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Solutions to problems related to the stability and time-step restriction of runge-kutta methods for numerical solutions of ordinary differential equations. It includes calculations for the two-stage runge-kutta method and the ode y' = -y/(1+y), as well as estimates and checks for the forward euler and 4th order runge-kutta methods.
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Jeffrey Hellrung Friday, November 09, 2005 Math 269A, Assignment 06
y∗^ = yn + αdtf (yn),
yn+1 = yn + dt (β 1 f (yn) + β 2 f (y∗)). Assume that the constants α, β 1 , and β 2 are chosen so that the method is second order (Assignment 3, Problem 1).
(a) Find the interval of absolute stability for the method. (b) Does the interval of absolute stability depend upon the coefficients α, β 1 , and β 2?
Solution From Assignment 3, Problem 1, we have that
β 1 + β 2 = 1, αβ 2 =
(a) Setting f (y) = λy, we obtain
y∗^ = yn + αhf (yn) = yn + αλhyn,
yn+1 = yn + h (β 1 f (yn) + β 2 f (y∗)) = yn + h
β 1 λyn + β 2 λyn + αβ 2 hλ^2 yn
1 + (β 1 + β 2 )λh + αβ 2 (λh)^2
1 + λh + (λh)^2 / 2
yn
hence we require that (^) ∣ ∣ ∣ ∣1 +^ λh^ +
(λh)^2 2
∣ <^1 ⇒^ λh^ ∈^ (−^2 ,^ 0).
(b) Clearly, the interval of absolute stability is independent of α, β 1 , and β 2 (assuming these constants are chosen such that the method is second order).
−β −α
y, y(0) =
(a) When α = 100.0 and β = 1.0 determine the eigenvalues of the Jacobian of this system of equations. (b) Estimate the time-step restriction so that i. Forward Euler ii. 4th^ Order Runge-Kutta yield a qualitatively correct solution. This is just an estimate, so you can do it graphically. (Attached are the plots of the regions of absolute stability for Runge-Kutta methods of orders 1-4.) (c) Use your program from Assignment 4 and check the validity of your estimate. For each method, determine if your timestep estimate is close to the timestep required to get a “reasonable looking” numerical solution.
Solution
(a) The Jacobian of the system is simply ( 0 1 −β −α
which has characteristic polynomial
P (λ) = λ(λ + α) + β = λ^2 + αλ + β
whose roots are λ+ = α 2
4 β α^2
λ− = α 2
4 β α^2
Notice that for α^2 ≫ β, these can be approximated as
λ+ ≈ − β α
, λ− ≈ −α.
Thus, with α = 100.0 and β = 1.0,
λ+ ≈ − 0. 01 , λ− ≈ − 100.
(b) i. The interval of absolute stability for the Forward Euler method is (− 2 , 0), hence, since λ+, λ− < 0, λ+h, λ−h ∈ (− 2 , 0) ⇒ h ∈ (0, 0 .02). ii. The interval of absolute stability for the 4th^ Order Runge Kutta method is (− 2. 785 , 0), hence, since λ+, λ− < 0, λ+h, λ−h ∈ (− 2. 785 , 0) ⇒ h ∈ (0, 0 .02785).
(c) i. The following table gives the approximated solution yN to t = 10 using Forward Euler for the given timestep h: h yN