BOUNDARY VALUE OF PROLEMS DIFFERENTIAL EQUATIONS

WEBSITE : http://maths-minhthe.violet.vn BOUNDARY VALUE PROBLEMS OF DIFFERENTIAL EQUATIONS (5390) The Minh Tran 1. Solution a) Because '' u (x) 1 = , so we have the genaral form of u(x) : ( ) 2 x u x ax b 2 = + + ' ' From the condition boundary : u(0) 2u (0) 1 ; u(1) 0 We have 2a b 1 3 a u(0) 2u (0) 1 2 1 a b u(1) 0 b 2 2 + = = + =    = + =   ⇔ ⇔    + = − =    = −   Thus, ( ) 2 x 3 u x x 2 2 2 = + − b) Solution : We have the equation : ( ) ( ) ( ) 1 N 0 u x G x, f d u (x) = ξ ξ ξ + ∫ We see that 2 2 1 N N N N N 1 d u Lu f and has two extra boundary conditions,so dx u L f u where u has to be in the null space of L We compute L to both sides of the equation so Lu f since Lu 0, u is det er mined by the boundary co ndition and u 2(1 x) Hence, u L f u − − = = = + = = = − − = + N 1 0 1 1 2 2 0 0 write out as: G(x, )f( )d 2(1 x) , compute L to this expression : d L G(x, )f( )d LG(x, )f( )d f(x), where L dx ξ ξ ξ− − ξ ξ ξ = ξ ξ ξ = = ∫ ∫ ∫ From the condition boundary, the Green’s function G(x, ) ξ satisfies : 2 ' 2 d G(x, ) (x ) , G(1, ) 0 , G(0, ) 2G (0, ) 1 dx ξ = δ − ξ ξ = ξ + ξ = We have ( ) a bx , x G x, c dx , x + < ξ  ξ =  + > ξ  The constant a, b,c,d are different Where x < ξ : ' G(0, ) 2G (0, ) 1 a 2b 1 a 1 2b ξ + ξ = ⇒ + = ⇒ = − Where x > ξ : G(1, ) 0 c d 0 c d ξ = ⇒ + = ⇒ = − Hence, ( ) 1 2b bx ,x (1) G x, d dx ,x − + < ξ  ξ =  − + > ξ  We continue to check at x = ξ ( ) ( ) ( ) (2) G , G , Derivativing of G x, from x to x dG dG (3) 1 dx dx x x From (1) and (2) we have : 1 2b b d d From (2) and (3) we have : d b 1 − + − + + − ξ ξ = ξ ξ ξ = ξ = ξ − = = ξ = ξ − + ξ = − + ξ − = Soving the equation system 1 2b b d d b 2 d b 1 d 3 − + ξ = − + ξ = − ξ   ⇒   − = = − ξ   ( ) ( )( ) ( )( ) Thus 1 2 x 2 ,x G x, 3 1 x ,x  − − ξ − < ξ  ξ =  − ξ − > ξ   The representation : ( ) ( ) ( ) 1 0 u x G x, f d 2(1 x) = ξ ξ ξ − − ∫ c) We have the equation : ( ) ( ) ( ) = ξ ξ ξ ∫ 1 0 u x G x, f d with 0 1 x < < We have the condition boundary : , ( ) ' ' ' 3 3 u (x) x u (0) u(0) 2u (0) 1 2 2 u(0) 2 1 3 u 1 2 0 2 2 = + ⇒ = ⇒ + = = − = α ⇒ = + − = = β We have the genarally form of u(x) : ( ) ( ) ( ) ( ) = ξ ξ ξ + β + α − ∫ 1 0 u x G x, f d x 1 x We have the function ( ) , G x ξ as the formula below : ( ) ( ) ( ) ( )( ) 1 2 x 2 ,x G x, 3 1 x ,x  − − ξ − < ξ  ξ =  − ξ − > ξ   ( ) ( ) '' 1 , 1 = = with u x f ξ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 0 x 1 0 x x 1 0 x 2 2 2 2 2 u x G x, f d x 1 x G x, f d G x, f d 2 1 x 1 2 x 2 d 3 1 x d 2 1 x x 1 2 x 2 3 1 x 2 1 x 2 0 2 x x 1 x x 2x x 2 (1 x) (3 ) (3x ) 2 1 x 2 2 2 x 3 1 x 2 2 2 = ξ ξ ξ + β + α − = ξ ξ ξ + ξ ξ ξ − − = − − ξ − ξ + − ξ − ξ − −       ξ ξ = ξ − ξ − − + ξ − − − −                   = − − − + − − − − − −             = + + ∫ ∫ ∫ ∫ ∫ Thus, ( ) 2 x 3 1 u x x 2 2 2 = + + We see two solutions from the above are the same 2. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 u u a ) sin5x t x We assum that 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 firstpartial derivetive of u(x,t) w ith respect to t and the fourth partial derivative with respect to x dG t u x t dt u x ∂ ∂ = + ∂ ∂ = φ φ ∂ = φ ∂ ∂ = ∂ ( ) ( ) 2 2 2 2 d G(t) dx dG t d x G(t) (1) dt dx φ φ ⇒ φ = ( ) ( ) 2 2 We can separate var iableby dividing both sides of (1) by x G(t) dG t 1 1 d sin5x G dt dx G We can not solve by the method of separation of var iables because the presence of a terms like sin5x is forbidden φ φ = + φ φ " 2 n n n n 1 ' t n n n 1 2 xx n n n 1 b) We have 0 We have the boundary condition (0) ( ) 0 2 and have the solution is (x) sinnx , n u(x,t) B (t) u (x,t) B (t) u (x,t) n B (t) ∞ = ∞ = ∞ = φ + λφ = φ = φ π = φ = λ = π ⇒ = φ ⇒ = φ ⇒ = − φ ∑ ∑ ∑ n n n 1 5 n n n 1 n Set q(x,t) sin5x C (t) 2 q(x,t) can written as q(x,t) (x) comparing with q(x,t) sin5x C (t) 2 n 5 We see that C (t) 0 n 5 ∞ = ∞ = = = φ = φ π = = φ  =  = π   ≠  ∑ ∑ The initial condition is 1 2 u(x,0) 0 sin0 (x) = = = φ π The fourier expantion is n n 0 n 1 n 2 u(x,0) B (0) (x) 0 (x) 2 n 0 B (0) 0 n 0 ∞ = = φ = = φ π  =  = π   ≠  ∑ t xx ' 2 n n n n n n n 1 n 1 n 1 ' 2 n n n ' 2 5 n n n From the equation u u sin5x and the formula from t he above B (t) n B (t) C (t) The differential equation : B (t) n B (t) C (t) We can solve : 2 C , n 5 B (t) n B (t) 0 ,n 5 2 n 0 with B (0) ∞ ∞ ∞ = = = = + ⇒ φ = − φ + φ ⇒ + =  = =  + = π   ≠  = = π ∑ ∑ ∑ 0 n 0     ≠  ' 0 0 0 Case 1: n 0 B (0) 0 2 B (0) 2 B (0) =  =  ⇒ =  π =  π  ( ) ' 25t 25t 5 5 5 5 Case 2 : n 5 2 B (t) 25B (t) 1 2 B (t) e e 1 25 B (0) 0 − =  + =  ⇒ = − π  π  =  ' 2 n n n n Case 3: n 5 and n 0 B (t) n B (t) 0 B (0) 0 B (t) 0 ≠ ≠  + =   =   ⇒ = ( ) n n n 1 25t 25t Finally ,the solution is : u(x,t) B (t) 1 2 2 u(x,t) e e 1 sin5x 25 ∞ = − = φ ⇒ = − π π ∑ 3. Solve the following two types of problems : ( ) ( ) " ' ' a) u u subject to u 0 0 u 1 − = λ = = We have the equation form : ( ) ( ) ( ) " 2 2 2 sx sx 2s 0 We have the auxiliary equation : r 0 (*) Case 1 : 0 (*) r 0 r 0 r s x Ae Be d 0 0 A B dx A B 0 and so d A Be 1 0 dx There are no − − φ + λφ = + λ = λ < ⇔ + λ = ⇔ = −λ > ⇔ = ± −λ = ± ⇒ φ = + φ  =  =   ⇔ ⇔ = =   φ =   =   λ ( ) ( ) ( ) ( ) 2 1 2 Case 2 : 0 (*) r 0 r r 0 x A Bx d 0 0 dx A 0; B 0 and so dim 1 d 1 0 dx x A const is an eigenfunction λ = ⇔ = ⇔ = = ⇒ φ = + φ  =   ⇔ ≠ = =  φ  =   φ = = ( ) ( ) ( ) ( ) 2 2 2 2 Case 3 : 0 (*) r 0 r 0 r i x A cos x Bsin x d 0 0 dx sin 0 d 1 0 dx n n n when n 0 dim 2 λ > ⇔ + λ = ⇔ = −λ < ⇒ = ± λ ⇒ φ = λ + λ φ  =   ⇔ λ =  φ  =   ⇔ λ = π ∈ ⇒ λ = π > ⇒ = » ( ) ( ) " ' ' b) u u subject to u 0 0 ; u 1 1 − = λ = = We have the equation form : ( ) ( ) ( ) " 2 2 2 sx sx s 2s sx sx 0 We have the auxiliary equation : r 0 (*) Case 1 : 0 (*) r 0 r 0 r s x Ae Be d 0 0 e dx A B where s 0 and so s(e 1)d 1 1 dx When 0 the solution is Ae Be ,where A,B is the same as the a − − − φ + λφ = + λ = λ < ⇔ + λ = ⇔ = −λ > ⇔ = ± −λ = ± ⇒ φ = + φ  =   ⇔ = = = −λ >  −φ  =   λ < + bove ( ) ( ) ( ) 2 1 2 Case 2 : 0 (*) r 0 r r 0 x A Bx d 0 0 A 0; B 0 dx d 1 1 A 0; B 1 dx There is no 0 λ = ⇔ = ⇔ = = ⇒ φ = + φ  = ⇒ ≠ =    φ  = ⇒ ≠ =   λ = ( ) ( ) ( ) ( ) 2 2 Case 3 : 0 (*) r 0 r 0 r i x A cos x B sin x d 0 0 B 0 B 0 dx d 1 1 A sin 1 dx 1 A sin 1 x A cos x cos x sin 1 When 0 the solution is cos x sin λ > ⇔ + λ = ⇔ = −λ < ⇒ = ± λ ⇒ φ = λ + λ φ  = ⇔ λ = ⇒ =    φ  = ⇔ − λ λ =   − = λ λ − ⇒ φ = λ = λ λ λ − λ > λ λ λ Condition : The problem a) is an eigenvalue prolem ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) i kx t tt xx 2 i kx t tt 2 i kx t xx tt xx 2 2 i kx t i kx t 2 2 1 tt xx 2 1 2 4. We have u(x,t) e and u cu u i e u ki e u cu i e c ki e ck k c Re turn the problem which is given c for x 0 We see that u cu where c(x) c for x 0 k c for x 0 k c for x −ω −ω −ω −ω −ω = = ⇒ = ω ⇒ = ⇒ = ⇔ ω = ⇔ ω = ⇒ ω = <  = =  >  < ω = > ( ) i kx t ikx ikx t 0 u(x,t) e u(x,0) e f(x) u (x,0) ( i)e g(x) −ω      = ⇒ = = ⇒ = −ω = ( ) ( ) 1 1 x ikx ikx ikx ikx 1 1 0 x ikx ikx ikx ikx 1 1 0 For x 0 u(x,t) F(x c t) G(x c t) 1 1 1 1 F(x) e ( i)e dx e 1 e 2 2c 2 2c k 1 1 1 1 G(x) e ( i)e dx e 1 e 2 2c 2 2c k < = − + + ω = − −ω = − − ω = + −ω = + − ∫ ∫ ( ) 1 1 1 1 1 1 tki c it 1 1 1 1 1 1 1 1 x ik c ( t ) ik( x c t ) c ik( x If x c t , u(x,t) f(x c t) x c t ,we have u(0,t) h(t) e e h(t) u(0,t) F( c t) G(c t) z z and h( ) F( z) G(z) G(z) h( ) F( z) ; z 0 c c x Thus, u(x,t) f(x c t) h( t) f (x c t) c e e e − −ω − + − − + < = − > = = = = = − + = − + ⇒ = − − > = − + + − − + = + − 1 c t) ( ) ( ) 2 2 x ikx ikx ikx ikx 1 1 0 x ikx ikx ikx ikx 1 1 0 2 2 2 2 2 For x 0 u(x,t) Q(x c t) P(x c t) 1 1 1 1 Q(x) e ( i)e dx e 1 e 2 2c 2 2c k 1 1 1 1 G(x) e ( i)e dx e 1 e 2 2c 2 2c k x c t : h(t) Q( c t) P(c t) where x 0 z Q(z) h( ) P( z) , z 0 c x u(x,t) h(t c > = − + + ω = − −ω = − − ω = + −ω = + − < = − + = − ⇒ = − − < = − ∫ ∫ 2 x i t c ) e   −ω −     = ( ) ( ) x x x ' ' ' ' 1 1 1 1 2 2 ' ' ' ' 1 1 1 2 2 1 1 2 We continue to use the continuity of u at x 0 u (0 ,t) u (0 ,t) 1 x 1 x f (x c t) h ( t) f (x c t) h (t ) c c c c Setting x 0 we have 1 1 f ( c t) h (t) f c t h (t) c c 2c f( c t) h(t) c c + − = =   − + + + − + = − −     =   − + + − = −     − ⇒ = + [...]... − x = 1 dt 2t dt 2t Multiplying e 1 − ln t 2 in two sides of the equation 1 − ln t 2 1 − ln t  dx s2 1  − x = e 2  dt 2t   1 1 1 − ln t dx − ln t 1 − ln t s2 e 2 −e 2 x=e 2 dt 2t 1 1 − ln t dx s2  − 2 ln t  e x s2  = e 2  dt   e − 1 − 1 t 2 x s2 = ∫ t 2 dt − 1 2 1 2 t x s2 = 2t + c ⇒ x s2 = 2t + c t We have int er sec tion of the equations : 4t = 3t + 1 ⇒ t = 1 with t = 1 then 3t + 1 = 2t... see that c = t Usin g the method of characteristics dx t2 t2 = −t ⇒ x = − + x0 ⇒ x 0 = x + dt 2 2 Generally, we have two cases dω If ω = 0 , =0 dt t2 ω(x,t) = p(x 0 ) = p(x + ) with ω(x,0) = p(x) = f(x) 2 2 t Thus, ω(x,t) = f(x + ) 2 dω ∂ω ∂ω = −t =ω If ω ≠ 0 then ∂t ∂x dt t2 When x 0 = x + > 0 2 2 t ⇒x>− 2 dω = ω ⇒ ω(x,t) = ω0 (x 0 ,t)et where ω0 (x 0 ,t) is the value of ω(x,t) at t = 0 dt t2 From the... 0 ) = f(x + ) 2 2 t Thus, ω(x,t) = f(x + )e t 2 When x 0 = x + ⇒x . http://maths-minhthe.violet.vn BOUNDARY VALUE PROBLEMS OF DIFFERENTIAL EQUATIONS (5390) The Minh Tran 1. Solution a) Because '' u (x) 1 = , so we have the genaral form of u(x) :. has two extra boundary conditions,so dx u L f u where u has to be in the null space of L We compute L to both sides of the equation so Lu f since Lu 0, u is det er mined by the boundary co ndition. separate var iableby dividing both sides of (1) by x G(t) dG t 1 1 d sin5x G dt dx G We can not solve by the method of separation of var iables because the presence of a terms like sin5x is forbidden φ φ =