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HW_2_solutions

# HW_2_solutions - HOMEWORK 2 SOLUTIONS Section 1.5#1 2 3...

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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 = 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 = with n an integer. Also, u ( L ) = A sin( BL + C ) = 0 BL + C = ± L + = L = So, if L = for some integer m , then u ( x ) = A sin( x + ) is a solution. If L = , then u ( x ) 0 is the only solution. #2) Consider the problem u ( 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 ( x ) + u ( x ) = 0 1

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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. (b)A solution does not necessarily exist unless f ( x ) satisfies the condition l 0 f ( x ) dx = 0 Proof. We integrate both sides of the ODE and use the conditions on u at x = 0 and at x = l .
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