Recitation 4

Existence and uniqueness theorems for 1st-order ODE

The general 1st-order initial value problem (IVP) is $$\begin{equation}\tag{*}y’=F(x,y), y(x_0) = y_0.\end{equation}$$ We are interested in the following questions:

1. Under what conditions can we be sure that a solution to (*) exists?
2. Under what conditions can we be sure that there is a unique solution to (*)?

Here are the answers.

Theorem (Existence and uniqueness). Suppose that $F(x,y)$ is a continuous function defined in some region $$R = (x_0-\delta, x_0+\delta)\times (y_0-\epsilon, y_0+\epsilon)$$ containing the point $(x_0, y_0)$. Then there exists $\delta_1 > 0$ so that a solution $y=f(x)$ to (*) is defined for $x\in (x_0-\delta_1, x_0+\delta_1)$. Suppose, furthermore, that $\frac{\partial F}{\partial y}(x,y)$ is a continuous function defined on $R$. Then there exists $\delta_2 > 0$ so that the solution is the unique solution to (*) for $x\in (x_0-\delta_2, x_0 + \delta_2)$.

Example 1 Consider the IVP $y'=x-y+1, y(1)=2$. In this case, both the $F(x,y)=x-y+1$ and $\frac{\partial F}{\partial y}(x,y)=-1$ are defined and continuous at all points. The theorem guarantees that a solution to the ODE uniquely exists in some open interval centered at 1. In fact, an explicit solution to this equation is $y(x) = x+e^{1-x}$. This solution exists for all $x$.

Example 2 Consider the IVP $y' = 1+y^2, y(0)=0$. Both $F(x,y)=1+y^2$ and $\frac{\partial F}{\partial y}(x,y)=2y$ are defined and continuous at all points, so by the theorem we can conclude that a unique solution exists in some open interval centered at 0. By separating variables and integrating, we derive a solution to this equation $y(x)=\tan x$. Remember that a solution to a DE must be a continuous function! In order for this function to be considered as a solution to this IVP, we must restrict the domain to $x\in (-\pi/2,\pi/2)$.
Example 3 Consider the IVP $y' = 2y/x, y(x_0)=y_0$. In this example, $F(x,y) = 2y/x$ and $\frac{\partial F}{\partial y}(x,y)=2/x$. Both functions are defined for all $x\neq 0$, so the theorem tells us that for each $x_0 \neq 0$ there exists a unique solution in an open interval around $x_0$. By separating variables and integrating, we derive solutions $y(x) = Cx^2$ for some constant $C$. Notice that all of theses solutions pass through $(0,0)$. So the IVP $y'=2y/x, y(0)=0$ has infinitely many solutions, but the IVP $y'=2y/x, y(0)=y_0\neq 0$ has no solutions.