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So In my research I stumbled upon a difficult ODE, It goes like this $$ y''(x)-[(\operatorname{Heaviside}(ax)+b]y(x)=0, $$ (a,b are the respective constants) I tried approximating Heaviside with analytical terms, which gave me these ODE's

$$ y''(x)-[b/2*(1+2/π * \arctan⁡(x/ϵ) )+a]y(x)=0, $$ or

$$ y''(x)-[a/(1+e^x\epsilon]+b)y(x)=0$$

The epsilon should multiply the power as well, it doesn't work for some reason. The mighty wolfram gives me answers that I barely understand, How would you suggest I go about it? Should I try a different approach? is it even solvable? Laplace transform also didn't help me. Thanks a lot!

jonan
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E.T.
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    Please use MathJax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Yuriy S Aug 27 '19 at 21:00
  • Use `\arctan(x)` to get $\arctan(x)$ and `\operatorname{Heaviside}(x)` to get $\operatorname{Heaviside}(x)$. – parsiad Aug 27 '19 at 21:17
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    If $a>0$, isn't $\operatorname{Heaviside}(ax)$ the same as $\operatorname{Heaviside}(x)$? – John Barber Aug 27 '19 at 21:25
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    JohnBarber is correct. OP: are you considering this as a **boundary** or an **initial** value problem? The latter is very easy and can be solved by integrating from $x_0$ (the initial point) to $0$ and then from $0$ to $x>0$ separately. – parsiad Aug 27 '19 at 21:29
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    I think this can be solved by a technique very similar to that used to solve the [Schrodinger equation for a step potential](https://en.wikipedia.org/wiki/Solution_of_Schr%C3%B6dinger_equation_for_a_step_potential). – John Barber Aug 27 '19 at 21:31

1 Answers1

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Assumptions. $a$ is positive and initial conditions $y(x_{0})$ and $y^{\prime}(x_{0})$ are specified at $x_{0}<0$.

As John Barber points out, $H(ax)=H(x)$ in this case. Therefore, the ODE is equivalent to $$ y^{\prime\prime}(x)=\begin{cases} \left(b+0\right)y(x) & \text{if }x<0\\ \left(b+1\right)y(x) & \text{if }x>0. \end{cases} $$ It follows that $$ y(x)=c_{1}e^{x\sqrt{b}}+c_{2}e^{-x\sqrt{b}}\qquad\text{for }x_{0}<x\leq0 $$ where $c_{1}$ and $c_{2}$ are obtained by solving the linear system $$ \begin{pmatrix}\phantom{\sqrt{b}}e^{x_{0}\sqrt{b}} & \phantom{-\sqrt{b}}e^{-x_{0}\sqrt{b}}\\ \sqrt{b}e^{x_{0}\sqrt{b}} & -\sqrt{b}e^{-x_{0}\sqrt{b}} \end{pmatrix}\begin{pmatrix}c_{1}\\ c_{2} \end{pmatrix}=\begin{pmatrix}y(x_{0})\\ y^{\prime}(x_{0}) \end{pmatrix}. $$ Similarly, $$ y(x)=C_{1}e^{x\sqrt{b+1}}+C_{2}e^{-x\sqrt{b+1}}\qquad\text{for }x>0 $$ where $C_{1}$ and $C_{2}$ are obtained by solving the linear system $$ \begin{pmatrix}1 & 1\\ \sqrt{b+1} & -\sqrt{b+1} \end{pmatrix}\begin{pmatrix}C_{1}\\ C_{2} \end{pmatrix}=\begin{pmatrix}y(0)\\ y^{\prime}(0) \end{pmatrix}. $$

parsiad
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