Solving ODEs: Piecewise Functions and Direction Fields - Homework Help

[STEMucator]

Homework Statement

So yeah, my first time playing with ODEs, how exciting. So my prof gave us a few suggested exercises and I want to know whether I'm actually doing these properly or not. The question and all relevant things will be included in the picture below :

http://gyazo.com/b3be5b5d56ce2201cf88bce5ea8d5838

Homework Equations

The Attempt at a Solution

(1a) So! I figured I should probably break this up into cases since I'm dealing with a piecewise function.

Case #1 : Suppose x>c, then y(x) = (x-c)^2 and y' = 2(x-c). Now we check that this satisfies our equation!

y' = 2$\sqrt{y}$
2(x-c) = 2$\sqrt{(x-c)^2}$
2(x-c) = 2(x-c)
1 = 1

Case #2 : Suppose x ≤ c, then y(x) = 0 and y' = 0. Now we check that this also satisfies our equation.

y' = 2$\sqrt{y}$
y' = 2$\sqrt{0}$
0 = 0

Check and check. ∴ y(x) satisfies our ODE for all x as desired.

Now for the figure, I've never drawn a direction field before, so I got the computer to do it for me for the first time. I know of course that the equilibrium point occurs at y' = 0 ( In this case it turns out after solving we get y = 0 ). Here's a picture of the direction field when we have the IVP y(0) = 0 :

http://gyazo.com/443b52dec28b53e61b3c7ffb3c0b4369

I'm not sure how this demonstrates that we have infinitely many solutions, perhaps someone could elaborate for me?

I'll save my attempt at (1b) for after I know what I'm doing with part (1a).

 

[jbunniii]

Your proof for case #1 is fine. Your proof for case #2 is really only valid for x < c. This is because x = c is an endpoint of the intervals where the two pieces of the function are joined. To see what can go wrong, consider f(x) = |x|, which has derivative equal to -1 for negative x, +1 for positive x, but is not differentiable at x = 0.

Therefore you need to provide a proof that y is differentiable at c, and that y'(c) = 0.

 

[jbunniii]

As for the figure, I'm not much into differential equations, so I'm not sure what it is you have plotted, nor how (or if) it illustrates that the differential equation has infinitely many solutions. Can you explain in words what the plot is depicting?

I'm also not sure why you need a plot in order to conclude that there are infinitely many solutions. If you show that your y(x) is a solution:
$$y(x) = \begin{cases} 0 & \text{for }x \leq c \\ (x - c)^2 & \text{for }x > c \end{cases}$$
then aren't there infinitely many values of $c$ that satisfy $y' = 2\sqrt{y}$ and $y(0) = 0$?

 

[STEMucator]

Your proof for case #1 is fine. Your proof for case #2 is really only valid for x < c. This is because x = c is an endpoint of the intervals where the two pieces of the function are joined. To see what can go wrong, consider f(x) = |x|, which has derivative equal to -1 for negative x, +1 for positive x, but is not differentiable at x = 0.

Therefore you need to provide a proof that y is differentiable at c, and that y'(c) = 0.

Ah, so my case #2 is really a case #2 and #3. So let me correct that :

Case #2 : Suppose x < c, then y(x) = 0 and y' = 0. Now we check that this also satisfies our equation.

y' = 2sqrt(y)
y' = 2sqrt(0)
0 = 0

Case #3 : Suppose x = c, then we must show that y is differentiable at c.

y'(c) = $lim_{h→0} \frac{y(c+h) - y(c)}{h} = lim_{h→0} \frac{h^2 - 0}{h} = 0$

Now would I be able to conclude the exact same thing as in Case 2 or am I misinterpreting what you said?

Also the graph I posted represents the direction field associated with the ODE as well as its equilibrium point at y = 0.

 

[STEMucator]

Also, here's a much better image of what's going on at the initial point y(0) = 0

http://gyazo.com/fc97b2c612ad68b1565d1749812aeeb4

 

[jbunniii]

Case #3 : Suppose x = c, then we must show that y is differentiable at c.

y'(c) = $lim_{h→0} \frac{y(c+h) - y(c)}{h} = lim_{h→0} \frac{h^2 - 0}{h} = 0$

You have to be slightly careful here. The expression $(h^2 - 0)/h$ is only correct for positive $h$. So what you have shown is
$$\lim_{h \rightarrow 0^+}\frac{y(c+h) - y(c)}{h} = 0$$
You also need to show that
$$\lim_{h \rightarrow 0^-}\frac{y(c+h) - y(c)}{h} = 0$$
in order to conclude that
$$\lim_{h \rightarrow 0}\frac{y(c+h) - y(c)}{h} = 0$$
Of course, this is obviously true for this function, but you definitely need to check the limits from both sides because for some functions they won't match, so the function won't be differentiable.

 

[jbunniii]

Also, here's a much better image of what's going on at the initial point y(0) = 0

http://gyazo.com/fc97b2c612ad68b1565d1749812aeeb4

So what is this plot telling me? How does it show that there are infinitely many solutions? I am not being pedantic here, I don't know what I am looking at. Explain in words what a direction field is, and why it demonstrates what you are trying to show.

 

[STEMucator]

You have to be slightly careful here. The expression $(h^2 - 0)/h$ is only correct for positive $h$. So what you have shown is
$$\lim_{h \rightarrow 0^+}\frac{y(c+h) - y(c)}{h} = 0$$
You also need to show that
$$\lim_{h \rightarrow 0^-}\frac{y(c+h) - y(c)}{h} = 0$$
in order to conclude that
$$\lim_{h \rightarrow 0}\frac{y(c+h) - y(c)}{h} = 0$$
Of course, this is obviously true for this function, but you definitely need to check the limits from both sides because for some functions they won't match, so the function won't be differentiable.

Yes of course, fair enough. So for argument sake saving some latex, since the limit would match from both sides, we could conclude that y(c) = 0 = y'(c). Then it would be literally a copy and pasta of the case 2 proof.

So onto the more important matter of the model.

The direction field represents the solutions of the differential equation which vary according to the constant c. So, depending on c, the graph will fit certain portions of the direction field better.

 

[jbunniii]

Yes of course, fair enough. So for argument sake saving some latex, since the limit would match from both sides, we could conclude that y(c) = 0 = y'(c). Then it would be literally a copy and pasta of the case 2 proof.

Yes.

The direction field represents the solutions of the differential equation which vary according to the constant c. So, depending on c, the graph will fit certain portions of the direction field better.

OK, I'm an analysis guy and not a differential equations guy. I like my proofs to be proofs instead of pictures. When I hear the word "illustrate" I think of it as a synonym for "demonstrate" or "prove", whereas to me, your plot merely SUGGESTS that there may be infinitely many solutions. However, you are in a better position than I am to know what your instructor expects for this question.

If someone asked me to demonstrate that there are infinitely many solutions satisfying $y' = 2\sqrt{y}$ and $y(0) = 0$, I would say: well, we just showed that
$$y(x) = \begin{cases} 0 & \text{for }x \leq c \\ (x - c)^2 & \text{for }x > c \end{cases}$$
satisfies $y' = 2\sqrt{y}$ for any $c$. Furthermore, it also satisfies $y(0) = 0$ for any $c \geq 0$. This gives us infinitely many solutions.

However, I would probably lose points because there is no picture in my above proof...

 

[STEMucator]

Yes.


OK, I'm an analysis guy and not a differential equations guy. I like my proofs to be proofs instead of pictures. When I hear the word "illustrate" I think of it as a synonym for "demonstrate" or "prove", whereas to me, your plot merely SUGGESTS that there may be infinitely many solutions. However, you are in a better position than I am to know what your instructor expects for this question.

If someone asked me to demonstrate that there are infinitely many solutions satisfying $y' = 2\sqrt{y}$ and $y(0) = 0$, I would say: well, we just showed that
$$y(x) = \begin{cases} 0 & \text{for }x \leq c \\ (x - c)^2 & \text{for }x > c \end{cases}$$
satisfies $y' = 2\sqrt{y}$ for any $c$. Furthermore, it also satisfies $y(0) = 0$ for any $c \geq 0$. This gives us infinitely many solutions.

However, I would probably lose points because there is no picture in my above proof...

I also prefer the rigor associated with analysis, this is a very new and... not so well defined topic for me.

I see what you're saying, and I see how it associates itself with the model, but I suppose that asking my professor about this would be a better idea.

Part b seems pretty obvious now though.

(1b) So we are given y' = 2sqrt(y), y(0) = b and obviously y'(0) = 1.

y' = 2sqrt(y)
1 = 2sqrt(b)
1/2 = sqrt(b)
1/4 = b

So for (i), we have no solutions if b < 0 and for (ii) If b = 1/4 we have a unique solution.

I think that's it.

 

[jbunniii]

(1b) So we are given y' = 2sqrt(y), y(0) = b and obviously y'(0) = 1.

Why is y'(0) obviously 1?

 

[STEMucator]

Why is y'(0) obviously 1?

Nice and early in the morning, best time to figure this out.

If y(0) = b, then consider :

y'(0) = $lim_{h→0^+} \frac{y(0+h) - y(0)}{h}$

and

y'(0) = $lim_{h→0^-} \frac{y(0+h) - y(0)}{h}$

Is this what you meant?

 

[jbunniii]

Nice and early in the morning, best time to figure this out.

If y(0) = b, then consider :

y'(0) = $lim_{h→0^+} \frac{y(0+h) - y(0)}{h}$

and

y'(0) = $lim_{h→0^-} \frac{y(0+h) - y(0)}{h}$

Is this what you meant?

No, you said we are given y' = 2sqrt(y), and y(0) = b. But why would this imply that y'(0) = 1? I don't think it does.

 

[STEMucator]

No, you said we are given y' = 2sqrt(y), and y(0) = b. But why would this imply that y'(0) = 1? I don't think it does.

Ohh wait a second here, I don't even need to know y'(0) because I have y(x) come to my rescue I believe.

 

[STEMucator]

My only question is about the algebra then.

So I'm given y' = 2sqrt(y), y(0) = b and I know from part (1a) that y(x) is a general solution to this equation.

So :

For x > c :
b = (0-c)^2
sqrt(b) = c

So we only have a solution if b ≥ 0

For x ≤ c :
b = 0

Thus we would have no solution if b < 0 and a unique solution only if b = 0.

 

[jbunniii]

Thus we would have no solution if b < 0 and a unique solution only if b = 0.

But didn't you just show in part (a) that there are infinitely many solutions when b = 0?

I agree that there are no solutions if y(0) = b < 0, for in that case y'(0) = 2sqrt(y(0)) = 2sqrt(b) is undefined. (Assuming we're working with real-valued functions.)

 

[STEMucator]

But didn't you just show in part (a) that there are infinitely many solutions when b = 0?

I agree that there are no solutions if y(0) = b < 0, for in that case y'(0) = 2sqrt(y(0)) = 2sqrt(b) is undefined. (Assuming we're working with real-valued functions.)

Okay, so there are no solutions if b < 0 since we need the sqrt to be defined. There are infinitely many solutions when b = 0 because y(0) = 0 = b from part (1a) showed that there are infinite solutions.

So (i) is answered and that leaves only one case left, that is when b>0. Though there could be far too many possibilities considering when b>0 for there to be a unique solution?

 

[jbunniii]

So (i) is answered and that leaves only one case left, that is when b>0. Though there could be far too many possibilities considering when b>0 for there to be a unique solution?

Do you have any theorems that can tell you what conditions will guarantee a unique solution?

 

[STEMucator]

Do you have any theorems that can tell you what conditions will guarantee a unique solution?

http://gyazo.com/84a902a929b20af89cc7b84d67a782eb

I believe that's the only thing I have that's close.

 

[jbunniii]

http://gyazo.com/84a902a929b20af89cc7b84d67a782eb

I believe that's the only thing I have that's close.

Unfortunately that doesn't apply because your differential equation is not of that form.

Do you have this theorem of Picard:

http://student.fizika.org/~aficnar/Korisno/Picard.pdf

 

[STEMucator]

Unfortunately that doesn't apply because your differential equation is not of that form.

Do you have this theorem of Picard:

http://student.fizika.org/~aficnar/Korisno/Picard.pdf

Sadly I do not have this theorem accessible to me at the moment.

What about if :

y' = 2sqrt(y)

So

y'(0) = 2sqrt(b)

Thus

y(0) = 2sqrt(b)x + b

Would this not be unique ^ Since from part a

y'(0) = 0
y(0) = 0

So for b < 0, we have no solutions. For b = 0 we have infinitely many, and then the only unique solution according to the IVP for part b would be y(0) = 2sqrt(b)x + b?

 

[jbunniii]

Sadly I do not have this theorem accessible to me at the moment.

What about if :

y' = 2sqrt(y)

So

y'(0) = 2sqrt(b)

Thus

y(0) = 2sqrt(b)x + b

How did you get that? It doesn't even make any sense - the right side depends on x, but the left side does not.

I'm not sure if you will be able to prove uniqueness without some kind of nontrivial theorem. Perhaps the question wants you to look at the plot and argue that it "suggests" uniqueness when b > 0? I recommend checking with your instructor to see what is intended here.

What you should be able to do, however, is recognize why only one function of the form
$$y(x) = \begin{cases} 0 & \text{for }x \leq c \\ (x - c)^2 & \text{for }x > c \end{cases}$$
can be a solution if y(0) = b > 0, versus infinitely many solutions of this form if y(0) = 0.

But this doesn't prove overall uniqueness because you haven't ruled out that there could be solutions that are not of the form indicated above.

 

[STEMucator]

How did you get that? It doesn't even make any sense - the right side depends on x, but the left side does not.

I'm not sure if you will be able to prove uniqueness without some kind of nontrivial theorem. Perhaps the question wants you to look at the plot and argue that it "suggests" uniqueness when b > 0? I recommend checking with your instructor to see what is intended here.

I got a bit desperate trying to wrap my head around this, but I suppose asking my professor what is best would be a good idea in this case.

 

[jbunniii]

I got a bit desperate trying to wrap my head around this, but I suppose asking my professor what is best would be a good idea in this case.

Yeah, that's what I would do.

I edited my post above, adding a partial step you can take, but I'm not sure if you have the machinery for a full proof for this problem.