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.

Theorem 1: 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 continuous 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.

Theorem 2: If $\frac{\partial f}{\partial y}$ is continuous as well, then the solution is unique on some interval J containing x=a.

So if $\frac{\partial f}{\partial y}$ has no discontinuity at x=a, and f(x,y) has no discontinuity around (a,b), there is a solution and the solution is unique. If $\frac{\partial f}{\partial y}$ has a discontinuity at x=a, there is at least 1 other solution. This solution is usually a trivial solution $y(x)=k$ where k is a constant.

Lets try a few examples.

Example 9

$y^{\prime }=ky,y(10)=500$

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

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

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

$y^{\prime }=\sqrt{y-1},y(1)=5$

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 $y(x)=1$.