ps7sol - ACM 95/100c Problem Set 7 Solutions Lei Zhang Each...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ACM 95/100c Problem Set 7 Solutions Lei Zhang May 22, 2006 Each Problem is worth 10 points. Each part of a multi-part problem is weighted equally. Problem 1 (No collaboration) Solve 2 u ∂x 2 + 2 u ∂y 2 = 0 for 0 < x < L, y > 0 subject to the following boundary conditions. If there is a solvability condition, state it and explain it physically. (a) ∂u ∂x (0 ,y ) = g 1 ( y ) , ∂u ∂x ( L,y ) = g 2 ( y ) , u ( x, 0) = f ( x ) (b) ∂u ∂x (0 ,y ) = 0 , ∂u ∂x ( L,y ) = 0 , ∂u ∂y ( x, 0) = f ( x ) Solution 1 (a) Let u v + w . Here, v solves the system 2 v ∂x 2 + 2 v ∂y 2 = 0 (1) with ∂v ∂x (0 ,y ) = g 1 ( y ), ∂v ∂x ( L,y ) = g 2 ( y ) and v ( x, 0) = 0. w solves the system 2 w ∂x 2 + 2 w ∂y 2 = 0 (2) with ∂w ∂x (0 ,y ) = 0, ∂w ∂x ( L,y ) = 0 and w ( x, 0) = f ( x ). For w , the boundary conditions at x = 0 and x = L imply that we should use a Fourier cosine series expansion, i.e., w ( x,y ) = X n =0 A n ( y )cos nπx L · . (3) Whether we plug in this series directly into the PDE or use the method of finite transforms, by orthogonality we get - L · 2 + d 2 A n dy 2 = 0 . (4) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Therefore, A n ( y ) = ( a 0 + b 0 y, n = 0 a n exp ± - nπy L ) + b n exp ± nπy L ) , n = 1 , 2 , 3 ,... . (5) Since w is regular as y → ∞ , b n = 0, so we may write w ( x,y ) = a 0 + X n =1 a n exp - nπy L · cos nπx L · . (6) Finally, applying w ( x, 0) = f ( x ) yields w ( x, 0) = f ( x ) = a 0 + X n =1 a n cos nπx L · , (7) so by orthogonality a 0 = R L 0 f ( x ) dx R L 0 1 2 dx = 1 L Z L 0 f ( x ) dx (8) and a n = R L 0 f ( x )cos ± nπx L ) dx R L 0 cos 2 ± nπx L ) dx = 2 L Z L 0 f ( x )cos nπx L · dx, n = 1 , 2 , 3 ,... . (9) For v , we assume that v ( x,y ) 0, g 1 ( y ) 0 and g 2 ( y ) 0 as y → ∞ . Since we have the condition v ( x, 0) = 0, we Fourier sine transform the PDE for v . Using the convention F s [ v ( x,y )] r 2 π Z 0 v ( x,y )sin( ωy ) dy V ( x,ω ) , (10) we have 2 v ∂x 2 + 2 v ∂y 2 = 0 -→ 2 V ∂x 2 + r 2 π v ( x, 0) - ω 2 V = 2 V ∂x 2 - ω 2 V = 0 . (11) Thus, V ( x,ω ) = c exp( ωx ) + d exp( - ωx ) = c 0 cosh( ωx ) + d 0 cosh[ ω ( L - x )] . (12) Now, ∂v ∂x (0 ,y ) = g 1 ( y ) -→ ∂V ∂x (0 ) = G 1 ( ω ) (13) and 2
Background image of page 2
∂x ( L,y ) = g 2 ( y ) -→ ∂V ∂x ( L,ω ) = G 2 ( ω ) , (14) where G 1 ( ω ) r 2 π Z 0 g 1 ( y )sin( ωy ) dy (15) and G 2 ( ω ) r 2 π Z 0 g 2 ( y )sin( ωy ) dy . (16) So, ∂V ∂x (0 ) = G 1 ( ω ) = c 0 ω sinh( ω 0) - d 0 ω sinh[ ω ( L - 0)] = - d 0 ω sinh( ωL ) , (17) implying that d 0 = - G 1 ( ω ) ω sinh( ωL ) . (18) Also, ∂V ∂x ( L,ω ) = G 2 ( ω ) = c 0 ω sinh( ωL ) - d 0 ω sinh[ ω ( L - L )] = c 0 ω sinh( ωL ) , (19) implying that c 0 = G 2 ( ω ) ω sinh( ωL ) . (20) Thus, V ( x,ω ) = G 2 ( ω ) ω sinh( ωL ) cosh( ωx ) - G 1 ( ω ) ω sinh( ωL ) cosh[ ω ( L - x )] , (21) so v ( x,y ) = r 2 π Z 0 V ( x,ω )sin( ωy ) = r 2 π Z 0 G
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '09
  • NilesA.Pierce
  • Sin, Boundary value problem, Boundary conditions, Sturm–Liouville theory, Sine and cosine transforms, nπx nπy

{[ snackBarMessage ]}

Page1 / 12

ps7sol - ACM 95/100c Problem Set 7 Solutions Lei Zhang Each...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online