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Initial value problem $dy/dx=y^{1/3}$, $y(0)=0$ has one of the following solution
I came across the following problem which says:
The initial value problem $y'=y^{1/3}$, $y(0)=0$ has:
(a) a unique solution,
(b) exactly two solutions,
(c)exactly three solutions,
(d)no solution.
The solution of the problem is given by: $2x=3y^{2/3}$. But I could not come to a conclusion. Clearly, (d) can not be true. But i am not sure about the other options. It will be helpful if someone throws light on it. Thanks in advance.
- ordinary-differential-equations
- $\begingroup$ The answer depends on your definition of $y^{1/3}$ when $y<0$. $\endgroup$ – Christian Blatter Nov 28, 2012 at 18:57
- 2 $\begingroup$ @ChristianBlatter No, it doesn't. The answer is "infinitely many solutions" regardless of what happens for $y<0$. Moreover, the only sensible interpretation in this context would be $y^{1/3}=-|y|^{1/3}$ for $y<0$. $\endgroup$ – Federico Dec 5, 2018 at 16:37
2 Answers 2
The answer is "none of the above". There are infinitely many solutions.
Pick any $\alpha > 0$ and define $f_\alpha (x) = 0$ for $x \le \alpha$ and $f(x) = (2/3)^{3/2} (x-\alpha)^{3/2}$ for $x > \alpha$. All these functions are solutions.
- 2 $\begingroup$ I see why this happened. This is because the function is ill defined on the negative axis. So hope the question asker takes note. $\endgroup$ – Gautam Shenoy Nov 29, 2012 at 6:26
- 1 $\begingroup$ i know the uniqueness criteria for first order differential equations and from that criteria i know that the above problem doesn't have a unique solution. so can we say either unique solution or infinitely many solutions? $\endgroup$ – PAMG Feb 16, 2016 at 4:03
- 3 $\begingroup$ @GautamShenoy I don't know what do you mean by "the function is ill defined on the negative axis", but it has nothing to do with the problem. The answer is "infinitely many solutions" regardless of what happens for $y<0$. Moreover, the only sensible interpretation in this context would be $y^{1/3}=-|y|^{1/3}$ for $y<0$. $\endgroup$ – Federico Dec 5, 2018 at 16:39
- 1 $\begingroup$ $y=0$ looks like a solution, corresponding to $\alpha=+\infty$ $\endgroup$ – Henry Oct 1, 2022 at 15:50
total $3$ solutions
1.$y=0$ is also a solution.
2.y=0 for $x \le 0$ and $y=(2/3x)^{(3/2)}$ (positive squ root)
3.similarly consider neg squ root.
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Boundary Value Problems
Boundary value problems (BVPs) are ordinary differential equations that are subject to boundary conditions. Unlike initial value problems, a BVP can have a finite solution, no solution, or infinitely many solutions. The initial guess of the solution is an integral part of solving a BVP, and the quality of the guess can be critical for the solver performance or even for a successful computation. The bvp4c and bvp5c solvers work on boundary value problems that have two-point boundary conditions, multipoint conditions, singularities in the solutions, or unknown parameters. For more information, see Solving Boundary Value Problems .
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This example shows how to solve a numerically difficult boundary value problem using continuation, which effectively breaks the problem up into a sequence of simpler problems.
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