I am struggling with obtaining y=0; however, Maple found it no problem.
Hi everyone,
This is final time, so I am reviewing all chapters ready for final exam. I get 1 problem that I do not know how to figure out. Could you guys please help. I appreciate it.
Problem:
Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0
Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5
a/ Show that f1 and f2 are both solutions to initial value problem.
b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?
Sorry, I got a typing mistake. The right problem is:
f1(x) ≡ 0 not f1(x) = 0
Problem:
Consider the initial value problem: y' = 5.y^(4/5) ; y(1) = 0
Consider the two functions f1: R --> R and f2: R --> R such that
f1(x) ≡ 0 , f2(x) = (x -1)^5
a/ Show that f1 and f2 are both solutions to initial value problem.
b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?
Writing the ODE in 'standard form'...
, (1)
... it has one and anly one solution if is 'Lipschitz continous' in . In Your case is which is not Lipschitz continous in because its partial derivative is unbounded. Furthemore in this case is function of the y alone and that means that if is solution of...
(2)
... the with arbitrary is also solution. In this case...
(3)
... satisfies (2) with initial condition and that means that with arbitrary also satisfies (2) with the same initial condition...
Kind regards
To "show" that a function satisfies a d.e. it is NOT necessary to integrate the equation. If y= f1 is identically 0 then, of course, it is constant and so y'= 0. The equation becomes 0= 5(0)^(4/5)= 0 which is true, and the initial condition, y(1)= f1(1)= 0 is true.
If y= f2= (x- 1)^5, then y'= 5(x- 1)^(4)= 5((x- 1)^5)(4/5) is true and y(1)= (1- 1)^5= 0 is true.
Well, what exactly does the "local existence and uniqueness theorem" say? Especially the hypotheses. Note that, with y'= F(x,y)= 5y^(4/5), F_y= 4y^{-1/5} which is not defined at x= 0.b/ From part a, we see that the initial value problem has more than one solution. why does this not violate the statement of local existence and uniqueness theorem?