Solve equation by shooting method

Question

I've got homework to numerical methods to solve this equation using shooting method

$ y'' - (1-e^{-x})y = 0, \quad y(0) = 1, \quad y(x\to+\infty) = 0 $

Hints I've got:

1. Easily you can find asymptotic solution, using this solution you can use functional substition, that will reduce functional values of solution

2. You can use substition of independent variable to make integration domain finite interval.

3. Hints above also say: Graph of function is unbounded in $x$ and $y$ axis. Using suitable substitution you can make graph bounded in $\mathbb{R}^2$

Can anyone help me with transformations? Thanks!

Answer 1

For part 1, they want you to note that if $y$ is unbounded, it grows asymptotically as $e^x$. Therefore, $e^{-x}y(x)$ will be bounded. Letting $z(x) = e^{-x}y(x)$, we find that $$ z''(x) +2 z'(x) +e^{-x}z(x)= 0\;\;\;;\;\;\;z(0) = 1\;\;\;;\;\;\;z(\infty) = 0. $$ For part 2, the infinite interval on $x$ needs to be reduced to a finite one. The equation itself suggests something like $u = e^{-x}$. Let's use $u = 1-e^{-x}$ so that the transform is increasing. Letting $z(x) = w(1-e^{-x})$ gives $$ (1-u) w''(u) + w'(u) +w(u) = 0\;\;\;;\;\;\; w(0) = 1\;\;\;;\;\;\;w(1) = 0. $$ Now you can try different initial conditions for $w'(0)$ and see which ones get $w(1)$ close to $0$. Of course, you can't actually do calculations at $u=1$ because of the singular point, but you can force $w(1-\epsilon) = 0$ for some small $\epsilon$. You can then recover $y'(0)$ using $y'(0) = w'(0)+w(0)$.

Answer 2

You can not integrate over infinite intervals. So you have to set an upper limit $b$ and establish some boundary condition there.

The limit point is a saddle point. There are exactly 2 paths that lead to it in the phase portrait, and any deviation from these will at some point exponentially diverge.

For $x>b$ the factor $e^{-x}$ becomes very small, for a suitably large $b$. The reduced equation $y''-y=0$ has known solutions, aka "far-field solutions". The one that remains bounded satisfies $y'+y=0$. Thus you can take $y'(b)+y(b)=0$ as upper boundary condition.

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To get higher accuracy, or the same accuracy with a smaller $b$, compute the next perturbation term in $y=c_0e^{-x}+c_1e^{-2x}+...$ to get $$ -3c_1e^{-2x}+...=c_0e^{-2x}+c_1e^{-3x}+... $$ Thus an improved boundary condition is $$(3-e^{-b})y'(b)+(3+2e^{-b})y(b)=0.$$