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Tutorial 6 - Heat equations

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Heat equation problems
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  IW242 TUT 6 24 OKTOBER 2014Handboek : Zill & Wright ( ZW ) ( 5de  Uitgawe)/ Textbook: Zill & Wright (ZW) (5th Edition) 13.3  Los op die volgende hittevergelykings met die substitusie  u ( x,t ) =  X  ( x ) T  ( t ):/ Solve the following heat equations with the substitution   u ( x,t ) =  X  ( x ) T  ( t ): 1.  3  ∂  2 u∂x 2  =  ∂u∂t ,  0  < x <  1 , t >  0  2.  ∂  2 u∂x 2  =  ∂u∂t ,  0  < x < π, t >  0 u (0 ,t ) = 0 =  u (1 ,t ) , t >  0  u (0 ,t ) = 0 =  u ( π,t ) , t >  0 u ( x, 0) =  x,  0  < x <  1.  u ( x, 0) = 1 ,  0  < x < π . 3.  ∂  2 u∂x 2  =  ∂u∂t ,  0  < x < π, t >  0  4.  2  ∂  2 u∂x 2  =  ∂u∂t ,  0  < x < π 2 , t >  0 u (0 ,t ) = 0 =  u ( π,t ) , t >  0  u (0 ,t ) = 0 =  u ( π 2 ,t ) , t >  0 u ( x, 0) = cos x,  0  < x < π .  u ( x, 0) =  x,  0  < x <  π 2 . 5(a)  ∂  2 u∂x 2  =  ∂u∂t ,  0  < x < π, t >  0 u (0 ,t ) = 0 =  u ( π,t ) , t >  0 u ( x, 0) =  x 2 ,  0  < x < π . (b)  Bepaal die waarde van die volgende  konvergente  p -reeks (  p  = 3)/ Find the value of the following   convergent  p -series (   p  = 3 ) : ∞  n =1 ( − 1) n +1 (2 n − 1) 3 ?[ Wenk:  Gebruik die oplossings van die hittevergelykings in vrae  5(a)  en  4  hierbo./ Hint: Use the solutions of the heat equations in questions   5(a)  and   4  above  .] Oplossings / Solutions  :  1.  u ( x,t ) = 2 π ∞  n =1 ( − 1) n +1 n e − 3 n 2 π 2 t sin nπx . 2.  u ( x,t ) = 2 π ∞  n =1 [1 + ( − 1) n +1 ] n e − n 2 t sin nx . 3.  u ( x,t ) = 2 π ∞  n =2 n [( − 1) n + 1] n 2 − 1  e − n 2 t sin nx .  4.  u ( x,t ) = ∞  n =1 ( − 1) n +1 n e − 8 n 2 t sin2 nx . 5(a)  u ( x,t ) = 2 π ∞  n =1  π 2 n  ( − 1) n +1 + 2(( − 1) n − 1) n 3  e − n 2 t sin nx  (b)  π 3 32. LW / Note  :  Tutoriaaltoets oor 13.3 begin om 12:20 / Tutorial test on   13.3  starts at 12:20  .1  15.4 Fourier transforme / Fourier transforms  : 1.  (a) Toon aan dat/ Show that  :    ∞ 0 e − x cos αxdx  = 11 +  α 2 .(b) Definieer ’n geskikte Fourier transform om die volgende hittevergelyking op te los:/ Define an appropriate Fourier transform to solve the following heat equation: k ∂  2 u∂x 2  =  ∂u∂t ,  −∞ < x < ∞ , t >  0 u ( x, 0) =  e −| x | ,  −∞ < x < ∞ . (c) Bepaal die oplossing van die hittevergelyking in (b) deur (a) se resultaat te gebruik./ Determine the solution of the heat equation in (b) by using the result in (a). 2.  Los op die volgende hittevergelyking deur ’n geskikte Fourier transform te gebruik:/ Solve the following heat equation by using an appropriate Fourier transform  : k ∂  2 u∂x 2  =  ∂u∂t ,  −∞ < x < ∞ , t >  0 u ( x, 0) =  0 , x < − 1 − 100 ,  − 1  < x <  0100 ,  0  < x <  10 , x >  1 . 3.  Los op die volgende hittevergelyking deur ’n geskikte Fourier transform te gebruik:/ Solve the following heat equation by using an appropriate Fourier transform  : ∂  2 u∂x 2  +  ∂  2 u∂y 2  = 0 ,  0  < x < π, y >  0 u (0 ,y ) = 0 , u ( π,y ) =  e − y , y >  0 u ( x, 0) = 0 ,  0  < x < π. Oplossings / Solutions  : 1.  u ( x,t ) = 2 π    ∞ 0 e − kα 2 t cos αx 1 +  α 2  dα  2.  u ( x,t ) = 200 π    ∞ 0 1 − cos αα e − kα 2 t sin αxdα 3.  u ( x,y ) = 2 π    ∞ 0 α sinh αx (1 +  α 2 )sinh απ  sin αydα . Voorspoed met die voorbereidings vir die finale eksamens ./ Good luck with the preparations for the final examinations. 2
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