# Maximal solution to ODE

Note: this has been edited to fix a typo.

## Homework Statement

The problem is to find the maximum solution on all of R of the differential equation dy/dt = t * y^(1/3) subject to the initial-value conditoin y(1) = -1.

## The Attempt at a Solution

This equation is not subject to the conclusion of Picard's standard existence and uniqueness theorem since t * y^(1/3) is not Lipschitz in the y variable. But the existence of a solution is guaranteed by Peano's theorem. Further, it's proved in the theory of ODEs that there will be a maximum solution--a solution that is greater than or equal to all others at all points.

By separation of variables it's easy to find a particular solution and then use the initial condition to find the value of the constant that emerges in the process. But as for finding the maximum solution I don't know where to begin.

Any help would be greatly appreciated!

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## Answers and Replies

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Mark44
Mentor

## Homework Statement

The problem is to find the maximum solution on all of R of the differential equation dt/dt = t * t^(1/3) subject to the initial-value conditoin y(1) = -1.
Don't you mean dy/dt = y*t^(1/3)?

## The Attempt at a Solution

This equation is not subject to the conclusion of Picard's standard existence and uniqueness theorem since t * y^(1/3) is not Lipschitz in the y variable. But the existence of a solution is guaranteed by Peano's theorem. Further, it's proved in the theory of ODEs that there will be a maximum solution--a solution that is greater than or equal to all others at all points.

By separation of variables it's easy to find a particular solution and then use the initial condition to find the value of the constant that emerges in the process. But as for finding the maximum solution I don't know where to begin.

Any help would be greatly appreciated!

Oh my, this was a bad typo. I meant dy/dt = t * y^(1/3) !

Mark44
Mentor
I found the solution, using separation. The graph looks something like two parabolas, one opening up, and one opening down. The one opening down goes through (1, -1). Being that this is the only solution that satisfies the initial condition and the differential equation, seems like it would also be the maximal solution. The other solution (the upward-opening parabola-like curve) is such that all of its y values are greater than those on the solution represented by the downward-opening parabola-like curve, but it doesn't pass through (1, -1).

Actually, I've realized I am somehow not solving it right, and your post verifies this since I got a different scenario.

Am I missing something here?

After separating variables and integrating one finds (3/2) * y^(2/3) = t^2/2 + const. But imposing the initial condition forces the constant to be complex as far as I can tell.

vela
Staff Emeritus
Homework Helper
Yeah, that confused me as well. You might have noted your equation will satisfy the initial condition if the constant is equal to 2/3. If you solve for y2, you get

$$y^2 = \left(\frac{t^2}{3}+\frac{2}{3}\right)^3$$

The solution you want is the negative root of that equation.

You are correct indeed.

Thanks very much to both of you!