# Ordinary Differential Equations/Existence

### Existence and uniqueness

So, does this mean that if we have an initial condition we will always have 1 and only 1 solution? Well, not exactly. Its still possible in some circumstances to have either none or infinitely many solutions.

We will restrict our attention to a particular rectangle for the differential equation ${\displaystyle y'=f(x,y)}$ where the solution goes through the center of the rectangle. Let the height of the rectangle be h, and the width of the rectangle be w. Now, let M be the upper bound of the absolute value of f(x,y) in the rectangle. Define b to be the smaller of w and h/M to ensure that the function stays within the rectangle.

Existence Theorem: If we have an initial value problem ${\displaystyle y'=f(x,y),y(a)=b}$, we are guaranteed a solution will exist if f(x,y) is bounded on some rectangle I surrounding the point (a,b).

Basically this means that so long as there is no discontinuity at point (a,b), there is at least 1 solution to the problem at that point. There can still be more than 1 solution, though.

Uniqueness Theorem: If the following Lipschitz condition is satisfied as well

For all x in the rectangle, then for two points ${\displaystyle (x,y_{1})}$ and ${\displaystyle (x,y_{2})}$, then ${\displaystyle |f(x,y_{1})-f(x,y_{2})| for some constant ${\displaystyle K}$,

then the solution is unique on some interval ${\displaystyle J}$ containing x=a.

So if the Lipschitz condition is satisfied, and, and ${\displaystyle f(x,y)}$ is bounded, there is a solution and the solution is unique. If the Lipschitz condition is not satisfied, there is at least 1 other solution[citation needed]. This solution is usually a trivial solution ${\displaystyle y(x)=k}$ where k is a constant.

We will use two different methods for proving these theorems. The first method is the Method of Successive Approximations and the second method is the Cauchy Lipschitz Method."

Lets try a few examples.

#### Example 9

${\displaystyle y'=ky,y(10)=500}$

Is the equation ${\displaystyle f(x,y)=ky}$ continuous? Yes.

Is the equation ${\displaystyle {\frac {\partial {f}}{\partial {y}}}=k}$ continuous? Yes.

So the solution exists and is unique.

#### Example 10

${\displaystyle y'={\frac {1}{x}},y(0)=5}$

Is the equation ${\displaystyle f(x,y)={\frac {1}{x}}}$ continuous? No. There is a discontinuity at x=0. If we used any other point it would exist.

So the solution does not exist.

#### Example 11

${\displaystyle y'={\sqrt {y-1}},y(1)=1}$

Is the expression ${\displaystyle f(x,y)={\sqrt {y-1}}}$ continuous? Yes.

Is the expression ${\displaystyle {\frac {\partial {y}}{\partial {x}}}={\frac {1}{2(y-1)^{\frac {1}{2}}}}}$ continuous for y(1)=1? No. It is discontinuous at y=1 (that is, it is not defined at ${\displaystyle y=1}$ and is unbounded as ${\displaystyle y\to 1}$), but continuous for all x.

Lifschitz condition is not satisfied, though the existence condition is satisfied. Hence, the solution exists but need not be unique.

The first solution is ${\displaystyle y(x)={\frac {(x-1)^{2}}{4}}+1,x\geq 1}$.

The other solution happens to be the trivial one, ${\displaystyle y(x)=1}$.