# Show IVP $u'(t)= -u(t)\ln(u(t))$ has unique solution

Consider the function $$f:[0,e^{-1}]\rightarrow \mathbb{R}, u\mapsto \begin{cases} -u\ln(u) & \text{if}\ u\in (0,e^{-1}] \\ 0, & \text{if}\ u=0 \end{cases}$$

The IVP is given as $$\begin{cases} u'(t)=f(u(t)) & \text{for}\ t\in[0,1] \\ u(0)=0 \end{cases}$$ For the first part of the problem I was supposed to show it is monotonically increasing and convex, which I did, furthermore I showed it satisfies $|f(u)-f(v)| \leq f(|u-v|) \quad \forall u,v \in [0,e^{-1}]$

Now I want to show that there exists a unique local solution. I tried the usual methods like separation of variables but the fact, that $u'(t)=-u(t)\ln(u(t))$ would not let me solve the integral in that process due to always having one integral which is not solvable.

So I thought I'd use Peano. As $f$ is a composition of continuous functions, $f$ is continous and Peano tells us there at least exists one solution in a local neighbourhood around $0$.

Now I want to show it's unique by using a hint I was given to "look at the difference of two solutions to the IVP".

So let's consider $u(t)-v(t)$ where both u and v are solutions to the IVP. I could either check if that is zero ( I would not know how ) or maybe I could differentiate the expression and show that is zero, but again, I don't know how.

Is the argument of existence correct? How can I show the uniqueness of the solution?

• Existence is very easy to see by observing that the constant function $u=0$ is a solution. Jun 1, 2017 at 19:12

Since $f$ is not Lipschitz continious you have to argue the following way:
$u\equiv 0$ is a solution of your IVP. Consider there exists and other solution $v\not\equiv 0$. W.l.o.g. we assume $v_1=v(t_1)\in(0,e^{-1})$ for some $t_1>0$ and $v(t)\in(0,e^{-1})$ for all $t\in(0,t_1]$. Then $v$ is a solution of $$\begin{cases} v'=f(v)\\ v(t_1)=v_1 \end{cases}.$$ We seperate the variables and integrate both sides and get $$\int_{t_1}^t\frac{v'(t)}{f(v(t))}~dt=\int_{t_1}^t1~ds\hspace{2cm}(*)$$ Since $v(t)\in(0,e^{-1})$ for $t\in(0,t_1]$ we have $f(v(t))=-v(t)\ln(v(t))$ and we get $$\int_{t_1}^t-\frac{v'(t)}{v(t)\ln(v(t))}~dt=\left[\ln(-\ln(v(t)))\right]_{t_1}^t=\ln(-\ln(v(t)))-\ln(-\ln(v_1)).$$ Now we use $(*)$ and for $t\in(0,t_1]$ we get \begin{align} \ln(-\ln(v(t)))-\ln(-\ln(v_1))=t-t_1&\Leftrightarrow \ln\left(\frac{\ln(v(t))}{\ln(v_1)}\right)=t-t_1\\ &\Leftrightarrow \frac{\ln(v(t))}{\ln(v_1)}=e^{t-t_1}\\ &\Leftrightarrow\ln(v(t))=\ln(v_1)e^{t-t_1}\\ &\Leftrightarrow v(t)=v_1^{e^{t-t_1}}. \end{align} But this yields $$0=v(0)=\lim_{\substack{t\to 0\\t>0}}v(t)=\lim_{\substack{t\to 0\\t>0}}v_1^{e^{t-t_1}}=v_1^{e^{-t_1}}\neq 0.$$ which is a contradition. Therefore $u\equiv 0$ is the only solution of your IVP.