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.
If you are interested in reviewing Calculus I or know someone interested, please refer to Center for Integral Development
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