Differential Equations Day 26 -- Revisiting the Picard-Lindelöf Theorem
Overview
In this activity we'll take a look back at the Picard-Lindelöf Theorem on the existence and uniqueness of first order ordinary differential equations with initial conditions from the previous activity. In particular, we'll take a closer look at the continuity condition in the Picard-Lindelöf Theorem.
A Closer Look at the Continuity Condition
It's common for a person just learning this material to be concerned about the continuity condition in the Picard-Lindelöf Theorem. You might wonder: "is it reasonable to expect and be continuous?" It turns out, it is a very reasonable condition to expect to be true; many differential equations one encounters satisfy this condition. Indeed, the Picard-Lindelöf Theorem applies almost every differential equation with initial condition we've seen throughout this course.
Let's take a look a specific example though so you can gain some intuition for yourself. Let's examine the continuity conditions in the case of the differential equation and initial condition
In this differential equation the functions that need to be studied vis-a-vis the Picard-Lindelöf Theorem are and . Intuitively, these two functions are continuous if small changes in the independent variables doesn't result in sudden ("jump") changes in the dependent variable.
Continuity is a famously tricky condition to check mathematically. To check if a function is continuous formally involves something called an -argument. We won't discuss this style of mathematical proof here. Instead, we'll appeal to the more intuitive definition of continuity: a function is continuous if its output can be drawn without any jumps, breaks, or tears.
For a function with a single independent variable this means that the curve can be drawn without any jumps. For instance is continuous, but and are not continuous (both and have a discontinuity at ). Checkout the applet below illustrating these three functions. Notice that has no jumps, and so is continuous, but and do have a jump, an so are not continuous.
Note that and are only discontinuous at one point, . Otherwise, both functions are continuous everywhere else except at .
This notion of looking for "jumps, breaks, or tears" can also also be used for functions of two independent variables such as and from our differential equation.
Let's look first at the graph of , paying special attention to the graph of the function at the point (0,1), corresponding with the initial condition.
As you can plainly see, the function has no "jumps, breaks or tears" at . Indeed, the graph of looks very much like a single warped piece of paper. This is fine. This means is continuous. Thus we are effectively visually observing the continuity of at , and indeed everywhere else too.
Now let's look at
Again, you can plainly see, the function has no "jumps, breaks or tears" at . Thus we can visually observe the continuity of at . Note that the notation is just shorthand for the partial derivative of with respect to .
Try out checking the continuity of other simple differential equations on your own, such as . Remember, if you can verify that and are continuous at , then you can be assured that there is a unique solution of the differential equation satisfying the initial condition.
A Non-Continuous Example
Let's take a look at
First, let's look at the slope field and the initial condition, and then let's look at the continuity. Try out the automated differential equation solver
SolveODE(sqrt(y), (0,0))Note that you get a non-zero solution! However, shouldn't also solve the differential equation and the initial condition? Yes it should, since its derivative is 0. Therefore there are non-unique solutions of this initial condition satisfying the differential equation. The reason why this doesn't violate the Picard-Lindelöf Theorem is because the continuity condition is not satisfied. Let's take a look.
First let's look at the graph of , paying careful attention to the graph at .
No problem here. the function isn't defined for y<0, but that doesn't cause a continuity problem at the initial condition.
The problem is really in the continuity of . Check it out:
Note that not only is discontinuous at (0,0), it isn't even defined! Being undefined is considered badly discontinuous! Hence the Picard-Lindelöf Theorem does not apply, and there may not be unique solutions. And in fact, there are not in this case as we saw with and
Conclusion
The issue with this topic from the point of view of the student is that it's reasonable to ask: "so what?" I totally get it. Who cares?
On this point, I'd just mention that the positive affirmation of the Picard-Lindelöf Theorem is much more useful than working out quirky equations where it doesn't hold. In other words, it's been far more likely to know that in many cases that a differential equation has a unique solution, than to know the quirks where uniqueness can't be guaranteed.
We unfortunately won't have a chance to dig into this topic any further in this course. But if you're interested in reading more, this is a great resources:
https://www.ma.imperial.ac.uk/~jswlamb/DynamIC/M2AA1-2010/M2AA1notes10Ch4.pdf