Deriving Equation (2) from Equation (1)

[Gu Jianjun]

recently, I study a research paper, and a problem confuse me, so i need help:

q(t,x)=z(t,x)+\int_t^T{k(\tau, t) z(\tau, x) d\tau } (1)

and q_t=z_t+\int_t^T{k_t(\tau, t) z(\tau, x) d\tau }- k(t,t)z(t) (2)

problem: how to according to (1) can get (2) ? please give me detailed process, thank you!

 

[HallsofIvy]

You are asking, from
q(t,x)= z(t,x)+ \int_t^T\left(k(\tau,t)z(\tau,x)\right)d\tau
how do you get
\frac{\partial q}{\partial t}= \frac{\partial z}{\partial t}+ \int_t^T\left(k_t(\tau,t)z(\tau,x)\right)d\tau - k(t,t)z(t)

I hope the \partial z/\partial x part at the front is obvious. The rest of it is from "Leibniz' rule"
\frac{d}{dx}\left(\int_{\alpha(x)}^{\beta(x)} F(t,x) dt\right)= \frac{d\beta(x)}{dx}F(\beta(x),x)- \frac{d\alpha(x)}{dx}F(\alpha(x),x)+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial F(x,t)}{\partial x} dt
The first two terms of the right hand side are derivable from the "fundamental theorem of calculus" together with the chain rule. The third term is just taking the derivative inside the integral.

In your case, the "\beta(x)" is a constant, T, so its derivative is 0. The "\alpha" is t so its derivative is 1. That's where "-k(t,t)z(t)" comes from.

 

[Gu Jianjun]

thank HallsofIvy very much , this is an important formula, I could not thought of it, oh, my god !