When does a differential equation have a solution?
When is there a unique solution?
I’m Matthew Wright, at St. Olaf College, and in this video, I will explain a simple
theorem that gives conditions under which a first-order ordinary differential equation
has a unique solution.
Consider the first-order ordinary differential equation dy/dt = f(t,y),
with initial condition y(t_0) = y_0.
Before we try to find a solution, we should answer two questions.
First, does a solution exist?
We would like to know that this initial value problem is actually solvable before we work
too hard to find a solution.
Second, if a solution exists, is it unique?
This question is important, because if this differential equation models a physical system,
we need to know that we have a unique solution before using a solution to make a prediction
about the system.
Otherwise, if there are multiple solutions, how would we know that we have chosen the
right one to make the prediction?
Fortunately, we have a theorem that gives simple conditions under which this initial-value
problem has a unique solution.
Informally, the theorem says that if f(t,y) is a continuous function “near” the point
(t_0, y_0), then a solution exists.
Additionally, if the partial derivative of f with respect to y is also continuous “near”
the point (t_0, y_0), then the solution is unique.
First, let's talk about existence of the solution.
Suppose f(t,y) is a continuous function in a rectangle containing all points (t,y) with
t between a and b, and y between c and d.
If (t_0, y_0) is a point in this rectangle, then there exists a positive real number epsilon
and a function y(t) defined for all t from t_0 minus epsilon to t_0 plus epsilon that
solves the given initial value problem.
In other words, as long as the function f(t,y) is continuous in a rectangle, even a tiny
rectangle, that contains the point (t_0, y_0) in its interior, then there exists a function
y(t) that solves the differential equation, that passes through the point (t_0, y_0),
and this function is defined in at least a tiny interval around that point.
Let's look at an example: dy/dt = y/t.
For any point (t_0, y_0) with t_0 not equal to zero, the function f(t,y) = y/t is continuous
in a rectangle containing (t_0, y_0).
Thus, the existence theorem tells us that a solution exists through (t_0, y_0).
However, the function f(t,y) = y/t is not continuous at points of the form (0, y_0)
-- it's not even defined for t=0 -- and thus the theorem doesn't tell us whether any solution
goes through these points.
Second, let's talk about uniqueness.
Suppose f(t,y) and the partial derivative of f with respect to y are both continuous
functions in a rectangle containing all points (t,y) with t between a and b and y between c and d.
If (t_0, y_0) is a point in this rectangle, then there is a unique solution y(t), defined
for all t from t_0 minus epsilon to t_0 plus epsilon, that solves the initial value problem.
In other words, if the hypotheses of the uniqueness theorem are met, and if you think there are
two functions y_1 and y_2 that both satisfy the initial-value problem on the interval
t_0 minus epsilon to t_0 plus epsilon, then in fact the two functions are the same, y_1= y_2, on this interval.
Since the conditions for the uniqueness theorem are stronger than those for the existence
theorem, the two theorems are often combined, and stated as one existence-and-uniqueness
theorem.
Returning to our example: dy/dt = y divided by t.
We said earlier that a solution exists through any point (t_0, y_0) with t_0 not equal to
zero.
Since the function f(t,y) = y/t and its derivative with respect to y, df/dy = 1/t, are both continuous
in a rectangle containing such a point, the uniqueness theorem tells us that the solution
through this point is unique.
However, as before, the theorem doesn't tell us what happens at points where t is zero.
What is this unique solution?
Either by separating variables or making a good guess, we can see that y = y_0 divided
by t_0 times t – this linear equation – satisfies the initial-value problem.
Thus, we now know that this line is the unique solution through any point (t_0, y_0) where
t_0 is not equal to zero.
What happens when t is zero?
Remember that the theorem doesn't tell us anything in this case.
In fact, the differential equation it is not defined when t equals zero (because it involves
division by t), so it doesn’t make sense to talk about solutions when t equals zero.
However, let’s multiply both sides of the differential equation by t, to obtain the
new differential equation t dy/dt = y.
Any solution of the old differential equation also satisfies this new differential equation,
at least when t is not equal to zero, but now it makes sense to ask about solutions
where t does equal zero.
In this example we see two things that can happen.
First, many solutions pass through the point (0,0).
Specifically, the differential equation t times dy/dt = y is satisfied by any line y=c
times t, and all of these lines pass through the origin.
However, no solution passes through any point (0,y_0) with y not equal to zero.
To see this, note that if t equals zero, then the left side of the differential equation
is zero, and so y, the right side, must be zero as well.
When using the existence and uniqueness theorem, keep two caveats in mind:
First, the theorem only makes a statement about what happens “near” (t_0, y_0) – in
some, possibly small, interval around this point.
It doesn't say anything about what happens far from this point.
It could be that there is a unique solution near (t_0, y_0), but multiple solutions elsewhere.
Second, the theorem is not an if-and-only-if statement.
If the hypotheses for the theorem are not satisfied, then the theorem does not make
any conclusion at all.
We conclude with one more example.
Consider the differential equation dy/dt = 3y^(2/3).
Since the right-hand side, f(t,y)=3y^(2/3), is continuous everywhere, a solution exists
through any point (t_0, y_0).
However, the partial derivative of f with respect to y is 2y^(-1/3), which does not
exist when y is zero, so the theorem does not say whether the solution through a point
(t_0, 0) is unique.
We find that the general solution is y(t) = quantity (t+c)^3 for some constant c, yet
there is also an equilibrium solution y=0.
Thus, there are two solutions through each point where y is zero, and so the solution
through any point on the t-axis is not unique.