Need analytical solution to this ODE

Question

I have two ordinary differential equations that I'm trying to solve analytically, if a solution exist. I'm not interested in numerical solutions. Here they are: \begin{align} \frac{\ddot{a}}{a} - \frac{\dot{a}^2}{a^2} = -\, a, \tag{A} \\[2ex] \frac{\ddot{b}}{b} - 2\frac{\dot{b}^2}{b^2} = -\, b^3. \tag{B} \end{align} The dot represents the derivative relative to a real variable $\theta \ge 0$, and $a$, $b$ are real and positive unknown functions satisfying (A) and (B): \begin{align}\tag{1} &a(\theta) \ge 0, &&b(\theta) \ge 0. \end{align} The initial conditions are arbitrary positive constants: \begin{align}\tag{2} &a(1) =1, &&b(1) = 1, &&\dot{a}(1) \ge 0, &&\dot{b}(1) \ge 0. \end{align} I'm not even sure that the equations (A) and (B) are individually consistent. I need to find an analytical solution and I currently don't see how to find one. The negative sign on the right hand side of (A) and (B) give me some troubles. Any help would be greatly appreciated. Anyone can find a solution to (A) or (B)?

Is there a way to show that these equations admit a solution?

Answer 1

$$\frac{\ddot{a}}{a} - \frac{\dot{a}^2}{a^2} = -\, a, \tag{A} \\[2ex]$$ $$\left (\frac{a'}{a} \right)'=-a$$ $$\frac {a'}{a}\left (\frac{a'}{a} \right)'=-a'$$ $$\frac {a'}{a}d\left (\frac{a'}{a} \right)=-da$$ Integrate to reduce the order of the DE. $$\left(\frac {a'}{a} \right)^2=-2a+C_1$$ This may not be easily integrable. (elliptic integral)

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$$\frac{\ddot{b}}{b} - 2\frac{\dot{b}^2}{b^2} = -\, b^3. \tag{B}$$ $$b^2(b'')-(b^2)'b'=-b^6$$ $$\left ( \frac {b'}{b^2} \right)'=-b^2$$ $$\frac {b'}{b^2}\left ( \frac {b'}{b^2} \right)'=-b'$$ You can reduce the oder but then it won't be integrable again easily.

Answer 2

Hint: As recommended in the second point here, you can substitute $v = a'$ to get a first order ordinary differential equation for $v$, which you can then solve for $a$, e.g. $$ a'' = \frac{(a')^2}{a} - a^2 $$ yields $$ \tag{1} v(a) v'(a) = \frac{v(a)^2}{a} - a^2, $$ which you can solve more easily and then get back $a$ via solving $a' = v(a)$.

Edit: How can we solve $(1)$?

The equation (1) is similar to a Bernoulli equation with $n = - 1$. Choosing $w = v^2$ transforms $(1)$ into $$ \frac{1}{2} w'(a) + \left(- \frac{1}{a}\right) w(a) = - a^2, $$ which is in standard form$^1$.

$^1$ The solution of $u'(t) + a(t) u(t) = f(t)$ is $$\exp\left(- \int_{t_0}^{t} a(s) \textrm{d}s \right) \cdot C + \int_{t_0}^{t} \exp\left(- \int_{s}^{t} a(\tau) \textrm{d}\tau \right) \cdot f(s) \textrm{d}s.$$

Answer 3

I've found the solution I was looking for equation (A): $$ a(\theta) = \frac{a_0}{\cosh^2 (\sqrt{\frac{a_0}{2}} \, \theta)}, $$ where $a_0 \equiv a(0)$.

For equation (B), well that's another story...