*This preview shows
pages
1–3. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **HOMEWORK 2 SOLUTIONS Section 1 . 5 : #1 , 2 , 3 , 4( a )- ( b ) • #1) Solve the boundary value problem: u xx + u = 0 u (0) = 0 and u ( L ) = 0 . Answer: The solution to the ODE is: u ( x ) = A sin( Bx + C ) If A = 0, then the solution is u ( x ) ≡ 0. If A 6 = 0, then we can solve for the variables by substituting the general solution into the ODE. u xx + u = ( A- AB 2 ) sin( Bx + C ) = 0 A (1- B 2 ) = 0 B = ± 1 Using the boundary conditions we have: u (0) = A sin( C ) = 0 Then C = nπ with n an integer. Also, u ( L ) = A sin( BL + C ) = 0 BL + C = mπ ± L + nπ = mπ L = mπ So, if L = mπ for some integer m , then u ( x ) = A sin( x + nπ ) is a solution. If L 6 = mπ , then u ( x ) ≡ 0 is the only solution. • #2) Consider the problem u 00 ( x ) + u ( x ) = f ( x ) u (0) = u (0) = 1 2 [ u ( l ) + u ( l )] (a) Does the following problem have a unique solution? (b) Does a solution necessarily exist? Answer: (a) The problem does not have a unique solution. Proof. Let u 1 and u 2 be solutions to the problem. Let v ( x ) = u 1 ( x )- u 2 ( x ). Then v ( x ) is a solution of u 00 ( x ) + u ( x ) = 0 1 2 u (0) = u (0) = 1 2 [ u ( l ) + u ( l )] Then v ( x ) = A + Be- x , and v (0) = v (0) = 1 2 [ v ( l ) + v ( l )]- B = A + B = 1 2 [- Be- l + A + Be- l ]- B = A + B = A 2 A =- 2 B v ( x ) = B ( e- x- 2) , for any real number B Then if u ( x ) is a solution of the initial problem, u 2 ( x ) = u 1 ( x ) + e- x- 2 is also a solution....

View
Full
Document