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I have the following exercise:

Consider the heat equation $$u_t = k \Delta u \quad (*)$$ on a domain $\Omega$ with boundary conditions $u _{|\partial \Omega} =0$ and initial data $\phi(x)$. Suppose that $k$ is unknown, but it can be established by experiment that the total temperature in $\Omega$ decays like $\int_{\Omega} u(\cdot , t) \, dx = F(t)$, where $F$ is strictly monotonically decreasing.

Suppose also that the solution of the heat equation $$v_t = \Delta v \quad (**)$$ with the same boundary and initial conditions decays like $\int_{\Omega} v( \cdot , t) \, dx=G(t)$, where $G$ is strictly monotonically decreasing. Can you determine $k$ in terms of $F$ and $G$?

My attempt: we have that $$\int_{\Omega} u_t \, dx = \frac{d}{dt} \int_{\Omega} u \, dx = F'(t)$$and $$\int_{\Omega} v_t \, dx = \frac{d}{dt} \int_{\Omega} v \, dx = G'(t).$$Now, if we set $w(x,t) :=u(x,kt)$ we observe that $w$ is a solution for $(**)$. Then by uniqueness we have $v(x,t)=u(x,kt)$. Therefore $$\underbrace{\int_{\Omega} v_t \, dx }_{=G'(t)} = k \underbrace{\int_{\Omega} u_t (x,kt) \, dx}_{=F'(kt)}$$ and thus we can conclude $$k = \frac{G'(t)}{F'(kt)}>0.$$ The problem is that $F'$ seems to "depend" on $k$ itself, but I don't know if this is actually a problem... any comment about that?

Thanks in advance!

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  • $\begingroup$ Do you mean to have $G,F$ swapped in the first two lines of equality? $\endgroup$ – charlestoncrabb Dec 4 '15 at 19:45
  • $\begingroup$ Oh, sorry! I'm going to fix it. $\endgroup$ – gangrene Dec 4 '15 at 19:47
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I don't think it's a problem: in the equation $(*)$, the role of $k$ is to essentially set a characteristic time scale for diffusion, thus it shouldn't be surprising that the total temperature at time $t$ should depend on this time scale. Moreover, by this reasoning I would be very suspicious if $F$ didn't depend on $k$.

The only thing that worries me is that since $k$ sets the time scale for $F$, by stretching $F$ in the $F(kt)$ term, we should eliminate the dependence on this time scale, but this is not obvious since $\partial_kF(kt)=\frac{t}{k}G'(t)$ which is necessarily nonzero by choice.

So, I agree there is something to be said here regarding the dependence on $k$ of these quantities, but overall I think your solution is perfectly reasonable.

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