caam336_practice_final_sol

# C we notice that the odes in the previous section

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(c) We notice that the ODEs in the previous section look a lot like the equations which define the eigenfunctions of the second derivative operator. The only difference is that we now solve an initial value problem instead of boundary value problem. We search for the solutions in the form a k ( t ) = A k cos λ k t + B k sin λ k t We use initial conditions to determine coefficients A k and B k b k = a k (0) = A k c k = da k dt (0) = - A k λ k sin 0 + B k λ k cos 0 = B k λ k Notice that the above formulas are valid for k = 1 , 2 , . . . , we consider the special case k = 0 separately d 2 a 0 dt 2 = - λ 0 a 0 = 0 so it must be that a 0 ( t ) = A 0 t + B 0 , where we determine A 0 , B 0 from a 0 (0) = B 0 = b 0 d 2 a 0 dt 2 (0) = A 0 = c 0 (d) Using the formulas from the previous two sections we derive an expression for the solution to one dimensional wave equation with the homogeneous Neumann boundary conditions u ( x, t ) = b 0 + c 0 t + k =1 ( b k cos λ k t + c k λ k sin λ k t ) cos πkx We observe that the solution has a term which grows or decreases (depending on the sign of c 0 ) linearly as t → ∞ as well as a term which keeps oscillating as the solution evolves in time. 5

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For the heat equation with homogeneous Neumann boundary conditions the solution has the form u ( x, t ) = b 0 + k =1 b k e - λ k t cos πkx, where b k has the same meaning as for the wave equation IBVP b k = ( ψ, φ k ) ( φ k , φ k ) = 2 1 0 ψ ( x ) cos ( πkx ) dx for k = 1 , 2 , . . . and b 0 = ( ψ, φ 0 ) ( φ 0 , φ 0 ) = 1 0 ψ ( x ) dx is an average of the initial profile over the interval [0 , 1]. In contrast to the solution to wave equation, the solution of the heat equation with the same bounudary conditions has a limit as t → ∞ , u ( x, t ) b 0 . The solutions flattens out to an average of the initial data. Problem 3. (a) Fredholm alternative tells us that for an equation Ax = b with a singular linear operator A we have exactly two possibilities If b Range( A ), then Ax = b has infinitely many solutions. If b / Range( A ), then Ax = b has no solutions. We see that the case when A is singular and Ax = b has a unique solution is impossible. Indeed, suppose that A is singular, hence Null A = . Let x 0 satisfy Ax 0 = b , then for y in the null space of A we have A ( x 0 + y ) = Ax 0 + Ay = b + 0, so x 0 + y also solves Ax = b . Thus, existence of solution of Ax = b implies that infinite number of solutions exist. In our case A is given by A = - 500 500 500 - 500 Ax = - 500 x 1 + 500 x 2 500 x 1 - 500 x 2 One can easily observe that Range( A ) = - α α , Null( A ) = β β , α R , β R Thus if b Range( A ) α R b = - α α . Then we can find solutions of Ax = b as x 1 x 2 satisfying x 1 - x 2 = α 500 . We can take x 0 = α 500 0 and generate a general solution by adding any vector from the null space of A. x = α 500 0 + β β (b) We compute matrix exponential e tA using diagonalization of A . The eigenvalues of A are λ 1 = 0, λ 2 = - 1000. Corresponding eigenvectors are q 1 = 1 2 1 1 , q 2 = 1 2 1 - 1 Since A is symmetric, we can represent A in factored form A = QDQ T , where Q = ( q 1 , q 2 ) is an orthogonal matrix and 6
D = 0 0 0 - 1000 We use the fact that for diagonalizable A we can represent matrix exponential as e tA = e tQDQ T = Qe tD Q T . This results in e tA = 1 2 1 + e - 1000 t 1 - e - 1000 t 1 - e - 1000 t 1 + e - 1000 t

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• Spring '10
• MOSKOW
• Ode, Boundary value problem, Boundary conditions, Bk, λk t

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