kontsyslos060531

kontsyslos060531 - DEAÖGA AEA ÖGA EGAYE5Ô 65 Ø u ( x, t...

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Unformatted text preview: DEAÖGA AEA ÖGA EGAYE5Ô 65 Ø u ( x, t ) Úa Öa ØeÑ Ô eÖa ØÙ ÖeÒ iÔÙÒk ØeÒ x Ú id Ø id eÒ t ≥ edhÒ ×ÝÒ Ø iÐÐa ØØd eÒÚÒ ×ØÖa Òd eÒ fö Öe t = 0 ÐÒg eÚa Ö iØg Öad igÓhd eÒhög ÖaÒd eÒ ÐÒg eha fØØeÑ Ô eÖa ØÙ ÖeÒ ◦ Óh Ñ edhÒ ×ÝÒ Ø iÐÐa ØØ×ØÒg eÒ ×Òabb ØÚÒ Ø×d t = 0 b ÐiÖ u ( x, 0) = 100(1 − x ) d < x < 1 ifÖ fö ÐjaÒd eÑ Ód e ÐÐ ∂u ∂t − ∂ 2 u ∂x 2 = 0 , < x < 1 , t > u (0 , t ) = 0 , u (1 , t ) = 100 , t > u ( x, 0) = 100(1 − x ) , < x < 1 . ihÓÑ Óg eÒ i×eÖa Öfö Ö×ØÖaÒdÚ iÐÐkÓ ÖeÒØØd Ö fö Ö v ( x, t ) = u ( x, t ) − 100 x D g ÐÐeÖa ØØ ∂v ∂t − ∂ 2 v ∂x 2 = 0 , < x < 1 , t > v (0 , t ) = 0 , v (1 , t ) = 0 , t > v ( x, 0) = 100(1 − 2 x ) , < x < 1 . igö ÖeÒ × iÒÙ ××eÖ ieaÒ ×a Ø×dÚ × v ( x, t ) = ∞ summationdisplay k =1 v k ( t ) sin kπx ØÝhÓÑ Óg eÒaD iÖ ih ÐeØÚ iÐÐkÓ ÖÒ ×ØØ Ò iÒg iDE Ðed eÖØ iÐÐ ∞ summationdisplay k =1 ( v ′ k ( t ) + k 2 π 2 v k ( t )) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = 0 sin kπx = 0 . D eØg eÖa ØØ v k ( t ) = c k e − k 2 π 2 t d ÖkÓÒ ×ØaÒ ØeÖÒa c k b e×ØÑ ×aÚb egÝÒÒ e Ð×eÚ iÐÐkÓ ÖeØ 100(1 − 2 x ) = v ( x, 0) = ∞ summationdisplay k =1 c k sin kπx, < x < 1 . i×eÖa ØØ c k ÖF ÓÙ Ö ieÖkÓeieÒ ØeÖE ÒÔa ÖØ ia ÐiÒ Øeg Öa Ø iÓÒ Ðed eÖØ iÐÐa ØØ c k = 2 1 integraldisplay 1 100(1 − 2 x ) sin kπx dx = 200(1 + ( − 1) k ) kπ . Úa Ö : u ( x, t ) = 100 x + 200 π ∞ summationdisplay k =1 1 + ( − 1) k k e − k 2 π 2 t sin kπx, < x < π, t > . d eÒÒaÙÔÔg ifØb ehöÚ eÖÚ ieÖagÒg eÖ iÒ Øeg Öa ÐeÒ integraldisplay ∞ e − ax dx = 1 a d Ö a > a ÖØÓgÓÒa ÐiØeØeÒb e×ØÑ ×aÚ ( ϕ 1 | ϕ 2 ) = integraldisplay ∞ e − x (2 e − 2 x − 1) e − x dx = integraldisplay ∞ 2 e − 4 x dx − integraldisplay ∞ e − 2 x dx = 2 4 − 1 2 = 0 . D ÖÑ ed ÖÓ ÖØÓgÓÒa ÐiØeØeÒÚ i×ad b Ò Øeg Öa ÐeÒkaÒ ×k Ö iÚa ××ÓÑ integraldisplay ∞ (1 − a 1 ϕ 1 ( x ) − a 2 ϕ 2 ( x )) 2 e − x dx = bardbl 1 − ( a 1 ϕ 1 + a 2 ϕ 2 ) bardbl 2 d ÖÒÓ ÖÑ eÒg e×aÚ ×ka ÐÖÔ ÖÓdÙk ØeÒ ia E fØeÖ×ÓÑ ϕ 1 Óh ϕ 2 ÖÓ ÖØÓgÓÒa ÐaeÖh ÐÐeÖÚ iiÒ Øeg Öa ÐeÒ × Ñ iÒ ×ØaÚ Öd eØdkÓÒ ×ØaÒ ØeÖÒa a 1 Óh a 2 Ú Ðj×eÒ Ðig ØÔ ÖÓ jek Ø iÓÒ ××a Ø×eÒ ×a Ø×4 dÚ × a 1 = ( ϕ 1 | 1) ρ 1 = 3 2 a 2 = ( ϕ 2 | 1) ρ 2 = − 5 7 d Ö ρ 1 = ( ϕ 1 | ϕ 1 ) = integraldisplay ∞ e −...
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This note was uploaded on 10/07/2011 for the course FMA 021 taught by Professor Pellepettersson during the Spring '11 term at Lund.

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kontsyslos060531 - DEAÖGA AEA ÖGA EGAYE5Ô 65 Ø u ( x, t...

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