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The nonlinear ODE: $y'(t)=y(t)^{1/2}$ with initial condition $y(0)=1$ has two solutions. Non-uniqueness is not surprising because of the failure of Lipschitz continuity in the $y$ term. While this is formally true, what if any, is the practical significance of the failure of uniqueness of solutions? For example, if this ODE (or a similar PDE) modelled some biological or physical phenomenon, would non-uniqueness mean anything?

Edit: Thanks for the responses so far! I found a passage in Fung and Tong's Classical and Computational Solid Mechanics in Chapter 21 on page 849 which reads:

One of the distinct characteristics of nonlinear problems is that the solution may not be unique. In the nonuniqueness, there lies much of nature’s secret. Examples in solid mechanics are the buckling of thin shells, self-equilibrating residual stresses and strains, three-dimensional solutions in bodies with apparently two-dimensional boundary conditions, and many problems in plasticity.

I am not so familiar with these models: perhaps someone wiser than myself might have some specific insight?

Josh
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A very good question. There is no reason why an equation would be forced to have unique solutions. But THEN the model of the problem should include the description of what happens when whatever natural object encounters a point of nonuniqueness!

Added: In other words, I feel that any possible nonuniqueness of an equation modeling a natural phenomenon is there either because we wanted it to be there (and perhaps there is then for example a probability of choosing a particular trajectory, but this should also be included in the model...) or simply because we missed it. In both cases, there is this no physical significance of the nonuniqueness, which according to these two possibilities would be feature of the model (to be solved in a way that the model itself should specify) or an incompleteness of the model.

John B
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Differential equations which are physically relevant tend to have the uniqueness property.

If $\gamma : I \rightarrow \mathbb{R}^3$ denotes the velocity of a particle moving through a fluid. Then in all likelihood $\gamma'(t) = F(\gamma(t))$ for some suitable function $F : \mathbb{R}^3 \rightarrow \mathbb{R}^3$. Now the velocity is uniformly Lipschitz continuous or the particle would be subject to a arbitrarily large acceleration/force. This is not something that occurs in real world. In other words, it is extremely likely that $F$ is Lipschitz continuous on its domain.

Carl Christian
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  • What about $\gamma'=\sqrt{\gamma}$? The two trajectories $\gamma=0$ and $\gamma=\frac{t^2}{4}$ have both smooth acceleration. – Miguel Apr 09 '22 at 21:41
  • @Miguel. Trajectories with unbounded velocities are not physically relevant. – Carl Christian Apr 09 '22 at 23:04
  • I am not sure I follow. These two trajectories have bounded velocities in a neighborhood of zero, where they intersect. – Miguel Apr 10 '22 at 08:06
  • @Miguel I current text is not making the point. I have several deadlines right now. I will notify you when I have edited the text to my satisfaction. – Carl Christian Apr 12 '22 at 19:21