Solve the differential equation $$ u'(t) = \frac{u^2(t) - u(t)}{t} $$ and then solve the initial value problem with $u(1) = \frac{1}{2}$.

I know the general solution is given by $u(t) = \frac{1}{c_1 t + 1}$ for all $t \in \mathbb{R}$ and some $c_1 \in \mathbb{R}$ determined by the initial condition, which is not always positive.

My problem is that when I integrate $\frac{1}{t}$ when using separation of variables, is my result $\ln|t|$ or just $\ln(t)$?

Here's what I've done so far: We exclude the constant solutions $u :\equiv 0$ and $u : \equiv 1$ Then our differential equation is equivalent to $$ \frac{u’(t)}{u^2(t) - u(t)} = \frac{1}{t}. $$ Integrating and substituting gives \begin{align*} \int_{u(t_0)}^{u(t)} \frac{1}{s(s - 1)} ds = \int_{t_0}^{t} \frac{1}{s} ds \implies & \int_{u(t_0)}^{u(t)} \frac{1}{s - 1} - \frac{1}{s} ds = \ln\left| \frac{t}{t_0} \right| \\ \implies & \ln\left|\frac{u(t_0) ( u(t) - 1 )}{u(t) ( u(t_0) - 1)} \right| = \ln\left| \frac{t}{t_0} \right| \\ \implies & \frac{| u(t) |}{|u(t) - 1|} = \frac{|t_0| |u(t_0)|}{|t| | u(t_0) - 1|} \end{align*} Do I now do a case distinction or how can I solve for $u$ now?

  • $\begingroup$ You can say $u/(u-1)=Ct$ where $C$ is a constant. If, instead, $C$'s sign changed with $t$, it would have to change only when $t=0$ in order to retain continuity, and then I suspect it would need to not change at all to be differentiable at $t=0$. $\endgroup$ – runway44 Jul 27 at 2:07
  • $\begingroup$ That was your original last equation without the absolute values and all the constants absorbed into $C$, except now it looks like your last equation got flipped upon editing. $\endgroup$ – runway44 Jul 28 at 20:48
  • $\begingroup$ My equation was just your equation with all the constants absorbed to be $C$. But my first comment should have also taken into consideration which side of $0$ and $1$ that $u$ was on. $\endgroup$ – runway44 Jul 28 at 22:32
  • $\begingroup$ @runway44 ok thanks I see it now but the argument still holds for $\frac{u - 1}{u} = C t$, right? $\endgroup$ – Viktor Glombik Jul 28 at 22:35
  • $\begingroup$ Yes, thanks to LutzL's observation about solutions in the three intervals $(-\infty,0)$, $(0,1)$ and $(1,\infty)$. $\endgroup$ – runway44 Jul 28 at 23:08

The absolute values in your calculation where applied properly where they are needed. However, in the next step you can remove them as follows:

As you observed, there are constant solutions at $u=0$ and $u=1$. Due to uniqueness of solutions any solution starting in $(0,1)$ stays inside that interval, the same for $(-\infty,0)$ and $(1,\infty)$. Thus the signs of the expressions under the absolute values are constant and can be merged into the integration constant. Resp. in a formulation with the initial condition, those fractions $\frac{u(t)}{u(t_0)}$ and $\frac{u(t)-1}{u(t_0)-1}$ are always positive, so that the absolute value is equal to the term within.

The ODE has a singularity at $t=0$, thus the whole vertical line there is excluded from the domain of the ODE. Thus the domain of the solution with initial condition at $t=1$ is $(0,\infty)$, not the whole of $\Bbb R$. So also $\frac{t}{t_0}$ is positive over the domain of the solution.

Then you continue with (with the correct signs in the partial fraction decomposition) \begin{align} \frac{u(t_0)(u(t)-1)}{u(t)(u(t_0)-1)}&=\frac{t}{t_0}\\[.8em] u(t)\Bigl(t_0u(t_0)-t(u(t_0)-1)\Bigr)&=t_0u(t_0)\\[.8em] u(t)=\frac{t_0u(t_0)}{t_0u(t_0)+(1-u(t_0))t} \end{align}

  • $\begingroup$ For your second part: The first task is to solve the differential equation without any initial condition. Do I then have to find two solutions, one for $t > 0$ and one for $t < 0$? $\endgroup$ – Viktor Glombik Jul 27 at 10:21
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    $\begingroup$ In principle yes. In practice the formulas stay the same, you just have to observe that you have 6 connected domains separated by the lines $u=0$, $u=1$ and $t=0$ and at $t=0$ you leave (trivially) the domain of the initial point. While you can continue the solution over the line $t=0$, this continuation is arbitrary, even if there is a preferred continuation keeping also the derivative continuous. $\endgroup$ – Dr. Lutz Lehmann Jul 27 at 10:29


$$ (t u)' = u^2 $$

now making $ v = t u$

$$ v' = \frac{v^2}{t^2}\Rightarrow \frac{dv}{v^2} = \frac{dt}{t^2} $$

or integrating

$$ \frac 1v = \frac 1t + C\Rightarrow v = \frac{t}{C t+1} = u t $$

  • $\begingroup$ As detailed in the other answer my approach is also correct and mandates the use of absolute values. By a uniqueness argument we can conclude that we can ignore these absolute values. Do you use that argument implicitly somewhere? I find it hard to believe that just by a substitution, we can completely get rid of the absolute values and not use that uniqueness argument. $\endgroup$ – Viktor Glombik Jul 27 at 12:41
  • $\begingroup$ The substitution can be considered as a coordinates change. Under the new coordinates, the tangent space can have more or less folds. There are many cases in which the same phenomena occur concerning ODEs. $\endgroup$ – Cesareo Jul 27 at 13:12
  • $\begingroup$ Please. Compare the stream plots for the $u$ and $v$ ODEs. $\endgroup$ – Cesareo Jul 27 at 13:54
  • $\begingroup$ I'm sorry. Im just not no familiar with the terminology tangent space (at least not in the ODE setting, I only know tangent spaces of manifolds.) Would you mind explaining that to me? $\endgroup$ – Viktor Glombik Jul 27 at 14:07
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    $\begingroup$ Perhaps not folds, but leaves of the foliation, branches of the covering etc. of the almost everywhere locally bijective coordinate transformation? This approach still needs to exclude the points with $v=0$ or $t=0$, and consider possible solutions through them separately. $\endgroup$ – Dr. Lutz Lehmann Jul 27 at 16:28

The maximal $(t,u)$-domain relevant to the given IVP is $\Omega:={\mathbb R}_{>0}\times{\mathbb R}$. Within $\Omega$ the standard existence and uniqueness theorem for ODEs is valid. By inspection one sees that there are the constant solutions $u_0(t)\equiv0$ and $u_1(t)\equiv1$ $(t>0)$. No other solution can cross the graphs of $u_0$ and $u_1$. Since the initial point $\bigl(1,{1\over2}\bigr)$ is given the solution $t\mapsto u(t)$ therefore stays in the $u$-interval $\>]0,1[\>$ for all $t>0$. Hence there are no case distinctions; you just have to select the primitive of $s\mapsto{1\over s^2-s}$ that is valid for $0<s<1$, namely $$\eqalign{\int{ds\over s^2-s}&=\int\left({1\over s-1}-{1\over s}\right)ds=\log|s-1|-\log |s| +C\cr &=\log{1-s\over s}+C\qquad(0<s<1)\ .\cr}$$

  • $\begingroup$ I get that the IVP domain is $\mathbb{R}_{>0} \times \mathbb{R}$ but when just solving the differential equation first (as the exercise wants it) why can't $\mathbb{R}_{<0} \times \mathbb{R}$ be the domain? $\endgroup$ – Viktor Glombik Jul 28 at 13:38
  • $\begingroup$ The set $\Omega':={\mathbb R}_{<0}\times{\mathbb R}$ is another domain for the given ODE, but the initial point $\bigl(1,{1\over2}\bigr)$ does not lie in $\Omega'$. $\endgroup$ – Christian Blatter Jul 28 at 14:46

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