esbagoar7kh

2022-04-23

Piece-wise second order differential equation

Suppose I have some function f(x) which is equal to kx when x is negative and -kx when x is positive, and I have the following differential equation :

$\frac{{d}^{2}x}{{dt}^{2}}=-k\left|x\right|=f\left(x\right)$

Suppose I have some function f(x) which is equal to kx when x is negative and -kx when x is positive, and I have the following differential equation :

belamontern9i

Beginner2022-04-24Added 19 answers

Step 1

For $k=0,x\left(t\right)={c}_{1}t+{c}_{2}.$

For $k\ne 0$, start by multiplying both sides of your differential equation by ${x}^{\prime}\left(t\right)$ and integrating in terms of t,

$\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\mathrm{d}\phantom{t}}{\mathrm{d}t}\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)$

$=-k\left|x\right|\frac{\mathrm{d}x}{\mathrm{d}t}\Rightarrow \frac{1}{2}(\frac{\mathrm{d}x}{\mathrm{d}t}{)}^{2}$

$=-k\int \left|x\right|\mathrm{d}x.$

Therefore,

$t+{c}_{1}=\pm \int \frac{\mathrm{d}x\text{'}}{\sqrt{-2k\int \left|x\right|\mathrm{d}x}}$

1) If $x>0,\int \left|x\right|\mathrm{d}x=\int x\mathrm{d}x=\frac{{x}^{2}}{2}+{c}_{2}$ Therefore,

$t+{c}_{1}=\pm \int \frac{\mathrm{d}x}{\sqrt{{c}_{2}-k{x}^{2}}}$

$=\pm \frac{1}{\sqrt{k}}{\mathrm{tan}}^{-1}\left(\frac{\sqrt{k}x}{\sqrt{{c}_{2}-k{x}^{2}}}\right)$

$\Rightarrow x\left(t\right)=\pm \sqrt{\frac{{c}_{2}}{k}}\mathrm{sin}\left(\sqrt{k}\right(t+{c}_{1}\left)\right)\mathrm{sgn}\left(\mathrm{cos}\right(\sqrt{k}(t+{c}_{1})\left)\right).$

2) If $x<0,\int \left|x\right|\mathrm{d}x=-\int x\mathrm{d}x=-\frac{{x}^{2}}{2}+{c}_{2}$

$t+{c}_{1}=\pm \int \frac{\mathrm{d}x}{\sqrt{{c}_{2}+k{x}^{2}}}$

$=\pm \frac{1}{\sqrt{k}}{\mathrm{tanh}}^{-1}\left(\frac{\sqrt{k}x}{\sqrt{{c}_{2}+k{x}^{2}}}\right)$

$\Rightarrow x\left(t\right)=\pm \sqrt{\frac{{c}_{2}}{k}}\mathrm{sinh}\left(\sqrt{k}\right(t+{c}_{1}\left)\right).$

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