This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Partial Differential Equations – Math 442 C13/C14 Fall 2009 Homework 6 Solutions 1. Let D = { ( x,y ) : x 2 + y 2 < 4 } , and solve Δ u = 0 , in D, u = 3 2 cos θ, r = 2 . Solution: Using the formula generated in class, we have that u ( r,θ ) = A 2 + ∞ summationdisplay n =1 r n ( A n cos( nθ ) + B n sin( nθ )) , for some unknown A n ,B n . Plugging in the boundary condition r = 2 gives u ( r, 2) = A 2 + ∞ summationdisplay n =1 2 n ( A n cos( nθ ) + B n sin( nθ )) = 3 2 cos θ, from which we can see that A = 6 ,A 1 = 1, and all the rest of the coefficients are zero, and thus the solution is u ( r,θ ) = 3 + 2 r cos( θ ) . 2. Let D be the square [0 , 1] 2 . Solve Δ u = 0 subject to the boundary conditions u (0 ,y ) = 0 , u x (1 ,y ) = 0 , u ( x, 0) = x 2 2 x, u y ( x, 1) = 0 . Solution: First make the Ansatz u ( x,y ) = A ( x ) B ( y ) , separating as usual gives A ′′ ( x ) + λA ( x ) = 0 , A (0) = A ′ (1) = 0 , B ′′ ( y ) λB ( y ) = 0 , B ′ (1) = 0 . Solving the first system gives A n ( x ) = sin( ω n x ) , ω n = ( n + 1 / 2) π,n = 1 , 2 , 3 ,...,λ n = ω 2 n . From this we obt ? 33 T E < 8 B > : D 2 ? 321331 WR bR K K q < 91 >< 9 C >< 9 D > : G < 95 > m < 9 F > < 9 D > N * ; < 9 D > < o < 3. In class, we derived difference approximations for the second derivative in3....
View
Full
Document
This note was uploaded on 01/15/2010 for the course MATH 346 taught by Professor S during the Spring '09 term at University of Victoria.
 Spring '09
 s
 Equations

Click to edit the document details