partial derivatives and integrals

Question

suppose I have a function $$v(x,t):=\int_0^t u(x,t;s) ds.$$

Why $$v_t(x,t)=u(x,t;t)+\int_0^t u(x,t;s) ds.$$

I do not know where does the $u(x,t;t)$ term come from ?

Answer 1

If you just had $ v ( t ) = \int _ { s = 0 } ^ t u ( s ) \, \mathrm d s $, then you'd say that $ \mathrm D _ t v ( t ) = u ( t ) $. Bump $ u ( s ) $ up to $ u ( x ; s ) $, and now you get $ u ( x ; t ) $. Bump it further to $ u ( x , t ; s ) $, and now you get $ u ( x , t ; t ) $.

Since $ u ( x , t ; s ) $ has an additional dependence on $ t $, you get another term, $ \int _ { s = 0 } ^ t \mathrm D _ t u ( x , t ; s ) \, \mathrm d s $. But you weren't confused about where that term came from. (However, as @Aruralreader mentioned in a comment, this does need a derivative with respect to $ t $ in it.)

Answer 2

Receipt: By linearity of multidimensional derivatives, supply an index to the symbol t at all places where it appears, do the differential and discard the indices by restriction to the diagonal $t=t_1=t_2$.

$$d \ \int_0^{t_1} u(x,t_2,s) ds = dt_1 \partial_{t_1}\int_0^{t_1} u(x,t_2,s) ds + dt_2 \partial_{t_2}\int_0^{t_1} u(x,t_2,s) ds$$ $$\to dt \left( u(x,t,t) + \int_0^t \partial_t u(x,t,s) ds \right) $$

This yields the same result as the conventional $1/h$-limits, again by linearity of linearizations. Even the high school analysis rules for $$(1*x)' =1'x + 1 x' , n x=x+x+\dots, x^n = x*x*\dots $$ are simple consequences of the basic rule: look for all linear occurences of the symbol, replace it by 1 and add.