Interval of validity of a non-linear differential equation

 In the following problem we are going to determine the interval of validity of a non-linear differential equation.

Problem. Determine the interval of validity of the initial value problem and give its dependence on the value of y₀.

y' = y²  y(0) = y₀

Let's see if f(t, y) = y² and δf/δy are continuous. It's obvious that f(t, y) =  y² is continuous. 

Let's calculate δf/δy:

δf/δy = 2y. It's continuous. Then there is a unique solution to the IVP in an interval that contains t₀ = 0

We have to determine this solution in order to find the interval of validity. 

Let's solve the differential equation

dy/dt = :y² 

dy = y² dt

dy/ y² = dt

y^-2 dy = dt

∫y^-2 dy = ∫dt

y^-2+1/-2 +1 = t + c

y^-1/-1 = t

-1/y = t + c

y = -1/t + c

Let's determine the value of c. Let's substitute t by 0 and y by  y₀ as it's given in the initial condition

y₀ = -1/c c = -1/y₀

Let's substitute c in y we have:

y = -y₀ /y₀ t -1

The interval of validity of the IVP is the interval where y is defined. 

Let's determine this interval :of validity.

The function y isn't defined if the denominator is equal to 0.

If y₀ t -1 = 0, t = 1/y₀ . Therefore y isn't defined for that value of t.

Y is defined in the following intervals: -∞<t<1/y₀  and 1//y₀<t<+∞

Let's determine the interval of validity for different values of y₀:

If y₀ = 0, then the interval of validity is -∞<t<+∞

If y₀<0  1/y₀<0 t = 0 doesn't belong to  -∞<t<1/y₀ but belongs to 1//y₀<t<+∞, which is an interval of validity

If y₀>0 then 1/y₀>0  -∞<t<1/y₀

The problem that we just solved demonstrates the difference between the interval of validity of linear differential equations and that of non-linear differential equations. The interval of validity of linear differential equations doesn't depend the value of y₀. The interval of validity of non-linear differential equations depends on the value of  y₀.

Intervals of validity (continued)

Theorem. Let's consider the Initial Value Problem (IVP) y' = f(t,y)  and y(t₀) = y₀. If f(t,y) and ẟf/ẟy are continuous in some rectangle α<t<β and  δ<y<ϒ containing the point (t₀,y₀) then there is a unique solution to the IVP in the interval t₀ - h<t< t₀ + h that is contained in α<t<β. 

This theorem doesn't allow to find the interval of validity. If its conditions are met, we know that the unique solution exists. This unique solution will be needed to determine the interval of validity.

In non-linear differential equations the value of y₀ can affect the interval of validity.

Example. Solve the initial value problem y' = y^1/3 y(0) = 0

Solution

Let's see  if the conditions are met

f(t,y) = y^1/3 is continuous on any interval.

Let's see if the derivative is continuous. The derivative of f is 

δf/δy = (1/3)y^1/3-1

         = (1/3)y^-2/3

         =  (1/3) 1/y^2/3

         =  1/3y^2/3

The function  δf/δy isn't continuous for y = 0. The theorem states that the function must be continuous in an interval that contains  y₀. Since the function isn't continuous for y = 0, it's not continuous in any interval that contains  y₀.= 0. The conditions aren't met. There isn't a unique solution to the differential equation in an interval that contains y = 0. Let's solve this equation n order to find its solutions.

dy/dt = y^1/3

dy =   y^1/3dt

Let's separate the variables;

dy/ y^1/3 = dt

 1/y^1/3dy = dt

y^-1/3dy = dt

Let's integrate both sides:

∫ y^-1/3dy = ∫dt

y^-1/3 + 1/-1/3 + 1 = t + c

y^2/3 /2/3 = t + c

3/2y^2/3 = t + c

For t = 0 y = 0. Let's substitute t and y to find c:

0 = 0 + c

c = 0. 

Let's substitute c in the last equation:

 3/2y^2/3 = t

 y^2/3 = t/3/2

   y^2/3  = 2/3t

Let's raise both sides to the third power:

 (y^2/3)³ = (2/3t)³

y² =  (2/3t)³

y = ∓⎷ (2/3t)³

y =  ∓ (2/3t)^3/2 

These solutions satisfy the initial condition t = 0 y = 0. y(t) = 0 is a solution since it verifies the initial condition t = 0 y = 0.

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