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 $y'=f(x,y)$ 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 $y'=f(x,y),y(a)=b$, 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 $(x,y_{1})$ and $(x,y_{2})$, then $|f(x,y_{1})-f(x,y_{2})|<K|y_{2}-y_{1}|$ for some constant $K$,

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

So if the Lipschitz condition is satisfied, and, and $f(x,y)$ 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. This solution is usually a trivial solution $y(x)=k$ 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

$y'=ky,y(10)=500$

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

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

So the solution exists and is unique.

Example 10

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

Is the equation $f(x,y)={\frac {1}{x}}$ 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

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

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

Is the expression ${\frac {\partial {y}}{\partial {x}}}={\frac {1}{2(y-1)^{\frac {1}{2}}}}$ continuous for y(1)=1? No. It is discontinuous at y=1 (that is, it is not defined at $y=1$ and is unbounded as $y\to 1$), 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 $y(x)={\frac {(x-1)^{2}}{4}}+1,x\geq 1$.

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