ps7sol

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

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

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

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

View Full Document
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
∂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

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

View Full Document
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
Ask a homework question - tutors are online