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Part CCSIR NET December 2023rk2-order-conditions

Rk2 order conditions

For the IVP dy/dx =f(x,y),y(x0)=y0= f(x, y), y(x_{0}) = y_{0}, with step size h, let the first iterate of a second-order scheme be y1=y0+y_{1} = y_{0} + Pk1+_{1} + Qk2_{2}, where k1=hf(x0,y0),k2=hf(x0+α0h,y0+β0k1)k_{1} = h f(x_{0}, y_{0}), k_{2} = h f(x_{0} + \alpha_{0}h, y_{0} + \beta_{0}k_{1}) and P,Q,α0,β0∈RP, Q, \alpha_{0}, \beta_{0} \in \mathbb{R}. Which of the following statements are correct?

  1. A.If α0=2\alpha_{0} = 2, then β0=2,P=3/4,Q=1/4\beta_{0} = 2, P = 3/4, Q = 1/4✓ Correct
  2. B.If β0=3\beta_{0} = 3, then α0=3,P=5/6,Q=1/6\alpha_{0} = 3, P = 5/6, Q = 1/6✓ Correct
  3. C.If α0=2\alpha_{0} = 2, then β0=2,P=1/4,Q=3/4\beta_{0} = 2, P = 1/4, Q = 3/4
  4. D.If β0=3\beta_{0} = 3, then α0=3,P=1/6,Q=5/6\alpha_{0} = 3, P = 1/6, Q = 5/6

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests numerical convergence.

See pricing

50 are analysed free — try those first.

The trap it tests

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this

Related counterexample: A convergent method is stable for any step size

More on this topic

The chapter behind this: Numerical methods for ODE — free to read

From Numerical Analysis › Numerical ODE: Euler, Runge–Kutta

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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