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In solving a homework problem, I have encountered the following DE ($v$ is a function of $x$).

$$v' = \frac{1}{x}\frac{1 + x^2}{1-x^2} v, \quad x \in (0,1)$$

I'd like to split the problem into two cases.

  1. $v=0$ somewhere
  2. $v=0$ nowhere

For Case 2, I can solve the problem by taking $v$ to the LHS and choosing an appropriate substitution.

However, Case 1 is giving me a bit of grief. I want to show that, assuming the hypothesis of Case 1, the only solution is the zero function.

Question A. Can I conclude that, if $v=0$ somewhere, then $v=0$ on the entire interval?

Question B. Is there a general principle that works for all DE's of the form $$v'(x) = f(x)v(x),\; x \in I$$

where $f$ is locally Lipschitz continuous and $I \subseteq \mathbb{R}$ is an open interval, allowing us to conclude that if $v=0$ somewhere, then $v=0$ everywhere?

Please keep answers as non-technical as possible; I have received answers to similar questions in the past, and had a lot of trouble understanding them.

goblin GONE
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  • If I understand correctly, your problem is a possible division by $0$. To deal with that I think you have no chance but to deal with technicalities. – Git Gud Aug 30 '13 at 09:31
  • @GitGud, yup that's the problem alright. – goblin GONE Aug 30 '13 at 09:33
  • @user18921 forget about separation of variables for a second here... just look at your DEqn. If we set $v(x)=0$ for all such $x$ in an interval where the DEqn is defined (see user37328 answer) then it is clear that is a solution. So, the zero solution is allowed. It may be extraneous to the separation technique, but that is pretty typical. – James S. Cook Aug 30 '13 at 09:56
  • @user18921 also, the existence and uniqueness theorem applies in this context $F(x,v)$ certainly has $\partial_v F = f(x)$ in your general part B. Therefore, the existence of the $v=0$ solution indicates it is in fact the only such solution. So, yes, if the solution is zero it stays zero at least to the next point where $f(x)$ is not defined. – James S. Cook Aug 30 '13 at 10:00

2 Answers2

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To solve such kind of equation you need to solve it on the intervals where $x(1-x^2)\neq 0$ which are $(-\infty,-1)$, $(-1,0)$, $(0,1)$ and $(1,+\infty)$. If you need a solution on $\mathbb{R}$, you have to try to find a $\mathcal{C}^1$ function that is a solution on each sub-intervals ie the form of a solution $f$ on $\mathbb{R}$ will be

$f(x)=\begin{cases} f_1(x) &\text{on } (-\infty,-1) \\ f_2(x) &\text{on } (-1,0) \\ f_3(x) &\text{on } (0,1) \\ f_4(x) &\text{on } (1,+\infty) \\ \end{cases}$

where $f_1$,$f_2$,$f_3$ and $f_4$ are solution on $(-\infty,-1)$, $(-1,0)$, $(0,1)$ and $(1,+\infty)$ respectively. You don't to consider if $v=0$ or not. You have a solution directly by integration : $v(x)=\exp(\int_{x_0}^x \frac{1+t^2}{t(1-t^2)}dt)$.

user37238
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  • I think your eye may have skipped a factor of $v$. The equation is of the form $v' = \mbox{fraction} \times v$. – goblin GONE Aug 30 '13 at 11:22
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If $v\ne0$, the equation is equivalent to $$ \frac{\mathrm{d}}{\mathrm{dx}}\log(v)=\frac1x+\frac1{1-x}-\frac1{1+x} $$ Integrate to get $$ \log(v)=\log\left(\frac{Cx}{1-x^2}\right) $$ so that $$ v=\frac{Cx}{1-x^2} $$ In this particular case, $v(0)=0$, yet $v$ is not necessarily $0$ everywhere.

Since $f(x)=\frac{v'}{v}=\frac{\mathrm{d}}{\mathrm{dx}}\log(v)$, if $v(x_0)=0$ at some point and $v$ is not identically $0$ in a neighborhood of $x_0$, $\log(v)$, and therefore, $\frac{\mathrm{d}}{\mathrm{dx}}\log(v)$, must blow up at $x_0$.

Thus, if $f$ is bounded in a neighborhood of $x_0$ and $v(x_0)=0$, then $v$ is identically $0$ in that neighborhood.

robjohn
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  • Hey how did you get the first line? – goblin GONE Aug 30 '13 at 11:57
  • Wait I think I get it.... – goblin GONE Aug 30 '13 at 12:06
  • Yeah okay, I get it. – goblin GONE Aug 30 '13 at 12:24
  • Okay I agree with everything up to but not including the following sentence: "In this particular case, v(0)=0, yet v is not 0 everywhere." The problem is, even just to do all that working, we had to assume $x \neq 0$. So you can't just suddenly plug in $x=0$. – goblin GONE Aug 30 '13 at 12:47
  • @user18921 While certainly the assumption that $x=0$ is required for the first part, by continuity ($v$ is continuous as $v$ is differentiable) you know that $v(0)=0$, so it doesn't end up being too much of a problem. – Andrew D Aug 30 '13 at 13:01
  • $f(0)$ does not exist, but the solution of the equation for $x\in(0,1)$ is $\frac{Cx}{1-x^2}$ and that has a limit of $0$ at $x=0$, extending by continuity, $v(0)=0$. The fact that $v(0)$ can be $0$ without $v$ being identically $0$ is allowed since $f$ is not bounded in a neighborhood of $0$. – robjohn Aug 30 '13 at 13:06
  • I'm just not sure why we're randomly extending the solution, when after all that undermines the property that either $v$ is everywhere nonzero, or else $v$ is everywhere vanishing. – goblin GONE Aug 30 '13 at 13:58
  • @user18921: We are continuously extending the function, not randomly extending it. In any case, this does not undermine the property, it shows that where $f$ is unbounded, $v$ can be $0$ without being identically $0$, and that is the result that I have described. To reiterate: If $v(x_0)=0$ and $f$ is bounded in a neighborhood of $x_0$, then $v$ is identically $0$ in the neighborhood of $x_0$. ($f$ is not bounded in a neighborhood of $0$). – robjohn Aug 30 '13 at 14:02
  • I think that in the presence of the assumption that $v : I \rightarrow \mathbb{R}$ solves the problem, we have that if $v(x_0)$ is well-defined, then $f : I \rightarrow \mathbb{R}$ cannot blow up at $x_0.$ That's because $f$ is continuous, and therefore can only blow up at points $x_0 \notin I$. Sure, we might be able to continuously extend $v$ to obtain a new function $v^$ that is defined at $x_0$, but then $v^$ will never be a solution the problem, since its domain is no longer $I$. So, I think that the possibility that $f$ is unbounded around some $x_0 \in I$ is a non-concern. Agree? – goblin GONE Aug 30 '13 at 15:08
  • @user18921: Everything I have said above says that if $v(x)=0$ for $x\in(0,1)$, then $v$ is identcally $0$, whether we restrict $v$ to $(0,1)$ or extend it to $(-1,1)$. This is because $f$ is bounded in a neighborhood of each $x\in(0,1)$. You were asking when a function $v$ satisfying $v'=f,v$ can be $0$ at a point and not be identically $0$. I was simply giving an example by continuously extending $v$ to $0$. – robjohn Aug 30 '13 at 15:24
  • Okay, I get the bits about how if $v : (0,1) \rightarrow \mathbb{R}$ solves the problem, then if $v(x) = 0$ for some $x \in (0,1),$ then we may conclude that $v(x) = 0$ for all $x \in (0,1)$. And, I thought you were talking about adding a single point to the domain of $v$, obtaining a new function $v^* : [0,1) \rightarrow \mathbb{R}.$ Which seems like a pretty silly thing to do. However, now I see that you're talking about a solution $v^\diamond : (-1,1) \rightarrow \mathbb{R}$ to the DE. I'm intrigued. How do we define the notion of a solution on $(-1,1)$, given that $f$ is undefined at $0$? – goblin GONE Aug 30 '13 at 16:26
  • @user18921: $u^\diamond$ satisfies the DE everywhere but $0$ and is continuous and differentiable everywhere. In fact $1/f$ is defined everywhere in $(-1,1)$ (if we set $1/f(0)=0$), and $u^\diamond$ satisfies ${u^\diamond}^\prime/f=u^\diamond$. – robjohn Aug 30 '13 at 17:19