I have a very simple linear first order ODE.

$$v(x) = c x + A - B x(1-x) v'(x)$$

$c, A ,B \in(0,1)$. The domain is $(\underline{x}, 1)$. where $\underline x > 0$.

I am guessing that for any initial condition $v(\underline x)$ there will be a unique solution. In particular, suppose I give an initial condition $v(\underline x)$ such that $0 < v(\underline x) < c \underline x + A$. I would like to argue that the solution must be convex. What are the sort of methods that I can use when an explicit closed form solution isn't available?



If I did not make any mistake, $v(x)$ need not be convex.

We find that $$B v'(x) = \frac{c x + A - v(x)}{x (1 - x)} ,$$ and therefore $$B v''(x) = \frac{(c - v'(x)) x (1 - x) - (c x + A - v(x)) (1 - 2 x)}{x^2 (1 - x)^2} .$$ Plugging in the expression for $v(x)$, we get $$B v''(x) = \frac{(c - v'(x)) x (1 - x) + B x (1 - x) v'(x) (1 - 2 x)}{x^2 (1 - x)^2} .$$ which leads to $$B v''(x) = \frac{c - (1 + B (1 - 2 x)) v'(x)}{x (1 - x)} .$$ This is positive a long as $(1 + B(1 - 2 x)) v'(x) < c$. This need not be satisfied at the initial condition if $\underline{x}$ is small enough: then $1 + B(1 - 2 \underline{x}) > 0$ and $v'(\underline{x})$ can be arbitrarily close to $$ \frac{c \underline{x} + A}{B \underline{x} (1 - \underline{x})} , $$ which can be arbitrarily large as $\underline{x} \to 0^+$.

  • $\begingroup$ My answer is slightly different but, I think, qualitatively that shouldn't change the main point of your argument. I get $ Bv'' = \frac{c- (1+B(1-2x) ) v'}{x(1-x)}$. This too suggests that close $x \approx 0$, $v$ should be concave. However, this also helps me in proving that at least near $x = 1$, for the same reasons $v$ must be convex, right? $\endgroup$ – avk255 Mar 17 '18 at 12:47
  • $\begingroup$ @Aditya: Thanks, I corrected the error. Regarding convexity near $x = 1$: a similar method may work, but I do not think this follows from the answer right away (one needs an upper bound for $v'$). $\endgroup$ – Mateusz Kwaśnicki Mar 17 '18 at 21:27
  • $\begingroup$ Thanks. Yeah, proving convexity is an issue even close to 1. Do you know if there are any uniqueness results though? $\endgroup$ – avk255 Mar 18 '18 at 8:39
  • $\begingroup$ @Aditya: Uniqueness (away from $x = 0$ and $x = 1$) follows from Picard's existence theorem. $\endgroup$ – Mateusz Kwaśnicki Mar 18 '18 at 9:21

Small clarification: According to Mathematica , your equation has solution $$ \frac{x^{\frac{1}{B}} e^{\frac{2 \tanh ^{-1}(1-2 x)}{B}} }{B+1} \left(x (A+c) \, _2F_1\!\!\left(1+\frac{1}{B},1+\frac{1}{B};2+\frac{1}{B};x\right)+ A (B+1) \, _2F_1\!\!\left(\frac{1}{B},\frac{1}{B};1+\frac{1}{B};x\right)\right)+c_1 e^{\frac{2 \tanh ^{-1}(1-2 x)}{B}}. $$ To me, this is "closed form". Of course, one has to be a consummate expert in hypergeometric functions to deduce useful information from that.

(Moderator: I am perfectly aware that mine is a (not very important) comment rather than an answer, but, as I am new on MO, my reputation does not allow me to post comments.)

  • $\begingroup$ Thanks. Yes, Mathematica does give this as an answer. But I haven't a slightest clue about how to work with hypergeometrics. The above answer suggests that close to $x \approx 1$, if $v'$ is large enough then the function may not be convex. The intuition about the problem that comes from elsewhere and may be wrong tells me that $v'$ would be large close to $1$. $\endgroup$ – avk255 Mar 18 '18 at 8:46
  • $\begingroup$ You write: "I haven't a slightest clue about how to work with hypergeometrics." Me either. $\endgroup$ – user539887 Mar 18 '18 at 9:04

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