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.