Asked by Ashley
Question:
Consider the IVP dy/dt=f(y,t) =(t^2)*y - t + 5 , y(0)=5
Show that the function satisfies the Lipschitz continuity condition on D={(t,y)|0<=y<=3, 0<=t<=4}
My attempt so far:
Following is a theorem we've taught in the class.
Theorem 5.3 :
Suppose f (t, y) is defined on a convex set D ⊆ R^2 . If a constant L > 0 exists with ∂f/∂y (t, y) ≤ L, for all (t, y) ∈ D, then f satisfies a Lipschitz condition on D in the variable y with Lipschitz constant L.
So I considered, | ∂f/∂t | = |2y-1| <= |2y| + |-1|
===> | ∂f/∂t | <= 2|y| + 1
===> | ∂f/∂t | <= 1
Can I conclude Lipschitz condition here is equal to 1?
Any help would be highly appreciated!
Consider the IVP dy/dt=f(y,t) =(t^2)*y - t + 5 , y(0)=5
Show that the function satisfies the Lipschitz continuity condition on D={(t,y)|0<=y<=3, 0<=t<=4}
My attempt so far:
Following is a theorem we've taught in the class.
Theorem 5.3 :
Suppose f (t, y) is defined on a convex set D ⊆ R^2 . If a constant L > 0 exists with ∂f/∂y (t, y) ≤ L, for all (t, y) ∈ D, then f satisfies a Lipschitz condition on D in the variable y with Lipschitz constant L.
So I considered, | ∂f/∂t | = |2y-1| <= |2y| + |-1|
===> | ∂f/∂t | <= 2|y| + 1
===> | ∂f/∂t | <= 1
Can I conclude Lipschitz condition here is equal to 1?
Any help would be highly appreciated!
Answers
Answered by
Ashley
Small typo in | ∂f/∂t |
It should be
| ∂f/∂t | = |2yt-1| <= |2yt| + |-1|
===> | ∂f/∂t | <= 2|yt| + 1
===> | ∂f/∂t | <= 1
Can I conclude Lipschitz condition here is equal to 1?
Any help would be highly appreciated!
It should be
| ∂f/∂t | = |2yt-1| <= |2yt| + |-1|
===> | ∂f/∂t | <= 2|yt| + 1
===> | ∂f/∂t | <= 1
Can I conclude Lipschitz condition here is equal to 1?
Any help would be highly appreciated!
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