# Ordinary Differential Equations/Existence

### Existence and uniqueness[edit]

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 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 , 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 and , then for some constant ,

then the solution is unique on some interval containing x=a.

So if the Lipschitz condition is satisfied, and, and 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 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[edit]

Is the equation continuous? Yes.

Is the equation continuous? Yes.

So the solution exists and is unique.

#### Example 10[edit]

Is the equation 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[edit]

Is the equation continuous? Yes.

Is the equation continuous? No. It is discontinuous at y=1

So the solution exists and is not unique. The other solution happens to be the trivial solution .