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Existence, uniqueness and stepping forward

1.[2p]

In Picard's theorem, what does the condition on ∂f/∂y provide?

Correct

The answer is: Uniqueness, since continuity of f alone gives existence but permits several solutions

The answer is: Uniqueness, since continuity of f alone gives existence but permits several solutions

The answer is: Uniqueness, since continuity of f alone gives existence but permits several solutions

2.[1p]

If f(t,y) is continuous near the initial point, the initial value problem has exactly one solution there.

The answer is: False

Correct

3.[3p]

For y′=y2 with y(0)=2.5, at what time does the solution cease to exist?

CorrectNot quite: 0.4

4.[2p]

Why does y′=y with y(0)=0 have infinitely many solutions?

Correct

The answer is: ∂f/∂y is unbounded at y=0, so the Lipschitz condition fails there

The answer is: ∂f/∂y is unbounded at y=0, so the Lipschitz condition fails there

The answer is: ∂f/∂y is unbounded at y=0, so the Lipschitz condition fails there

5.[3p]

Apply Euler's method to y′=t+y with y(0)=1 and step h=0.2. What is the estimate after two steps, at t=0.4?

CorrectNot quite: 1.48

6.[2p]

Euler's method gives an error of 0.28 at step size h=0.25 on a certain problem. What error does its first order accuracy predict at h=0.0625?

CorrectNot quite: 0.07

7.[2p]

Put these methods in order of increasing accuracy, from the largest error at a given step size to the smallest.

  1. Improved Euler (Heun)

  2. Classical fourth order Runge-Kutta

  3. Euler

Show the answer

c, a, b

8.[3p]

Which statements about numerical solutions are correct?

Select all that apply

Correct
Correct
Correct

The answer is: A solver keeps printing values after the true solution has blown up, Runge-Kutta of fourth order costs four evaluations of f per step, A stiff equation can force a tiny step for stability even where accuracy would allow a large one

9.[2p]

Why does a first order linear equation need no local hedge in its existence statement?

Correct

The answer is: Its right side has y-derivative −p(t), so the hypotheses hold wherever the coefficients are continuous

The answer is: Its right side has y-derivative −p(t), so the hypotheses hold wherever the coefficients are continuous

The answer is: Its right side has y-derivative −p(t), so the hypotheses hold wherever the coefficients are continuous