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 {f} }{\partial {y} }=\frac{1}{2(y-1)^{\frac{1}{2}}} 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 y(x)=1.