# Solving a system of separable ODE's

Mathematics Asked by costard on December 4, 2020

I am attempting to solve the system of first-order ODE’s:

$$frac{dy_1(x)}{dx} = g(y_1)frac{y_2(x)}{int_{a}^b y_2(z) dz}$$
$$frac{dy_2(x)}{dx} = frac{y_2(x)}{f(y_1(x),x)} frac{partial f(y_1(x),x)}{partial x}$$

with boundary conditions $$y_1(a) = c$$, $$y_1(b) = d$$ for constants $$a,b,c,d$$.

My problem is conceptual, and very likely silly: I’m not sure whether I can break apart this problem apart in order to solve it. I would like to exploit separability to obtain a general solution for the second equation:

$$y_2(x) = e^N f(y_1(x),x)$$

with N a constant, and use this to reduce the first equation to

$$frac{dy_1(x)}{dx} = g(y_1)frac{f(y_1(x),x)}{int_{a}^b f(y_1(z),z) dz}$$

which I would then solve numerically, presumably by treating the integral as a constant to be determined. I’m not seeing the problem with this approach, but on the other hand it strikes me as "too good to be true." My questions:

1. Am I mistaken in thinking that $$y_2(x) = e^N f(y_1(x),x)$$ is in fact the general solution for $$y_2$$, irrespective of the dependence of $$y_1$$ on $$y_2$$?
2. When numerically solving a differential equation (say IVP) of the form $$y_1′(x) = hleft(y_1,x,int_{a}^b f(y_1(z),z) dzright)$$ with $$a,b$$ constant, is it problematic to treat the integral as a constant $$A$$, use an initial guess for $$A$$ + e.g Euler’s method to solve for $$y_1$$, and then iterating until the guess and actual value of $$A$$ converge?

Any input would be much appreciated.

Your system does not define $$y_2(x)$$ uniquely. If $$(y_1(x), y_2(x))$$ is a solution then also $$(y_1(x), Cy_2(x))$$ where $$C neq 0$$ is a constant also is a solution.

Thus you may impose an extra condition on $$y_2(x)$$, the simplest to me is $$int_a^b y_2(x) dx = 1.$$ After that the system reduces to $$y_1'(x) = g(y_1(x)) y_2(x)\ y_2'(x) = y_2(x) F(x, y_1(x))$$ with conditions $$y_1(a) = c, y_1(b) = d, int_a^b y_2(x) dx = 1$$. Here I've denoted $$F(x, y) = frac{f_x(y, x)}{f(y, x)}$$.

Having a system of two first-order ODE's with three conditions makes me think that the problem does not have a solution in general (but may have for some particular values of $$c$$ or $$d$$, like an eigenproblem).

The general solution to the second equation is $$y_2(x) = C_1 expleft(int_a^x F(s, y_1(s)) dsright)$$ with $$C_1$$ determined by $$int_a^b y_2(x) dx = 1$$ condition. Practically it is easier to put $$y_2(a) = C_1 = 1$$ and normalize $$y_2(x)$$ after integration.

Summarizing, the following algorithm may solve the problem:

1. Take some initial $$y_1(x)$$, e.g. linear approximation $$y_1(x) = c + frac{x-a}{b-a} (d - c)$$.
2. Having $$y_1(x)$$ compute the $${tilde y}_2(x)$$ from the following Cauchy problem $$begin{cases} tilde y_2'(x) = tilde y_2(x) F(x, y_1(x)), &quad x in [a, b]\ I'(x) = tilde y_2(x), &quad x in [a, b]\ tilde y_2(a) = 1\ I(a) = 0 end{cases}$$ Here $$I(x) = int_a^x y_2(s) ds$$, so $$I(b) = int_a^b y_2(x) dx$$.
3. Compute $$y_2(x)$$ by $$y_2(x) = frac{tilde y_2(x)}{I(b)}.$$
4. Plug obtained $$y_2(x)$$ into the first equation and solve the new Cauchy problem $$begin{cases} tilde y_1'(x) = g(tilde y_1(x)) y_2(x)\ tilde y_1(a) = c\ end{cases}$$
5. Now, for sure, $$tilde y_1(b) neq d$$, so we need to correct $$y_1(x)$$ and start from the item 2. And this is quite tricky. As I said, the problem almost certainly won't have a solution. If it was a boundary problem with three conditions and the functions we might have use the residual $$delta = d - tilde y_1(b)$$ to correct some free parameter (see shooting method), but there is no such parameter in this problem, at least as it was stated. You may try combining new approximation $$y_1(x)$$ as $$y_1(x) := (1-omega) y_1(x) + omega left[tilde y_1(x) + frac{x-a}{b-a}deltaright]$$ and try to deduce how $$delta$$ behaves in iterations depending on different $$omega$$ values. The extra term in brackets is needed to correct the right boundary conditions in new $$y_1(x)$$ before restarting from item 2.

Answered by uranix on December 4, 2020