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I'm reading a PDE book, and the authors use this often, for $u = u(x,t)$, and V an arbitrary volume on in the interior of a solid:

$${\frac{d}{dt} \int_V u \space dx}$$

$$= {\int_V u_t \space dx}$$

How is differentiating under the integral w.r.t. to time $t$ justified?

I imagine it must be so simple that I'm simply rusty with my introductory analysis coursework and don't see it right away.

My attempt:

Writing out the difference quotient and using linearity of the integral, we have that:

$$ \lim_{h \to 0} \int_V \frac { u(x,t+h) - u(x,t)} {h} \space dx $$

and assuming I can use the dominated convergence theorem, we have that:

$$ \lim_{h \to 0} \int_V \frac { u(x,t+h) - u(x,t)} {h} \space dx = \int_V \lim_{h \to 0} \frac { u(x,t+h) - u(x,t)} {h} \space dx $$

$$ = {\int_V u_t \space dx} $$

So, there seem to be some nice conditions on $u$ that the authors are assuming.

Thanks,

user685689
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2 Answers2

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The integral is a linear operator, so if we call it $L$, it satisfies $${1\over h}(L(v)-L(w))=L({v-w\over h})$$

Now take $v=u(t+h,x)$ and $w=u(t,x)$, pick a small $x$ and write

$$\eqalign{{d\over dt}\int udx&\approx {\int u(t+h,x)dx - \int u(t,x)dx\over h}\cr &=\int {u(t+h,x)-u(t,x)\over h} dx\cr &\approx \int u_tdx}$$

Of course these approximations to the derivatives make sense only if the derivatives exist, so there are things to check, but this is the idea.

Asura Path
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  • Nice work here, especially using the linearity. Swapping the limit is likely gotten by the dominated convergence theorem (suggested in Kavi's answer). Thanks! – user685689 Jul 08 '19 at 06:25
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The interchange is not always justified. A typical argument is to apply MVT to write $\phi(x, t+h)-\phi (x,t)$ as $h \frac {\partial} { \partial \,t } \phi (x,s)$ for some $s$ between $t$ and $t+h$. If the function $\phi$ is nice enough you can show that $\frac {\phi(x, t+h)-\phi (x,t)} {h}$ is dominated by an integrable function in which case DCT tells you that you can interchange the integral and the limit.

Kavi Rama Murthy
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  • Thanks Kavi, I wrote it out (see my edit), but am wondering what integrable function the integrand is dominated by. I'm guessing it's so obvious that the authors don't bother explaining. I'll move on and assume that $u$ is nice enough. – user685689 Jul 08 '19 at 06:26
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    @Jay Finding the dominating function depends on the context. For example, if $V$ is a bounded domain and the partial derivative of $\phi$ is bounded then there is a constant function which serves the purpose. – Kavi Rama Murthy Jul 08 '19 at 06:30