# Differential Equations dy/dx=sqrt(x^2+y^2-9)

just a couple of questions:
given dy/dx=sqrt(x^2+y^2-9)
and y(a)=b
i)for what values of a and b would a local solution exist and why
ii)for what values of a and b would a unique solution exist and why

for i) i got all values such that a^2+b^2<=9
for ii) i got all values such that a^2+b^2>9
i'm not 100% sure this is correct and would be happy if someone could confirm/show otherwise these results.

secondly
how would one show that the only solution of
y''+ty'+(1+t^2)y^2=0
that touches the t axis at some point (t0,0) is the identically zero solution

matt grime
Homework Helper
If i) were correct then you're allowing the derivative at (a,b) to have (possibly) imaginary values. Is that permitted?

Not sue what you mean by local, and why a local solution apparently can't also be unique, and is this really a question about Lipschitz and Peano Conditions?

HallsofIvy
Homework Helper
Basically, this is a problem about "existance and uniqueness" and involves the "Lipschitz and Peano conditions" as Matt Grime said.

The d.e. dy/dx= f(x,y) will have a solution (satisfying y(a)= b) as long as f is continuous in some neighborhood of (a,b). It will have a unique solution as long as f is also "Lipschitz" in y in some neighborhood of (a,b).

In this case f(x,y)= sqrt(x2+ y2- 9) and that will be continuous as long as x2+ y[/sup]2[/sup]>= 9. That is, (a,b) must be outside or on the circle x2+ y[/sup]2[/sup]= 9.
(In the case that (a,b) is on that circle, the solution would only exist for (x,y) outside it.)

A function f(x,y) is "Lipschitz" in y on a set as long as |f(x,y1)- f(x,y2)|<= C |y1- y2| for some constant C and (x,y1), (x,y2) in the set. In particular, if the function is differentiable with respect to y on a set, then it is Lipshchitz on that set. Here, f'= y/(x2+y2-9)1/2 which exists as long as (a,b) is strictly outside the circle x2+ y2= 9.

"how would one show that the only solution of
y''+ty'+(1+t^2)y^2=0
that touches the t axis at some point (t0,0) is the identically zero solution?"

One wouldn't. It's not true. The fact that this was paired with the first question should give you a clue. If we write the equation as y"= -ty'+ (1+t2)y2= f(x,y,y') then f is a "nice" function (infinitely differentiable in all three variables) and so this equation has a unique solution for all "y(a)= b, y'(a)= c" initial conditions. In particular, there must exist a unique solution such that y(t0)= 0 and y'(t0)= 1 so y is NOT identically equal to 0.

firstly thank you for your responses
however regarding HallsofIvy's reply i didnt really understand the following

"If we write the equation as y"= -ty'+ (1+t2)y2= f(x,y,y') then f is a "nice" function (infinitely differentiable in all three variables) and so this equation has a unique solution for all "y(a)= b, y'(a)= c" initial conditions. In particular, there must exist a unique solution such that y(t0)= 0 and y'(t0)= 1 so y is NOT identically equal to 0."

firstly y''=-ty'-(1+t^2)y^2=f(t,y,y') (*) but i dont think that would change things too much anyway.
(*) can be written as the following system of equations if we let
x1=y
x2=y'
so x1'=x2=f1
and x2'=-tx2-(1+t^2)x1^2=f2
now partial derivative of fi/xj for i=1,2 and j=1,2 exists and are infintely differentiable so we have the lip****z condition and f is continuous everywhere so we must have only 1 solution to (*) namely y=0 and we have proven it.

sounds ok to me but it goes against what HallsofIvy said, what do you guys think

HallsofIvy
Homework Helper
There exist only one solution to the differential equation with given initial values. If you write the equation as two first order equations (or as a single first order vector equation) then you are solving for x1 and x2. The condition "y(t0)= 0" is saying that x1(t0)= 0 but you are still free to assign whatever value you want to x2(t0). In particular, there exist a unique solution to the differential equation with x1(t0)= 0 and x2(t0)= 1. Since x2= y', y'(t0)= 1 means that y is NOT identically 0.

when you say "there exists only one solution to the differential equation with given initial values" are you saying that for this particular question you have to have y'(t0)=0 as a given initial value?
i am not sure where or how you plucked the value y'(t0)=1 from either

thnx

matt grime