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I have an ODE such as the one below,

$$y"=a^{2}y$$

The analytical solution is clearly

$$y=C_{1}e^{ax} + C_{2}e^{-ax}$$

To apply boundary conditions $y(x=x_{1})=A$ and $y(x=x_{2})=B$, I recast into a matrix:

$\left( \begin{array}{ccc}e^{ax_{1}} & e^{-ax_{1}} \\ e^{ax_{2}} & e^{-ax_{2}} \end{array} \right)$$\left( \begin{array}{ccc}C_{1}\\ C_{2}\end{array} \right)=\left( \begin{array}{ccc}A\\ B\end{array} \right)$

My issue is that when $x_{1}$ and $x_{2}$ are far apart from one another (say $a=1,x_{1}=-10^{4},x_{2}=10^{4}$ or when $a$ is large (say $a=10^{4},x_{1}=-1,x_{2}=1$), the matrix inversion cannot be solved within machine precision. The issue is the exponential, which varies so widely. I was wondering what steps I could take to find the constants to satisfy my ODE solution.

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Perhaps it might help to write the solution as

$$ y(x) = \frac{\sinh(a (x-x_2))}{\sinh(a (x_1 - x_2))} A - \frac{\sinh(a (x - x_1))}{\sinh(a (x_1 - x_2))} B$$

There will still be problems with overflow/underflow in numerical computation, though. Of course, when $t$ is large and positive, $\sinh(t) \sim e^t/2$, while when $t$ is large and negative, $\sinh(t) \sim -e^{-t}/2$. These can be used to get useful approximations in various regions.

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  • $\begingroup$ This works, but I was wondering if there was a more robust method such as transforming my matrix into one with a better conditioning number. For example, dividing both sides of the equation $y(x=x_{1})=A$ would allow me to change the order of the first row. Similarly scaling the constant $C_{1}$ could allow me to scale the first column of the matrix. Is there a systematic algorithm I can use to implement such steps of rescaling? $\endgroup$ – Geoffrey Xiao Jul 13 '17 at 21:36
  • $\begingroup$ This is the exact solution. I don't see how you can do better. $\endgroup$ – Robert Israel Jul 13 '17 at 23:47
  • $\begingroup$ Search the topic of ill-conditioned matrices. There are iterative methods for solving $Ax=b$ when $\lambda_{max} / \lambda_{min}$ of the matrix $A$ is large. $\endgroup$ – player100 Jul 15 '17 at 1:31

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