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I have a system of two linear differential equations with non-constant coefficients (which involve hypergeometric functions),

$$ u_1'(y) = \alpha_1(y) u_1(y) + \beta_1(y) u_2(y) \\ u_2'(y) = \alpha_2(y) u_1(y) + \beta_2(y) u_2(y) $$

that I want to integrate numerically in the interval $y\in[0,1]$. The coefficients are singular at the boundary of the interval (i.e., at $y=0$ and $y=1$).

As we have a linear system of 2 equations, the solutions conform a 2-dimensional vector space of functions, and a general solution can be written as

$$ u(y) = C_\text{I} u^\text{I}(y)+C_\text{II}u^\text{II}(y) $$

where $u=(u_1,u_2)$. By construction, every particular solution (that is, every choice of $C_\text{I}$ and $C_\text{II}$) vanishes when $y\to1$, but there's only a 1-dimensional subspace of solutions that gives finite value for $u_1$ and $u_2$ when $y\to0$.

I have the asymptotic expansion of the solution around $y=0$ and $y=1$. These series give us $u_1$ and $u_2$ around $y=0$ depending on some constants $C_\text{I}^{(0)}$ and $C_\text{II}^{(0)}$, and around $y=1$ with another set of constants $C_\text{I}^{(1)}$ and $C_\text{II}^{(1)}$.

I know a relation between $C_\text{I}^{(0)}$ and $C_\text{II}^{(0)}$, $G(C_\text{I}^{(0)}$,$C_\text{II}^{(0)})=0$, such that the solution doesn't explode on $y=0$, but I want to know the equivalent choice of $C_\text{I}^{(1)}$ and $C_\text{II}^{(1)}$ for the expansion around $y=1$, that allows me to integrate the system from $y=1$, ensuring that the solution will be finte at $y=0$.

Any insight on how could I achieve this in Mathematica? I have been looking about the Shooting Method, so I could shot solutions from $y=1$ and pick the one (up to a multiplicative constant factor) that is finite at $y=0$, but I am not sure on how could I implement this.

dpravos
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  • Perhaps my brain is stuck in the mud this morning: If a particular solution is a linear combination of two given solutions at $y=0$, why isn't it the same linear combination at $y=1$? In other words, why doesn't $G=0$ apply at $y=1$ and indeed throughout the interval? – Michael E2 Dec 07 '16 at 11:49
  • Somewhat related: http://mathematica.stackexchange.com/questions/91854/nonlinear-differential-equation-numerical-solution/97733#97733 – Michael E2 Dec 07 '16 at 13:14
  • The particular solutions at $y=0$ and $y=1$ are not the same. – dpravos Dec 07 '16 at 14:54
  • Can you show us a specific example, in the form of Mathematica code? I'm afraid currently the question is a bit too abstract. – xzczd Jan 06 '17 at 06:44

1 Answers1

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An alternative may to use Eliminate to convert your system of first-order equations to a single second order equation (see e.g., http://www.mathematica-journal.com/issue/v7i3/tricks/contents/html/Links/index_lnk_4.html). You will still need to use the asymptotic expansion of the solution around $y=0$ and $y=1$ and likely apply the shooting method.

TheDoctor
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