ExercisesSolutions

Substituting this into the given equation gives u y

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Substituting this into the given equation, gives u = y . So the system of two first order ODEs is given by y = u u = y. (b) The corresponding initial conditions for the system of ODEs in part (a) are: y (0) = 1 and u (0) = 2. (c) We find the solution of this system and consider the behaviour as t increases: bracketleftbigg y u bracketrightbigg = bracketleftbigg 0 1 1 0 bracketrightbigg bracketleftbigg y u bracketrightbigg = α bracketleftbigg 1 1 bracketrightbigg e t + β bracketleftbigg 1 - 1 bracketrightbigg e t So we find that y = α e t + β e t u = α e t - β e t . Using the initial conditions, α = 3 2 and β = 1 2 , so the solution of the system is y = 3 2 e t - 1 2 e t . From this, it can be seen that the solutions of the system are unstable because as t → ∞ , y → ∞ . 21
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(d) Euler’s method state that, given y ( t 0 ) = y 0 , t 0 and h , then y ( t 1 ) = y ( t 0 ) + hf ( t 0 , y 0 ). In this case, we have that f : Y = bracketleftbigg y u bracketrightbigg , y ( t 0 ) : Y(0) = bracketleftbigg 2 1 bracketrightbigg = Y 0 , t 0 : t 0 = 0, h = 0 . 5. Now since Y(0) = bracketleftbigg 2 1 bracketrightbigg , then Y 1 = Y 0 + h Y 0 = bracketleftbigg 2 1 bracketrightbigg + 0 . 5 bracketleftbigg 2 1 bracketrightbigg = bracketleftbigg 3 2 3 bracketrightbigg , so y (0 . 5) = 3 and u (0 . 5) = 3 2 . (e) With the step size h = 0 . 5, Euler’s method takes the form Y k +1 = Y k + 0 . 5Y k = 3 2 Y k . In this case, the amplication factor is 3 2 , which is greater than one so Euler’s method will be unstable with h = 0 . 5. (f) For backward Euler, Y k +1 = Y k + 0 . 5Y k +1 = - 2Y k = ( - 2) k +1 Y 0 . Since | ( - 2) | > 1, then backward Euler will be unstable for this problem with h = 0 . 5. 5. Purpose: Identifying properties of methods. From [Hea02, pp.417–418,#9.9] . Exercise : For each property listed below, state which of the following two ODE methods has or have the given property: y k +1 = y k + h 2 (3 f ( t k , y k ) - f ( t k 1 , y k 1 )) (1) y k +1 = y k + h 2 ( f ( t k , y k ) - f ( t k +1 , y k +1 )) (2) Properties: (a) Second-order accurate (b) Single-step method (c) Implicit method (d) Unconditionally stable (e) Good for solving stiff ODEs 6. Purpose : Determining the accuracy of a method through Taylor expansions. Note to Instructor: You may need to help the students by giving them a bit more background on how to Taylor expand about points other than t n +1 . For example, y ( t n 1 ) = y ( t n ) - hy ( t n ) + h 2 2 y ′′ ( t n ) - h 3 3! h ′′′ ( t n ) + O ( h 4 ) From [Hea02, pp.418,#9.12] . 22
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Exercise : The centered difference approximation y y k +1 - y k 1 2 h leads to the two-step leapfrog method y k +1 = y k 1 + 2 hf ( t k , y k ) for solving the ODE y = f ( t, y ) . Determine the order of accuracy for this method. Solution: Write expressions for y k +1 and y k 1 as follows, y k 1 = y k - hy k + h 2 2 y ′′ k - h 3 3! y ′′′ k + O ( h 4 ) y k +1 = y k + hy k + h 2 2 y ′′ k + h 3 3! y ′′′ k + O ( h 4 ) Thus we can compute y using the leapfrog method as shown, y = y k +1 - y k 1 2 h = 2 hy k + h 3 3!
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