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 | = |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!
To show that the function f(y,t) = (t^2)*y - t + 5 satisfies the Lipschitz continuity condition on the set D = {(t,y) | 0 <= y <= 3, 0 <= t <= 4}, we need to find a Lipschitz constant L such that the partial derivative of f with respect to y is always less than or equal to L for all (t,y) in D.
You correctly calculated the partial derivative of f with respect to t as ∂f/∂t = 2yt - 1. However, to satisfy the Lipschitz condition, we need to find the maximum value of this derivative over the set D.
Since 0 <= y <= 3 and 0 <= t <= 4, we have |∂f/∂t| <= 2|yt| + 1. To find the maximum value of this expression, we consider the extreme values of y and t in the set D.
When y = 3 and t = 4, we have |∂f/∂t| = 2(3)(4) + 1 = 25.
When y = 0 and t = 0, we have |∂f/∂t| = 2(0)(0) + 1 = 1.
Therefore, the maximum value of |∂f/∂t| over the set D is 25. This gives us a Lipschitz constant L = 25, which means that f(y,t) satisfies the Lipschitz condition on D with a Lipschitz constant of 25.
In conclusion, the function f(y,t) = (t^2)*y - t + 5 satisfies the Lipschitz continuity condition on the set D = {(t,y) | 0 <= y <= 3, 0 <= t <= 4} with a Lipschitz constant of 25.