Refer to following question- Determine the *interval* in which the solution is defined?

In context of the solution by Florain, I have some doubts.

By uniqueness and existence theorem, we can say that for $y'=f(x,y)$, an initial value problem on the region where $f(x,y)$ is continuous, we will have a unique solution. But what is asked in the question is that solution exist, uniqueness is not required. Does this means that a solution is not valid only because it is not unique? Consider the interval $(-1,2)\cup (4,\infty)$ for the same solution. It seems to me that it satisfies the equation at all points in the interval. Is this solution not valid in this interval only because it is not unique?

Please correct me if any of my statement is incorrect.


Uniqueness is a desirable property of a solution but not required.

But continuity is much more important.

From a theoretical perspective

It is possible to distinguish between solutions based only on their domain.

For example, if you have two solutions

$\{0,1\} \rightarrow u(t)$

$\{-1,2\}\rightarrow v(t)$

with $u(t)=v(t)$

that only differ in the domain, then $v(t)$ is called an extension of $u(t)$. Nevertheless, some mathematicians would call them separate solutions.

But this is often not done because then we would always have an infinite amount of possible solutions.

The same happens with your approach. Let us look at your specific problem. In an initial value problem our initial value would lie in $(-1,2)$ or $(4,\infty)$ but not in both.

So we could assign the interval where the initial value is not located the stationary solution.

Not very nice (but from a theoretical standpoint it would be feasible).

The derivative itself can be non continuous on the other hand. For example


is totally valid, even for $u(t)=0$.

From a numerical perspective

Many numerical approximation schemes rely on Taylor expansion of the form $u(t+\Delta t)=u(t)+hu'(t)+u''(\xi)$.

If your function is discontinuous this property does not automatically hold anymore. A smaller step size would not automatically improve the solution anymore (if the algorithm is consistent) because a smaller $h$ could land on undefined points.


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