Time integral of divergence of matrix-vector product ;)

Question

I have a formula: $$ \nabla \cdot (D\,\nabla \varphi(r,t) ) $$ and I want to get the time integral of it. I know that $\int_a^b \varphi \, dt=0$, so the whole thing is probably equal to zero as well, but how to show it? I can't make head or tail of it ;)

Answer 1

To simplify notation in the matrix product let's say $\vec r=(x_1,x_2,x_3)$.

When you don't know what to do, try using the definitions of each thing. Maybe there's a more direct way, but in this case if you have patience, you arrive to the result.

In this case $$\nabla\varphi(\vec r,t)=\big(\varphi_{x_1}(\vec r,t),\varphi_{x_2}(\vec r,t),\varphi_{x_3}(\vec r,t)\big)^T,$$ and $$D\nabla\varphi(\vec r,t)= \left(\sum_{j=1}^3 d_{1j}\varphi_{x_j}(\vec r,t),\sum_{j=1}^3 d_{2j}\varphi_{x_j}(\vec r,t),\sum_{j=1}^3 d_{3j}\varphi_{x_j}(\vec r,t)\right)^T .$$

So the divergence results $$\nabla\cdot\big(D\nabla\varphi(\vec r,t)\big)=\frac{\partial}{\partial x_1}\sum_{j=1}^3 d_{1j}\varphi_{x_j}(\vec r,t)+\frac{\partial}{\partial x_2}\sum_{j=1}^3 d_{2j}\varphi_{x_j}(\vec r,t)+\frac{\partial}{\partial x_3}\sum_{j=1}^3 d_{3j}\varphi_{x_j}(\vec r,t)=$$ $$=\sum_{i=1}^3\sum_{j=1}^3 d_{ij}\varphi_{x_jx_i}(\vec r,t)$$

and it's clear that when you integrate, all you'll need to know is the value of the integrals $$\int_a^b \varphi_{x_ix_j}(\vec r,t)\,dt.$$

But assuming that Leibniz's integral rule apply, you have $$\int_a^b \frac{\partial}{\partial x_j}\frac{\partial}{\partial x_i}\varphi(\vec r,t)\,dt=\frac{\partial}{\partial x_j}\int_a^b \frac{\partial}{\partial x_i}\varphi(\vec r,t)\,dt=\frac{\partial}{\partial x_j}\frac{\partial}{\partial x_i}\int_a^b \varphi(\vec r,t)\,dt=0.$$ and so the answer is $\nabla\cdot\big(D\nabla\varphi(\vec r,t)\big)=\vec 0$.