Simple system of nonlinear ordinary differential equations

Question

I'm trying to solve a system of ODEs of the form:

$$\frac{d^2a}{dt^2} = \frac{-1}{(a-b)^2}$$

$$\frac{d^2b}{dt^2} = \frac{+1}{(a-b)^2}$$

and with the following boundary conditions:

$$a'(0) = 0$$

$$b'(0) = 0$$

$$a(0) = a_0$$

$$b(0) = b_0$$

where a₀ and b₀ are just arbitrary real constants representing the initial values of a(t) and b(t), with a₀ > b₀.

I'm working on a problem in partial differential equations (with some irrelevant coefficients, that I've taken out of the problem here) that relies on being able to solve a system of ODEs of this type. I remember doing this problem in my introductory ODE class a couple of years ago, but I don't remember what the solution was or how to solve it.

If this helps, I've figured out that the values of a(t) and b(t) will converge at some time in the future (t > 0) such that $$a(t) = b(t) = \frac{a_0+b_0}{2}$$ although I don't know the earliest such value of t.

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Edit: I may be mistaken, but it appears that this system of ODEs is essentially equivalent through a simplifying substutition of variables to a single ODE:

$$\frac{d^2c}{dt^2} = \frac{1}{(c-d_0)^2}$$

$$c'(0) = 0$$

$$c_0 < d_0$$

Thanks to Hans Ludmark for the suggestion.

Edit 3: Moved solution to a separate answer.

Answer 1

Note that $a''(t)+b''(t)=0$ for every $t$ hence $a'(t)+b'(t)=a'(0)+b'(0)=0$ for every $t$ hence $a(t)+b(t)=a_0+b_0$ for every $t$. Plugging $a(t)=a_0+b_0-b(t)$ in the differential equation for $b''(t)$ yields $$2b''(t)b'(t)=\frac{2b'(t)}{(a_0+b_0-2b(t))^2}=\left(\frac1{a_0+b_0-2b(t)}\right)',$$ hence, using $b'(0)=0$, $$b'(t)^2=\frac1{a_0+b_0-2b(t)}-\frac1{a_0-b_0}=\frac{2(b(t)-b_0)}{a_0+b_0-2b(t)}.$$ Since $b'(0)=0$ and $b''(t)\gt0$ for every $t$, $b'(t)\geqslant0$ and $b(t)\geqslant b_0$ for every $t$, hence $$b'(t)=+\sqrt{\frac{2(b(t)-b_0)}{a_0+b_0-2b(t)}}.$$ Thus, $b'$ is increasing on its interval of definition $[0,T)$ until $b'(t)\to+\infty$ and $b(t)\to\frac12(a_0+b_0)$ when $t\to T$, where $$T=\int_{b_0}^{(a_0+b_0)/2}\sqrt{\frac{a_0+b_0-2x}{2(x-b_0)}}\mathrm dx=\frac{a_0-b_0}2\int_0^1\sqrt{\frac{1-s}s}\mathrm ds=\frac\pi4(a_0-b_0).$$ Note that $b(t)$ solves the implicit equation, valid for every $t$ in $[0,T]$, $$t=\frac{a_0-b_0}2\int_0^{2(b(t)-b_0)/(a_0-b_0)}\sqrt{\frac{1-s}s}\mathrm ds.$$ Finally, all these functions are different scalings of one unique solution, to wit, consider the change of variable $$b(t)=b_0+(a_0-b_0)\beta\left(\frac{2t}{a_0-b_0}\right),$$ then $\beta$ does not depend on $(a_0,b_0)$ and solves the differential equation $$\beta(0)=\beta'(0)=0,\quad\beta'(\tau)=\sqrt{\frac{\beta(\tau)}{1-\beta(\tau)}},\quad0\leqslant\tau\lt\frac\pi2,$$ whose implicit solution is $$\tau=\sqrt{\beta(\tau)(1-\beta(\tau))}+\arcsin\sqrt{\beta(\tau)}.$$ A plot of the function $\beta$:

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Answer 2

Thanks, Hans and Did! I've got a solution to the problem.

Since d^2a/dt^2 + d^2b/dt^2 = 0 and da/dt = db/dt = 0 at t=0, we know that for all t, a + b = a_0 + b_0. Therefore, after rearrangement, we can combine the following two equations:

$$\frac{d^2b}{dt^2} = \frac{+1}{(a-b)^2}$$

$$a = a_0 + b_0 - b$$

to get

$$\frac{d^2b}{dt^2} = \frac{1}{(a_0+b_0-2b)^2}$$

and since a₀ and b₀ are just arbitrary constants, define c = a₀ + b₀ to simplify the equation. The system has then been reduced to a single nonlinear ODE. WolframAlpha can solve this; the answer is here (stated implicitly). Then, having solved for b(t), a(t) can be obtained by trivial substitutions.