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.

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 [math]f\left(x,y\right)[/math] and [math]\frac{\partial}{\partial y}f\left(x,y\right)[/math] 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.[br][br]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[br][br][math]\frac{dy}{dx}=x^2y;y\left(0\right)=1[/math][br][br]In this differential equation the functions that need to be studied [i]vis-a-vis[/i] the Picard-Lindelöf Theorem are [math]f(x,y)=x^2y[/math] and [math]\frac{\partial}{\partial y}f\left(x,y\right)=\frac{\partial}{\partial y}\left(x^2y\right)=x^2[/math]. Intuitively, these two functions are [b]continuous [/b]if small changes in the independent variables doesn't result in sudden ("jump") changes in the dependent variable. [br][br]Continuity is a famously tricky condition to check mathematically. To check if a function is continuous [i]formally[/i] involves something called an [math]\epsilon,\delta[/math]-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. [br][br]For a function with a single independent variable this means that the curve can be drawn without any jumps. For instance [math]h(x)=2x^2-2x+1[/math] is continuous, but [math]g(x)=\frac{\left|x-1\right|}{x-1}[/math] and [math]k(x)=1/(x-1)[/math] are not continuous (both [math]g[/math] and [math]k[/math] have a discontinuity at [math]x=1[/math]). Checkout the applet below illustrating these three functions. Notice that [math]h(x)[/math] has no jumps, and so is continuous, but [math]g(x)[/math] and [math]k(x)[/math] do have a jump, an so are not continuous.

Note that [math]g(x)[/math] and [math]k(x)[/math] are only discontinuous at one point, [math]x=1[/math]. Otherwise, both functions are continuous everywhere else [i]except [/i]at [math]x=1[/math]. [br][br]This notion of looking for "jumps, breaks, or tears" can also also be used for functions of two independent variables such as [math]f(x,y)=x^2y[/math] and [math]\frac{\partial}{\partial y}f\left(x,y\right)=\frac{\partial}{\partial y}\left(x^2y\right)=x^2[/math] from our differential equation.[br][br]Let's look first at the graph of [math]f(x,y)[/math], 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 [math]f\left(x,y\right)[/math] has no "jumps, breaks or tears" at [math](0,1,f(0,1))=(0,1,0)[/math]. Indeed, the graph of [math]f(x,y)[/math] looks very much like a single warped piece of paper. This is fine. This means [math]f(x,y)[/math] is continuous. Thus we are effectively visually observing the continuity of [math]f\left(x,y\right)[/math] at [math]x=0,y=1[/math], and indeed everywhere else too.[br][br]Now let's look at [math]\frac{\partial}{\partial y}f\left(x,y\right)=\frac{\partial}{\partial y}\left(x^2y\right)=x^2[/math]

Again, you can plainly see, the function [math]\frac{\partial}{\partial y}f\left(x,y\right)=f_{y\left(x,y\right)}[/math] has no "jumps, breaks or tears" at [math](0,1,f_y(0,1))=(0,1,0)[/math]. Thus we can visually observe the continuity of [math]f_y\left(x,y\right)[/math] at [math]x=0,y=1[/math]. Note that the notation [math]f_y[/math] is just shorthand for the partial derivative of [math]f[/math] with respect to [math]y[/math].[br][br]Try out checking the continuity of other simple differential equations on your own, such as [math]\frac{dy}{dx}=2x\left(y+1\right);y\left(0\right)=3[/math]. Remember, if you can verify that [math]f(x,y)[/math] and [math]\frac{\partial}{\partial y}f\left(x,y\right)=f_{y\left(x,y\right)}[/math] are continuous at [math](0,3)[/math], then you can be assured that there is a unique solution of the differential equation satisfying the initial condition.

Let's take a look at [br][br][math]\frac{dy}{dx}=\sqrt{y};y\left(0\right)=0[/math][br][br]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[br][br][code]SolveODE(sqrt(y), (0,0))[/code]

Note that you get a non-zero solution! However, shouldn't [math]y(x)=0[/math] 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.[br][br]First let's look at the graph of [math]f(x,y)=\sqrt{y}[/math], paying careful attention to the graph at [math](0,0)[/math].

No problem here. the function isn't defined for y<0, but that doesn't cause a continuity problem at the initial condition. [br][br]The problem is really in the continuity of [math]\frac{\partial}{\partial y}f\left(x,y\right)=\frac{\partial}{\partial y}\sqrt{y}=\frac{1}{2\sqrt{y}}[/math]. Check it out:

Note that not only is [math]\frac{\partial}{\partial y}f\left(x,y\right)[/math] discontinuous at (0,0), it [i]isn't even defined! [/i]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 [math]y\left(x\right)=0[/math] and [math]y\left(x\right)=0.25x^2[/math]

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? [br][br]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. [br][br]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: [br][br][url=https://www.ma.imperial.ac.uk/~jswlamb/DynamIC/M2AA1-2010/M2AA1notes10Ch4.pdf ]https://www.ma.imperial.ac.uk/~jswlamb/DynamIC/M2AA1-2010/M2AA1notes10Ch4.pdf [/url]