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a) How to solve, or at least to prove the existence of a solution to differential equation for given initial condition $y(s)=y_0>0$ and $y'(s)=y_1$, $s<0$, $$y''+(2-n)\coth(t) y'=\frac{(n-1)\sinh(2y)}{2}, t<0.$$ Here $n$ is an integer $>2$.

b) Can the previous equation have two different solution (with different initial conditions) in $(-2,-1)$, such that one is bounded and the second is not bounded?

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@Shahrooz And which is the solution? Can you spell it? Thanks. –  djoke Dec 29 '12 at 20:59
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@Shahrooz Probably you have "ask matlab" to solve slightly different equation with the right hand side containing sinh(2t)! –  djoke Dec 30 '12 at 6:52
    
Sorry, you are right. –  Shahrooz Dec 30 '12 at 16:19

2 Answers 2

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$\coth(t)$ has a singularity at $t=0$, so the hypotheses of the existence and uniqueness theorems are not satisfied there. In fact if $\lim_{t \to 0} y(t) = y_0$ and $\lim_{t \to 0} y'(t) = y_1$ exist, $y''(t) \sim (n-2) y_1 t^{-1}$ as $t \to 0$. If $y_1 \ne 0$, this is impossible, as $t^{-1}$ is not integrable at $0$. So there are no solutions with such an initial condition.

EDIT: For some initial conditions at $t=-2$, the solution will "blow up" before $t=-1$. It suffices to prove, e.g., that on any interval $[-2,a)$ where the solution exists we have $\dfrac{dy}{dt} \ge y^2$ for $t \ge -2$ with $y(-2) > 1$, as then $$a - (-2) \le \int_{y(-2)}^{y(a)} \dfrac{dy}{y^2} < \int_{1}^\infty \dfrac{dy}{y^2} = 1$$

Now note that if $f = \dfrac{dy}{dt} - y^2$ we have $$ \dfrac{df}{dt} = \dfrac{d^2y}{dt^2} - 2 y \dfrac{dy}{dt} = ((n-2) \coth(t) - 2 y)(y^2 + f) + \dfrac{n-1}{2} \sinh(2y) $$ Given $n$, there is some $Y$ such that for all $f \in [0,1]$, $t \in [-2,-1]$ and $y \ge Y$, the right side is positive. So if $y(-2) > Y$ and $y'(-2) > y(-2)^2$, we will have $y' > y^2$ for $t \in [-2,a]$.

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I updated the formulation. The question is can we find a global solution for example in (-2,-1). –  djoke Jan 3 '13 at 18:17
    
Why you can assume that $f\in[0,1]$. You should prove that $f>0$? –  djoke Jan 4 '13 at 16:38
    
You start with $f > 0$ at $t=-2$. $f$ is continuous. In order for $f$ to get to $0$, it would have to pass through the interval $(0, f(-2))$, which it can't because $df/dt > 0$ whenever $0 \le f \le 1$. –  Robert Israel Jan 4 '13 at 19:17
    
Thanks! One more question. If the initial speed $y'(-2)=0$, can you have a similar conclusion? –  djoke Jan 4 '13 at 21:16
    
Presumably. E.g. for $n=3$, numerical methods indicate the solution blows up before $t=-1$ for $y(-2) \ge .902$ approximately. –  Robert Israel Jan 6 '13 at 7:10

The result you want is true because of existence and Uniqueness for second order non-linear ODE, for example see Boyce and Diprima.

Writing $y^{\prime\prime}=f(t,y,y^\prime)$ and verifying that $f$, $f_y$, and $f_{y^\prime}$ are continuous you can guarantee existence and uniqueness in a small interval of the initial condition. Here you need to see that the functions you have in the problem ($coth$ and $sinh$) are smooth.

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Thanks, but I need to prove more than the existence. –  djoke Dec 30 '12 at 13:17

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