How to justify taking the derivative operator inside of the integral?

Question

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,

Answer 1

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.

Answer 2

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.