In a paper the differential equation $-y'(t)=\frac{1}{2}\alpha y(t)^2+y(t)^{3/2}$ occured. First of all, it does not satisfy the usual Lipschitz-condition. But there are probably weaker criteria. The solution was given as: $$ y(t)=\frac{4}{\alpha^2}\left(1+W(-C_1 e^{-\frac{t}{\alpha}})\right)^{-2}, $$ where $W$ is the Lamber-$W$-function. Mathemica/WolframAlpha give nothing. I have not yet seen such a differential equation. How can one solve this?

  • $\begingroup$ Maple, though, finds an implicit solution. Plus, this is just a separable equation, which should be straightforward to solve. $\endgroup$ – Artem Apr 27 '12 at 15:10
  • $\begingroup$ It's usually a good idea when referring to Wolfram|Alpha results to link to them. In the present case, it seems you entered the equation either incorrectly or in an inconvenient form; when I enter it, I get an implicit solution. $\endgroup$ – joriki Apr 27 '12 at 15:51
  • $\begingroup$ Ouch, that is actually easy to see. I was so puzzled by this $W$-function. However, this was my input: WA-Input. To my knowledge, this is standard mathematica input. $\endgroup$ – user13655 Apr 27 '12 at 16:22
  • $\begingroup$ The right side of the equation is locally Lipschitz (on $[0,\infty)$). To avoid problems with negative $y$, replace the $y^{3/2}$ on the right side with $|y^{3/2}|$. Since this modified equation satisfies the hypotheses of the existence and uniqueness theorem, and $y=0$ is a solution, the solution with any initial value $> 0$ will never hit $0$, so you don't have to worry about negative $y$. $\endgroup$ – Robert Israel Apr 27 '12 at 16:26

The standard way to solve such a separable first-order ordinary differential equation is to divide through by the function of $y$ to get separate integrals with respect to $t$ and $y$:

$$ -\frac{y'}{\frac12\alpha y^2+y^{3/2}}=1\;, \\ \int-\frac{y'}{\frac12\alpha y^2+y^{3/2}}\mathrm dx=\int1\,\mathrm dt\;, \\ \int-\frac1{\frac12\alpha y^2+y^{3/2}}\mathrm dy=\int1\,\mathrm dt\;. $$

Wolfram|Alpha solves the integral on the left (and shows steps if you click "Show Steps"):

$$\frac2{\sqrt y}-\alpha\log\left(\alpha+\frac2{\sqrt y}\right)=t+C\;.$$

With $u=\alpha+2/\sqrt y$, this becomes

$$u-\alpha\log u=t+C'\;.$$

Rescaling by $u=-\alpha s$ to get rid of the factor $-\alpha$, we get

$$s+\log s=-\frac t\alpha+C''\;.$$

Then exponentiating yields

$$s\mathrm e^s=C'''\mathrm e^{-t/\alpha}\;,$$

which is solved using the Lambert W function by

$$s=W\left(C'''\mathrm e^{-t/\alpha}\right)\;.$$

Then resolving the substitutions yields

$$y=\left(\frac2{u-\alpha}\right)^2=\left(\frac2\alpha\right)^2\left(1+s\right)^{-2}=\frac4{\alpha^2}\left(1+W\left(C'''\mathrm e^{-t/\alpha}\right)\right)^{-2}\;.$$

  • $\begingroup$ Thanks for this extensive answer! $\endgroup$ – user13655 Apr 27 '12 at 16:42
  • $\begingroup$ @user13655: You're welcome! $\endgroup$ – joriki Apr 27 '12 at 16:43
  • $\begingroup$ @user13655 I think joriki has other standards for extensive $\endgroup$ – Pedro Tamaroff Apr 28 '12 at 16:35

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