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 $$ y(x) = \frac{(x - 1)^2}{4} + 1, x \ge 1$$.

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