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,&amp;\mbox{ if }<br /> 0\leq t&lt; t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, &amp; \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 &lt; 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&#039; = 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,&amp;\mbox{ if }<br /> 0\leq t&lt; t_0\\ \pm [\frac{2}{3}(t - t_0)]^{3/2}, &amp; \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 &lt; 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.