Nonlinear differential equation (quick)

[DivGradCurl]

I need some help with the problem that follows. Any help is highly appreciated.

Problem:

"Consider the initial value problem y^{\prime}=y^{1/3}, \mbox{ }y(0)=0 from Example 3 in the text.
(a) Is there a solution that passes through the point (1,1)? If so, find it.
(b) Is there a solution that passes through the point (2,1)? If so, find it.
(c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at t=2."

Example 3:

y^{\prime}=y^{1/3}, \qquad y(0)=0 \quad \qquad (1)

y^{-1/3} \: dy = dt

\int y^{-1/3} \: dy = \int dt

\frac{3}{2} y^{2/3} = t + \mathrm{C}

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

The initial condition is satisfied if \mathrm{C}=0, so

y=\phi _1 (t) = \left( \frac{2}{3}t \right) ^{3/2}, \qquad t \geq 0 \quad \qquad (2)

On the other hand, the function

y=\phi _2 (t) = -\left( \frac{2}{3}t \right) ^{3/2}, \qquad t \geq 0 \quad \qquad (3)

is yet another solution. For an arbitrary positive t_0 the functions

y=\left\{ \begin{array}{ll} 0, &\mbox{ if } 0 \leq t < t_0 \\ \pm \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2} , &\mbox{ if } t \geq t_0 \end{array} \right. \quad \qquad (4)

are solutions of Eq. (1).

Answers:

"(a) No. (b) Yes; set t_0=\frac{1}{2} in Eq. (4) from Example 3 (c) \left| y \right| \leq \left( \frac{4}{3} \right) ^{3/2}"

My work:

(a) (Why is this wrong?)

y^{\prime}=y^{1/3}, \qquad y\left( 1 \right)=1

y^{-1/3} \: dy = dt

\int y^{-1/3} \: dy = \int dt

\frac{3}{2} y^{2/3} = t + \mathrm{C}

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

The initial condition is satisfied if \mathrm{C}=\frac{1}{2}, so

y= \left[ \frac{2}{3} \left( t + \frac{1}{2} \right) \right] ^{3/2}

which seems to work.

(b) (This one seems to be ok.)

y^{\prime}=y^{1/3}, \qquad y\left( 2 \right)=1

y^{-1/3} \: dy = dt

\int y^{-1/3} \: dy = \int dt

\frac{3}{2} y^{2/3} = t + \mathrm{C}

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

The initial condition is satisfied if \mathrm{C}=-\frac{1}{2}, so

y= \left[ \frac{2}{3} \left( t - \frac{1}{2} \right) \right] ^{3/2}

(c) (This one seems to be ok.)

Considering all possible solutions of the given initial value problem, the maximum value at t=2 is given by the solution

y = \left( \frac{2}{3}t \right) ^{3/2} \Rightarrow y(2) = \left( \frac{4}{3} \right) ^{3/2}

 

[StatusX]

I'm a little confused. Is the y(0)=0 condition applied through the whole problem? If so, there is one unique solution, since this is a first order differential equation. Unless you're supposed to consider the cube root as a multivalued function, in which case the other solutions will be complex.

 

[saltydog]

I don't believe the Lipschitz condition is met with:

\frac{\partial}{\partial y} y^{1/3}

That is why we obtain two solutions passing through (0,0), that is:

y(x)=[\pm\sqrt{(2/3)x}]^3

Edit: I deleted the earlier post I made. Misinterpreted the problem. Just easier to remove it.

Edit: Also, y(x)=0 is a solution.

 

[saltydog]

One more Thiago:

For the initial value equation:

y^{'}=y^{1/3};\qquad y(0)=0

We find three solutions:

y(x)=0

y(x)=[\pm\sqrt{(2/3)x}]^3

So, for a and b I say no.

 

[StatusX]

Those solutions do appear to work. It seems strange though, because if you start at (0,0), the initial slope is 0, so y doesn't change, and so the slope stays 0. I don't understand how the function starts to curve away from the x-axis.

 

[DivGradCurl]

I'm a little confused. Is the y(0)=0 condition applied through the whole problem? If so, there is one unique solution, since this is a first order differential equation. Unless you're supposed to consider the cube root as a multivalued function, in which case the other solutions will be complex.

That annoyed me at first too. If the initial condition were different for parts (a) and (b), then it'd be so nice! I assumed it worked this way, but I am wrong. In fact, it seems that I should focus on Eq. (4), in particular:

y = \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}

Frankly, I don't see how t_0 comes into play, and why t_0 should equal \frac{1}{2} for part (b). Could you please clarify that?

Those solutions do appear to work. It seems strange though, because if you start at (0,0), the initial slope is 0, so y doesn't change, and so the slope stays 0. I don't understand how the function starts to curve away from the x-axis.

I think that's because t increases, which implies that y increases.

Thank you so much.

 

[OlderDan]

Those solutions do appear to work. It seems strange though, because if you start at (0,0), the initial slope is 0, so y doesn't change, and so the slope stays 0. I don't understand how the function starts to curve away from the x-axis.

Lot's of functions have zero slope somewhere, and all power functions with exponent > 1 have zero slope at (0,0), but they change. With a properly chosen coefficient, the derivative of these power functions is a root of the function.

y = \left(\frac {t}{n}\right)^n = \frac {t^n}{n^n} \ \rm{,} \ n>1 \ \rm{,} \ t \ge 0

y' = n\frac {t^{n-1}}{n^n} = \frac{n}{t}y = \frac{ny}{ny^{1/n}} = y^{(n-1)/n}

One of the solutions to this problem is the case n = 3/2

 

[saltydog]

Well I have some questions:

1. How come we're getting 3 solutions for this particular initial-value problem? Why not just two or a handfull or even an infinite number like for this one:

ty^{''}+(t-1)y^{'}+y=t^2;\qquad y(0)=0,\quad y^{'}(0)=0

2. What determines how many solutions we get?

3. Just how do you pronounce "Lipschitz" anyway?

4. What is the connection between the Lipschitz condition and the number of solutions?

5. How about a plot?

Alright, one more:

6. Can we be guaranteed a unique solution in any neighborhood not including the origin?

 

[saltydog]

This one too has an infinite number of solutions:

y^{&#039;}=\left\{<br /> \begin{array}{rc1}<br /> \frac{4x^3y}{x^4+y^2} &amp;\mbox{for} &amp; x,y\neq 0 \\<br /> 0 &amp;\mbox{for} &amp; x,y=0<br /> \end{array}\right<br />

satisfying the initial conditions:

y(0)=0

 

[DivGradCurl]

Folks, I finally figured it out. Thank you very much!

This is what "Example 3" SHOULD be:

y^{\prime}=y^{1/3}, \qquad y\left( t_0 \right)= 0

y^{-1/3} \: dy = dt

\int y^{-1/3} \: dy = \int dt

\frac{3}{2} y^{2/3} = t + \mathrm{C}

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

The initial condition is satisfied if \mathrm{C}=-t_0, so

y = \eta _1 = \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}, \qquad t \geq t_0

On the other hand, the function

y = \eta _2 = - \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}, \qquad t \geq t_0

is yet another solution. Furthermore, the function

y = \eta _3 = 0, \qquad 0 \leq t &lt; t_0

also is a solution.

(a) We are given the point (1, 1). Then, we use \eta _1:

y = \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}

1 = \left[ \frac{2}{3} \left( 1 - t_0 \right) \right] ^{3/2} \Rightarrow t_0 = -\frac{1}{2}

Thus, t_0 lies outside the domain, which implies that the solution does not pass through the point (1, 1).

(b) We are given the point (2, 1). Then, we use \eta _1:

y = \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}

1 = \left[ \frac{2}{3} \left( 2 - t_0 \right) \right] ^{3/2} \Rightarrow t_0 = \frac{1}{2}

Thus, t_0 lies within the domain, which implies that the solution passes through the point (2, 1). We have

y = \left[ \frac{2}{3} \left( t - \frac{1}{2} \right) \right] ^{3/2}

(c) Post #1 (seems to be ok).

 

[saltydog]

Thiago . . . I don't want to be a pain really because I enjoy working the problems you propose but for the record, I disagree with (a). As I see it,

y(t) = \left[ \frac{2}{3} \left( t+1/2 \right) \right] ^{3/2}

Is a solution satisfying:

y^{&#039;}=y^{1/3}\qquad y(-1/2)=0 \quad\text{and}\quad y(1)=1

As I understand it, the solution y(t)=0 is only when t<-1/2.

 

[DivGradCurl]

As I understand it, the solution y(t)=0 is only when t<-1/2.

Consider the following

y = \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2}, \qquad t \geq t_0 \qquad (1)

y = 0, \qquad 0 \leq t &lt; t_0 \qquad (2)

As you can see, each solution has a different domain, but collectively we can say that the domain consists of nonnegative real numbers. In part (a) of my previous post, I let t=1 and y=1. Solving Eq. (1) for t_0 gives t_0 = -\frac{1}{2}. As a result, it invalidates the statement. Then, it follows that the solution does not pass through the point (1, 1). It seems to me that we have the same thing.

I appreciate your input.