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Consider the model $$\partial _{t}\mu + \partial_{x}(b(t,\mu)\mu)+c(t,\mu) \mu=0,~~~\mu \in \mathcal{M}^{+}(\mathbb{R}^{+}), t \in [0,T], x \in \mathbb{R}^{+} $$ $$ \mu(0)=\mu_{0} $$ where $ \mu (t)$ is the measure determining the distribution of the population with respect to the structural variable $x$ and $b,c$ are vital rates.

I want to know what it means for a solution to be measure-valued i.e. solution is a measure? If the solution (in weak sense) is measurable function then it makes sense to me but solution as a measure is really confusing me, because measure is defined on a set and solution we usually need at every point of domain.

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    $\begingroup$ Perhaps you are just confused by the terminology? The domain of your solution is a set, $\mathbb{R}^+$ in this case. While pointwise values $\mu(t,x)$ do not make sense for a measure, what does make sense is any integral of the form $\int_a^b w(x)\mu(t,x)$, giving you a weighted average of $w(x)$ over the interval $x\in[a,b]$ with respect to $\mu(t)$, a "population distribution" in your case. Perhaps such weighted averages are all that you need from your solution. $\endgroup$ Commented Mar 3, 2020 at 10:51
  • $\begingroup$ @IgorKhavkine yes, that is what I was missing, because solution here in my case is a solution of the population model. $\endgroup$ Commented Mar 3, 2020 at 11:36
  • $\begingroup$ I don't know this concrete equation but I know many similar. The thing that you should remember (and @IgorKhavkine explained it well) is that if you work on PDE problems of this kind, you can't only have pointwise solutions (that is kinda rare). You'll usually have some weak solutions such as measure-valued solutions, solutions in the distribution sense, etc. $\endgroup$
    – Mark
    Commented Mar 3, 2020 at 16:06

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