The Existence and Uniqueness Theorem for Solutions to ODEs - Rhea

Before starting this tutorial, you are supposed to be able to:

· Find an explicit solution for $ \frac{dy}{dt}=f(t) $. This is the same thing as finding the integral of $ f(t) $ with respect to $ t $.

· Know the difference between a general solution and a particular solution satisfying the initial conditions.

· Check one function is a solution to an ODE.

· Distinguish ODE and PDE, know the usual notations.

· Know the basic concepts of ODEs (order, linearity, homogeneity, etc).

2.1 Concept

From the first example from 1.1, here we still suppose that we had a linear equation $ ax+b=0 $ with respect to $ x $.

· When $ a=0 $, $ b≠0 $, there is no solution.

· When $ a≠0 $, there is one solution $ x=-\frac{b}{a} $.

· When $ a=b=0 $, there are infinitely many solutions to this linear equation.

Similarly, an ODE may also have no solution, a unique solution or infinitely many solutions. The existence theorem is used to check whether there exists a solution for an ODE, while the uniqueness theorem is used to check whether there is one solution or infinitely many solutions.

2.2 Existence Theorem

First we are going to define the continuity of a function. Similar with what we have learnt in Calculus 1 but replaced by a 2-variable function, $ f(t,y) $ is continuous at the point $ (t=t_0,y=y_0) $ if here, it is defined, its limit exists and the function value equals to the limit value.

Now let's make an extension. Imagine we are looking at part of a graph of a function by a magnifier, then suppose $ f(t,y) $ is continuous in a closed bounded area (seeing the graph from above the magnifier) in coordinates. So $ f(t,y) $ will be continuous on $ R^2 $ if it's continuous at each point in $ R^2 $. To understand this, just imagine we are smoothly moving our magnifier over the graph, and we can see the graph is always continuous, without any exception.

The geometric meaning of a function with one variable is a line or a curve. We know a dot moves to form a line, and a line moves to form a surface. So, the geometric meaning of a function with two variables is going to be a surface in three-dimensional coordinates. Since $ f(t,y) $ is continuous at every point in $ R^3 $, it will always touch the coordinates, geometrically. Hence, we know there exist solution(s) to this ODE.

Briefly speaking, An ODE will have solution(s) in one particular area, if the initial function $ f(x,y) $ is continuous there.

2.3 Uniqueness Theorem

Now we apply the same consideration to the graph of $ \frac{∂f}{∂t} $. We know $ f(t,y) $ is graphed as a surface. Similar with the tangent line of a curve, if we are doing the partial derivative to a surface function, we will have a tangent plane for it. The tangent plane is parallel to $ y $-axis and can slide back and forth, if we are treating $ y $ as a constant. So basically, the partial derivative $ \frac{∂f}{∂t} $, $ \frac{∂f}{∂y} $ is the gradient of a surface $ f(t,y) $. It shows how the surface changes (going up and down, being wide and narrow, etc.) while we are holding one variable as a constant and sliding the tangent plane.

If the partial derivative is continuous at the point $ (t=t_0,y=y_0) $, it means the gradient is continuous. Just imagine the unceasingly flowing hills where you can never see the end within one particular area. Since the hills are unceasing and can't have two ridges, (Mathematically, $ x $ must only corresponds to one value of $ f(x) $ to make it a valid function) the solution is going to be unique when touching the coordinates.

Briefly speaking, An ODE will have a unique solution in one particular area, if the partial derivative \frac{∂f}{∂t}</math>, $ \frac{∂f}{∂y} $ is continuous here.

2.4 Exercises

Test the existence and uniqueness of the solution of the differential equations:

· $ \frac{dy}{dx}+y=x+3 $