 2 1 0 2 1 ( 0) 2 2 Case r r r i x A x B x A A B d d B A B dx dx A B Cos Cos Because A B n n n λ λ λ λ φ λ λ φ φ π π λ π λ φ φ π π λ π λ π λ π λ π λ π λ > ⇔ + = ⇔ = − < ⇒ = ± ⇒ = +  =  = +   ⇔   = = − −     ⇔ + − = ⇔ − + ≠ ⇔ = ∈ ⇒ = » 2. Explicitly show there are no negative eigenvalues for ( ) ( ) 2 2 0 0 , 0 d d d subject to L dx dx dx φ φ φ λφ = − = = Solution : We have equation form : ( ) ( ) ( ) " 2 2 2 0 : 0 (*) : 0 0 0 0 0 0 2 0 , sx sx We have auxiliary equations r When r r r s x Ae Be d dx d dx Thus there are no negative eigenvalues φ λφ λ λ λ λ λ φ φ λ φ π − + = + = < ⇔ + = ⇔ = − > ⇔ = ± − = ± ⇒ = +  =   ⇔ =   =   3. Solve the heat equation : ( ) ( ) 2 2 , 0 , 0 0, 0 0 , , 0 0 u u x L t subject to t x u u t t L t t x x ∂ ∂ = < < > ∂ ∂ ∂ ∂ = > = > ∂ ∂ Solution : ( ) 3 ) ,0 6 4cos x a u x L π = + We have infinite series solution ( ) 2 2 2 1 , cos kn t L n n n x u x t A e L π π ∞ − =   =     ∑ So ( ) 1 0 0 0 0 3 0 3 6 4cos ,0 cos 0 2 3 6 4cos cos 1 3 6 4cos 6 2 3 3 6 4cos cos 4 0 n n L n L L x n x u x A with t L L x n x A dx L L L x A dx L L A x x A dx L L L Others π π π π π π π ∞ =   + = = =         ⇒ = +           ⇒ = +     ⇒ =     ⇒ = + =         = ∑ ∫ ∫ ∫ ( ) 8 ) ,0 3cos x b u x L π = − We have infinite series solution ( ) 2 2 2 1 , cos kn t L n n n x u x t A e L π π ∞ − =   =     ∑ So ( ) 1 0 0 0 0 8 0 8 0 8 3cos ,0 cos 6 8 cos .cos 1 8 cos 0 6 8 8 cos .cos 3 16 (1 cos .) 3 0 n n L n L L L x n x u x A L L x n x A dx L L L x A dx L L A x x A dx L L L x A dx L L Others π π π π π π π π ∞ =   − = =       ⇒ = −     ⇒ = ⇒ =   ⇒ = −     = = − + = − = ∑ ∫ ∫ ∫ ∫ ( ) ) ,0 2sin x c u x L π = − We have infinite series solution ( ) 2 2 2 1 , cos kn t L n n n x u x t A e L π π ∞ − =   =     ∑ So ( ) 1 0 0 0 2sin ,0 cos 4 sin .cos 1 4 2sin . n n L n L x n x u x A L L x n x A dx L L L x A L L π π π π π π ∞ =   − = =       ⇒ = −       ⇒ = − = −     ∑ ∫ ∫ 4. For the following PDEs, what ODEs are implied by the method of separation of variables ? Solution : We can use the method of separation of variables to find the following ODEs ( ) ( ) ( ) ( ) ( ) ( ) 4 4 4 4 4 ) ( , ) ( , ) u u a t x From the equation u x t x G t where x is only a function of x and G t is a function of t We will get first partial derivetive of u x t with respect to t and the fourth partial derivative with respect to x dG t u x t dt u d x φ φ φ ∂ ∂ = ∂ ∂ = ∂ = ∂ ∂ = ∂ ( ) ( ) 4 4 4 ( ) ( ) (1) G t dx dG t d x G t dt dx φ φ φ ⇒ = ( ) ( ) 4 4 var (1) ( ) 1 1 We can separate iableby dividing both sides of b y x G t dG t d G dt dx φ φ φ = We can give the both sides equal the same constant ( ) ( ) 4 4 4 4 1 1 , , ( ) ( ) dG t d where is an arbitrary constant G dt dx Thus we have ODEs one for G t and one for x dG t G dt d dx φ λ λ φ φ λ φ λφ = = − = − = − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 ) ( , ) sec ( , ) sec u u b c t x From the equation u x t x G t where x is only a function of x and G t is a function of t We will get ond partial derivetive of u x t with re spect to t and the ond partial derivative with respect to x d G t u x t d t φ φ φ ∂ ∂ = ∂ ∂ = ∂ = ∂ ( ) ( ) 2 2 2 2 2 2 2 2 2 ( ) ( ) (1) u d G t x dx d G t d x c G t d t dx φ φ φ ∂ = ∂ ⇒ = ( ) ( ) 2 2 2 2 2 var (1) ( ) 1 1 We can separate iableby dividing both sides of b y x G t d G t d c G d t dx φ φ φ = We can give the both sides equal the same constant ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 1 , , ( ) ( ) d G t d c where is an arbitrary constant G d t dx Thus we have ODEs one for G t and one for x d G t G d t d c dx φ λ λ φ φ λ φ λφ = = − = − = − ( ) ( ) ( ) ( ) ) var , ( , ) ( , ) u k u c r t r r r Inthe method of separation of iables we will fin d solutions in the product form u r t r G t where r is only a function of r and G t is a function of t We will get the first partial derivetive of u r t with respec φ φ ∂ ∂ ∂   =   ∂ ∂ ∂   = ( ) ( ) ( ) ( ) ( ) ( ) sec (1) t to t and the ond partial derivative with respect to r dG t u r t dt u d d r G t r r r dr dr dG t kG t d d r r dt r dr dr φ φ φ φ ∂ = ∂ ∂ ∂     =     ∂ ∂       ⇒ =     ( ) ( ) var (1) ( ) 1 1 We can separate iableby dividing both sides of b y k r G t dG t d d r kG dt r dr dr φ φ φ   =     We can give the both sides equal the same constant ( ) 1 1 dG t d d r where is an arbitrary constant kG dt r dr dr φ λ λ φ   = = −     ( ) , , ( ) ( ) Thus we have ODEs one for G t and one for r dG t kG dt d d r r dr dr φ λ φ λφ = −   = −     .  ∑ ∫ ∫ 4. For the following PDEs, what ODEs are implied by the method of separation of variables ? Solution : We can use the method of separation of variables to find the following ODEs. THE METHOD OF SEPARATION OF VARIABLES AND THE EIGENVALUE PROBLEM FALL 2011 “ The universe has always been presented by mathematical problems “ The Minh Tran . u c r t r r r Inthe method of separation of iables we will fin d solutions in the product form u r t r G t where r is only a function of r and G t is a function of t We will get the first partial