Problem with Nonunique Solutions

[mkkrnfoo85]

Hello all,

I am just having a problem understanding a problem in my textbook on nonunique solutions. Let me get to the problem:

So, consider the initial value problem:

$$y' = y^{1/3}\mbox{, } y(0) = y_0 = 0$$
$$\mbox{for t}\geq 0$$

So, solving for the differential equation, I get:

$$y = [\frac{2}{3}(t + C)]^{3/2}$$

So, satisfying initial condition, $$0 = [\frac{2}{3}(0 + C)]^{3/2}$$

So, $$C = 0$$

$$y = [\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

So, that's all understandable to me.
But the answer in the book goes on to say that:

$$y = -[\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

is also a solution. And:

$$y = 0$$
, for $$t\geq 0$$

is also a solution. Finally, the answer says you can generalize the solultion to:

$$y = \chi (t) =\left\{\begin{array}{cc}0,&\mbox{ if } 0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{ if } t\geq 0\end{array}\right$$

This last part is very confusing for me. If someone could explain it, it would be very helpful. For example, if the value of $$t_0$$ was given, and it followed the generalization above, wouldn't values for $$0 \leq t < t_0$$ not be 0, but instead be undefined? Since, you can't do $$(negative number)^{3/2}$$
Right?

Thanks in advance for all the help.
-mk

 

[HallsofIvy]

Hello all,

I am just having a problem understanding a problem in my textbook on nonunique solutions. Let me get to the problem:

So, consider the initial value problem:

$$y' = y^{1/3}\mbox{, } y(0) = y_0 = 0$$
$$\mbox{for t}\geq 0$$

So, solving for the differential equation, I get:

$$y = [\frac{2}{3}(t + C)]^{3/2}$$

So, satisfying initial condition, $$0 = [\frac{2}{3}(0 + C)]^{3/2}$$

So, $$C = 0$$

$$y = [\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

So, that's all understandable to me.
But the answer in the book goes on to say that:

$$y = -[\frac{2}{3}(t)]^{3/2}$$
, for $$t\geq 0$$

is also a solution.

Yes, it is.
$$(-[\frac{2}{3}t]^{3/2})= -\frac{2}{3}[\frac{3}{2}t^{\frac{1}{2}$$
while $$(-[\frac{2}{3}t]^\frac{3}{2})^\frac{1}{3}]$$ is exactly the same thing. Also, of course, y(0)= 0.

And:

$$y = 0$$
, for $$t\geq 0$$

is also a solution.

Absolutely: (0)'= 0 and (0)^1/3= 0.

Finally, the answer says you can generalize the solultion to:

$$y = \chi (t) =\left\{\begin{array}{cc}0,&\mbox{ if } 0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{ if } t\geq 0\end{array}\right$$

This last part is very confusing for me. If someone could explain it, it would be very helpful. For example, if the value of $$t_0$$ was given, and it followed the generalization above, wouldn't values for $$0 \leq t < t_0$$ not be 0, but instead be undefined? Since, you can't do $$(negative number)^{3/2}$$
Right?

Well, yes. Perhaps that is why they specifically said $$t\geq 0$$?

Thanks in advance for all the help.
-mk

You'er welcome.

 

[tacman]

Hello mk,

Thank you for sharing your confusion about nonunique solutions. It can definitely be a tricky concept to understand.

First of all, let's define what a nonunique solution is. In mathematics, a solution is considered nonunique if there is more than one possible answer that satisfies the given conditions. In the context of differential equations, this means that there can be multiple functions that can be considered solutions to the same differential equation.

Now, let's take a closer look at the problem you mentioned. The differential equation y' = y^{1/3} is a separable equation, which means that we can separate the variables and solve for y. However, when we do this, we end up with a constant C that can take on any value. This is where the nonunique solutions come into play.

When we solve for y, we get y = [\frac{2}{3}(t + C)]^{3/2}. This is the general solution to the differential equation. However, we also need to consider the initial condition y(0) = 0. In this case, we can see that when C = 0, the function becomes y = 0. But, as you pointed out, this is not the only solution.

If we go back to the general solution, we can see that when C = -\frac{2}{3}t, the function becomes y = -[\frac{2}{3}(t)]^{3/2}. This is also a valid solution that satisfies the initial condition. In fact, any value of C that satisfies C = -\frac{2}{3}t will result in a solution that satisfies the initial condition.

Now, let's look at the last part of the answer that you mentioned. This is a generalization that takes into account all possible values of C. The function \chi(t) is defined as follows:

\chi (t) =\left\{\begin{array}{cc}0,&\mbox{if } 0\leq t< t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, & \mbox{if } t\geq 0\end{array}\right

As you can see, when t is greater than or equal to 0, we have two possible solutions - one with the positive sign and one with the negative sign. This is because when t