Existence, uniqueness and stepping forward
1.[2p] In Picard's theorem, what does the condition on provide?
In Picard's theorem, what does the condition on provide?
The answer is: Uniqueness, since continuity of alone gives existence but permits several solutions
The answer is: Uniqueness, since continuity of alone gives existence but permits several solutions
The answer is: Uniqueness, since continuity of alone gives existence but permits several solutions
2.[1p] If is continuous near the initial point, the initial value problem has exactly one solution there.
If is continuous near the initial point, the initial value problem has exactly one solution there.
The answer is: False
3.[3p] For with , at what time does the solution cease to exist?
For with , at what time does the solution cease to exist?
4.[2p] Why does with have infinitely many solutions?
Why does with have infinitely many solutions?
The answer is: is unbounded at , so the Lipschitz condition fails there
The answer is: is unbounded at , so the Lipschitz condition fails there
The answer is: is unbounded at , so the Lipschitz condition fails there
5.[3p] Apply Euler's method to with and step . What is the estimate after two steps, at ?
Apply Euler's method to with and step . What is the estimate after two steps, at ?
6.[2p] Euler's method gives an error of 0.28 at step size on a certain problem. What error does its first order accuracy predict at ?
Euler's method gives an error of 0.28 at step size on a certain problem. What error does its first order accuracy predict at ?
7.[2p] Put these methods in order of increasing accuracy, from the largest error at a given step size to the smallest.
Put these methods in order of increasing accuracy, from the largest error at a given step size to the smallest.
Improved Euler (Heun)
Classical fourth order Runge-Kutta
Euler
Show the answer
c, a, b
8.[3p] Which statements about numerical solutions are correct?
Which statements about numerical solutions are correct?
Select all that apply
The answer is: A solver keeps printing values after the true solution has blown up, Runge-Kutta of fourth order costs four evaluations of 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?
Why does a first order linear equation need no local hedge in its existence statement?
The answer is: Its right side has -derivative , so the hypotheses hold wherever the coefficients are continuous
The answer is: Its right side has -derivative , so the hypotheses hold wherever the coefficients are continuous
The answer is: Its right side has -derivative , so the hypotheses hold wherever the coefficients are continuous