To illustrate the problem that I'm having, consider the following ODE:

$$y' = \frac{y}{1+x}$$

To solve this ODE, my professor had done the followings:

Assuming $y(x) \not = 0$, and $x \not = -1$ (actually, he never stated these), then we can arrange the ODE as

$$\frac{dy}{y} = \frac{dx}{1+x} \\ \ln|y| = \ln|x+1| + \ln|c| \\ y = c(x+1). \quad \text{(The General solution)}$$

And later, while introducing some initial values to this ODE, he gave the case where $y(-1) = 0$, and claimed that since when $x = -1$, for any values of $c\in \mathbb{R}$, we get different solutions from the "general solution", there exists infinitely many solutions satisfying this initial condition.

However, (even if we skip how we found $y = c(x+1)$) to verify that $y = c(x+1)$ is indeed a solution (for some $c$) to this ODE, when we plug it in to the ODE, in order to cancel the terms $1+x$ in the denominator and in the numerator, we need to assume that $x \not =-1$, otherwise it would also mean we are calcelling zero's, which is not valid since, in the algebra that we are doing, zero do not have any multiplicative inverse.Therefore, any solution of the form $y = c_0(x+1)$ cannot contain the the point $x = -1$, so how can we say that "there are infinitely many solution of this ODE satisfying the initial condition $y(-1) = 0$" ?

With my logic, there exists no such solution satisfying this initial condition, so which one of us is right, and why?

A Further Question:

As we have done it in this particular case, to reach some solutions of a given ODE, we need to do some arrangements to the form of the ODE, such as collection same variables to the same side, but this generally requires some assumptions, such as $y \not= 0$, so in such cases, the result that we get should not be applicable to the cases, for example, in the above case, it there was any other point other than $x = -1$ where $y = 0$, say $y(x_1) = 0$ the solutions that we got from the equation $y = c(x+1)$ should not contain that point $(x_1, 0)$ also, right ?

  • 3
    $\begingroup$ I agree with you, the ODE is not defined for $x = -1$ $\endgroup$ – GBes Feb 15 '18 at 7:03
  • 1
    $\begingroup$ The original equation is not singular for $y=0$, one can easily confirm that $y(x)=0$, $x>-1$ is a solution. Always check that the exclusions due to solution methods/techniques do not lose valid solutions. $\endgroup$ – LutzL Feb 15 '18 at 7:20
  • $\begingroup$ @LutzL Of course, but my point in there is that the solution we got should not contain the point $(x_1, 0)$ if we do not have any extra observation. $\endgroup$ – onurcanbektas Feb 15 '18 at 7:28
  • 1
    $\begingroup$ Not if the task is to find the solution of the ODE. Then you also have to identify the solutions that are excluded by the solution method. The usual case is that separation of variables leads to a constant factor $C=\pm e^c$ where $C=0$ is excluded by the method, but nevertheless produces a valid solution. $\endgroup$ – LutzL Feb 15 '18 at 7:34
  • $\begingroup$ Of course, if the original equation is singular for $y=0$, like $y'=\frac1y$, then any solution, $y=\sqrt{2(x-c)}$ ends just before that value is reached, $x\in(c,\infty)$. $\endgroup$ – LutzL Feb 15 '18 at 7:35

I fully agree that, already when writing an ODE in the form $$ y' = \frac{y}{1+x}, \tag{1} $$ you're already implicitly assuming that $x \neq -1$, otherwise the fraction is not defined. However, since we're dealing with an ODE (and hence implicitly assume that $y$ is continuously differentiable), the concept of a limit is intimately involved.

So, let's look at equation $(1)$ on $I := \mathbb{R} \backslash \{-1\}$. It's easy to that on $I$, the function $\eta_c(x) = c(1+x)$ is a solution of $(1)$, for any $c \in \mathbb{R}$. Now, even though $\eta_c$ is not defined at $x=-1$, the limit $\lim_{x \to -1} \eta_c(x)$ exists. That is, both left and right limits (to $x=-1$) of exist and are equal. Furthermore, by the same reasoning, the limit of the derivative ($x \to -1$) also exists.

Therefore, it's a small step to extend $\eta_c$ to $\mathbb{R}$ by including the definition $\eta_c(-1) := \lim_{x \to -1} \eta_c(x)$. This new extended function -- let's call it $y_c$ -- is continuously differentiable on the entire real line. Since the procedure sketched above is straightforward and unambiguous, people tend to call $y_c : \mathbb{R} \to \mathbb{R}, x \mapsto c (1+x)$ 'the solution' to $(1)$, without worrying too much about the point $x=-1$.

Be that as it may, there's still something going on at $x=-1$, of course. To see what's the problem, I think it's instructive to look at the slightly rewritten ODE $$ (1+x) y'=y. \tag{2} $$ It's clear that on $I$, equations $(1)$ and $(2)$ are equivalent, so they give rise to the same solutions. However, if we are looking for continuously differentiable solutions to $(2)$ on the entire real line, we take the limit $x \to -1$ on both sides of equation $(2)$. As we have assumed $y$ to be continuously differentiable on $\mathbb{R}$, both the limit $\lim_{x \to -1} y(x)$ and the limit $\lim_{x \to -1} y'(x)$ exist (and are finite), so we obtain $$ 0 \times y'(-1) = y(-1), $$ i.e. $y(-1) = 0$. Hence, we can't choose the function value at $x=-1$; every continuously differentiable solution to $(2)$ obeys $y(-1)=0$. Of course, you can still take your initial condition $y(x_0)=y_0$ anywhere else (i.e. $x_0 \in I$), and you'll get a unique, continuously differentiable solution to $(1)$.

To summarise: equation $(2)$ is singular at $x=-1$. This has the effect that every continuously differentiable solution to $(2)$ obeys $y(-1)=0$.

  • $\begingroup$ How the derivative of $\eta_c$ is defined at $x= -1$, in which the function is not defined ? I mean by defining the derivative of $\eta_c$ as the limit of right and left limit, you are basically defining a new concept, say differentiablity v2, which is different from what we call normally. $\endgroup$ – onurcanbektas Feb 16 '18 at 13:37
  • $\begingroup$ I have slightly edited the answer to address your concern. $\endgroup$ – Frits Veerman Feb 16 '18 at 17:08
  • 1
    $\begingroup$ Solution domains with gaps can, in general, give rise to problems related to continuity and differentiability of the solution. For your specific example, I believe my answer is reasonably complete. If you still have questions about this, as I believe you do, I recommend that you post another question. This allows more users to give their opinion, which would benefit you, I think. The question what one either mathematically or physically 'can' or 'cannot' do, is closely related with the type of solutions you're looking for, in terms of differentiability. [...] $\endgroup$ – Frits Veerman Feb 21 '18 at 14:45
  • 1
    $\begingroup$ [...] Mathematical concepts such as open and closed sets, continuity, differentiability, and even analytic continuation have been introduced precisely to deal with these kind of issues. $\endgroup$ – Frits Veerman Feb 21 '18 at 14:46
  • 1
    $\begingroup$ Actually, I have posted another question 7 hours ago related with this, see math.stackexchange.com/questions/2660014/… $\endgroup$ – onurcanbektas Feb 21 '18 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.