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I'm reading nonlinear control systems book. The author provides this example $$ \dot{x} = r + x^2, \quad r < 0. $$ I would like to compute the analytical solution for the proceeding ODE. My attempt is

$$ \begin{align} \frac{dx}{dt} &= r + x^2 \\ \frac{dx}{r+x^2} &= dt \\ \int^{x(t)}_{x_0} \frac{1}{r+x^2} dx &= \int^{t}_{t_0} d\tau \\ \frac{\tan^{-1}\left(\frac{x}{\sqrt{r}}\right)}{\sqrt{r}} \Big|^{x(t)}_{x_0} &= (t-t_0) \end{align} $$

Now the problem with the assumption that $r<0$, how I can handle the substitution for the left side? I need to reach the final step where $x(t)$ is solely in the left side.

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    $\begingroup$ Substitute $u = \frac{x}{\sqrt{-r}}$. You'll get an inverse hyperbolic tangent. $\endgroup$
    – Christoph
    Commented Jan 30, 2019 at 6:16
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    $\begingroup$ So it is $x^2 - a^2$ at the bottom. Could use partial fractions. $\endgroup$ Commented Jan 30, 2019 at 6:16
  • $\begingroup$ @Christoph, why did you put the minus ? Also, how to handle the square root of negative number? $\endgroup$
    – CroCo
    Commented Jan 30, 2019 at 6:21
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    $\begingroup$ If $r<0$ you have $-r>0$, so you can take the square root of that. $\endgroup$
    – Christoph
    Commented Jan 30, 2019 at 6:22
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    $\begingroup$ $-a^2=r$, so $a$ is the square root of the positive number $-r$. $\endgroup$ Commented Jan 30, 2019 at 6:23

4 Answers 4

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Let us consider

$$\dot x=x^2-1$$ for convenience.

When $|x|<1$, we solve the separable equation with

$$\frac{dx}{1-x^2}=-dt$$ and

$$\text{artanh }x-\text{artanh }x_0=t_0-t,$$

i.e.

$$x=\tanh(t_0-t+\text{artanh }x_0).$$

When $|x|>1$, we solve with

$$\text{arcoth }x-\text{arcoth }x_0=t_0-t,$$

i.e.

$$x=\coth(t_0-t+\text{arcoth }x_0).$$

Notice that this solution has a vertical asymptote at $t=t_0+\text{arcoth }x_0$.

Finally, $x=\pm1$ are two valid solutions.

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Solve it as Riccati equation by setting $x=-\frac{u'}{u}$. Then $$ u''-a^2u=0 $$ has the solution $u(t)=Ce^{at}+De^{-at}$ and thus $$ y(x)=-a\frac{Ce^{at}-De^{-at}}{Ce^{at}+De^{-at}} $$ with some redundancy in the parameter pair $(C,D)$.

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Let $r=-a^2$. Then $$\int\frac{dx}{x^2-a^2}=\int dt$$ $$-\frac{\log{\left( x+a\right) }-\log{\left( x-a\right) }}{2 a}=t+c$$ Take $c=\frac{\log C}{2a}$. $$\log{\left( \frac{x-a}{C\, \left( x+a\right) }\right) }=2 a t,$$ $$\frac{x-a}{x+a}=C\, {{e}^{2 a t}}.$$ General solution is $$x=\frac{a(1+Ce^{2at})}{1-Ce^{2at}}.$$

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We use separation of variables to solve the ODE $\dot{x} = r + x^2$, $r < 0$:

  1. We first find the constant solutions $x \equiv \pm \sqrt{-r}$. In the following steps we shall assume $r+x^2 \not \equiv 0$.

  2. We use the Leibniz notation and we separate the variables: $\frac{dx}{dt} = r + x^2$, $\frac{1}{r+x^2} dx = dt$.

  3. We integrate on both sides: $\int \frac{1}{r+x^2} \, \mathrm{d}x = \int 1 \, \mathrm{d}t$.

  4. We solve the integral on the left-hand side using the substitution $x = \sqrt{-r} u$, $dx = \sqrt{-r} du$: \begin{eqnarray} \int \frac{1}{r+x^2} \, \mathrm{d}x &=& \int \frac{1}{r-ru^2} \sqrt{-r} \, \mathrm{d}u = \frac{\sqrt{-r}}{r} \int \frac{1}{1-u^2} \, \mathrm{d}u = \frac{\sqrt{-r}}{r} \left\{ \begin{array}{ll} \operatorname{artanh}(u) + \tilde{C}, & |u| < 1\\ \operatorname{arcoth}(u) + \tilde{C}, & |u| > 1 \end{array} \right.\\ &=& \frac{\sqrt{-r}}{r} \left\{ \begin{array}{ll} \operatorname{artanh}\left(\frac{x}{\sqrt{-r}}\right) + \tilde{C}, & |x| < \sqrt{-r}\\ \operatorname{arcoth}\left(\frac{x}{\sqrt{-r}}\right) + \tilde{C}, & |x| > \sqrt{-r} \end{array} \right., \quad \tilde{C} \in \mathbb{R}. \end{eqnarray}

Thus we now obtain the nonlinear equations \begin{eqnarray} \operatorname{artanh}\left(\frac{x}{\sqrt{-r}}\right) = C -\sqrt{-r} t, \quad |x| < \sqrt{-r},\\ \operatorname{arcoth}\left(\frac{x}{\sqrt{-r}}\right) = C -\sqrt{-r} t, \quad |x| > \sqrt{-r}, \end{eqnarray} with a constant $C \in \mathbb{R}$.

Finally, we obtain 4 possible solutions of the ODE, depending on the initial value $x(t_0) = x_0 \in \mathbb{R}$: \begin{eqnarray} x(t) &=& \sqrt{-r} \coth\left(\operatorname{arcoth}\left(\frac{x_0}{\sqrt{-r}}\right)-\sqrt{-r}(t-t_0)\right), \quad |x_0| > \sqrt{-r},\\ x(t) &=& -\sqrt{-r}, \quad x_0 = -\sqrt{-r},\\ x(t) &=& \sqrt{-r}, \quad x_0 = \sqrt{-r},\\ x(t) &=& \sqrt{-r} \tanh\left(\operatorname{artanh}\left(\frac{x_0}{\sqrt{-r}}\right)-\sqrt{-r}(t-t_0) \right), \quad |x_0| < \sqrt{-r}. \end{eqnarray}

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