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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.

One Answer

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

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