# Existence and Uniqueness

This note contains some theorems that refer to existence and uniqueness of solution of ODE.

Theorem 1. Consider the linear differential equation of n-th order: y^((n))+p_1(t)y^((n-1))+p_2(t)y^((n-2))+...+p_n(t)=f(t) . If all coefficients p_1(t), p_2(t), ..., p_n(t) and f(t) are continuous in interval (a,b) then equation has unique solution, which satisfies given initial conditions y(t_0)=y_0 , y'(t_0)=y_0^' , ..., y^((n-1))(t_0)=y_0^((n-1)) , where t_0 belongs to the interval (a,b) .

Note that there are no restrictions on y_0,y_0^',...,y_0^((n-1)) .

Theorem 2. Consider the IVP y'=f(t,y) y(t_0)=y_0 . If f(t,y) and (partial f)/(partial y) are continuous in some rectangle (t_1,t_2) , (y_1,y_2) that contains point (t_0,y_0), then there exists unique solution to the IVP in some interval t_0-epsilon<t<t_0+epsilon (epsilon>0 ) that is contained in interval (t_1,t_2) .

These theorems are very similar, however there is one big difference: for the linear ODE there are no restrictions on initial function values (y_0,y_0^',...,y_0^((n-1)) ), unlike the case with non-linear ODE (theorem 2).

Example 1. Determine all sollutions to the following IVP: y'=y^(1/5) , y(0)=0.

Here f(t,y)=y^(1/5) and it is continuous, but (partial f)/(partial y)=1/5 y^(-4/5) is not continuous at y=0 so theorem 2 doesn't hold an there is more than solution. Note, that first solution is y=0.

To find second solution note that (dy)/(dt)=y^(1/5)->(dy)/(y^(1/5))=dt

Integrating both sides gives 5/4 y^(4/5)=t+C.

Plugging initial conditions yields 5/4*0^(4/5)=0+C ->C=0

So, 5/4 y^(4/5)=t

Or

y^(4/5)=4/5t

y^4=(4/5t)^5

y=+-(4/5t)^(5/4)

Thi gives two other solutions. So, together with y=0 there are three different solutions.