Thanks for the interesting introduction of Continuity Equation from a new perspective. The weak version is easy to understand. But in the first step (at 6:45) of the (sketchy) proof of the strong version, it is not clear for me why the integral on the right side in the weak version (with gradient of \phi (x) inside the integral) can be replaced by the negative of an integral with the divergence of q_tv_t.
Kirill said "by parts" :), this is true. Some steps to just make you believe that it is correct, not to derive. You use s: scalar, v: vector div(sv) = s div(v) + . Now you recognise (up to integration) left side of this as second line and right side as something that we need to make as first line (you can put q inside as it is scalar). So to then you integrate this around any closed curve and then enforce some boundary condition. May be the most simple way is to just to do everything in coordinates, then you can apply usual 1-dim integration by parts.
thank you! as @iesh00a said, this is just integration by parts (the product rule of differentiation rearranged) and the fact that the density vanishes at infinity. i didn't put the integration limits but they are minus and plus infinity for every coordinate.
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Thanks for the interesting introduction of Continuity Equation from a new perspective. The weak version is easy to understand. But in the first step (at 6:45) of the (sketchy) proof of the strong version, it is not clear for me why the integral on the right side in the weak version (with gradient of \phi (x) inside the integral) can be replaced by the negative of an integral with the divergence of q_tv_t.
Kirill said "by parts" :), this is true. Some steps to just make you believe that it is correct, not to derive. You use s: scalar, v: vector div(sv) = s div(v) + . Now you recognise (up to integration) left side of this as second line and right side as something that we need to make as first line (you can put q inside as it is scalar). So to then you integrate this around any closed curve and then enforce some boundary condition. May be the most simple way is to just to do everything in coordinates, then you can apply usual 1-dim integration by parts.
thank you! as @iesh00a said, this is just integration by parts (the product rule of differentiation rearranged) and the fact that the density vanishes at infinity. i didn't put the integration limits but they are minus and plus infinity for every coordinate.