Ordinary Differential Equations/Existence

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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 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[citation needed]. 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[edit]

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[edit]

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[edit]

y'=\sqrt{y-1}, y(1)=5

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

Is the equation \frac{\partial {y} }{\partial {x} }=\frac{1}{2(y-1)^{\frac{1}{2}}} continuous for y(1)=5? No. It is discontinuous at y=1, but continuous for all x.

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

The other solution happens to be the trivial solution y(x)=5.