Special case of Leibniz formula

Question

I am wondering if I understand it right. Considering the following partial derivative: $$ \partial_tF(t, x) = \partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds $$ It looks to me as Leibniz should be used. So we should get $$ \partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds \\ = \int_{a(t)}^{b(t)} \partial_t f(t-s, x)ds - f(t-a(t), x)(1-a'(t)) + f(t-b(t), x)(1-b'(t)) $$ Is this correct, or did I get something wrong?

Answer 1

Applying the Leibniz integral rule we obtain \begin{align*} \color{blue}{\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f\left(u(t,s),x\right)\,ds\right)} &\color{blue}{=f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)}\\ &\qquad\color{blue}{-f(u(t,a(t)),x)\,\frac{d}{dt}a(t)}\\ &\qquad\color{blue}{+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds}\tag{1} \end{align*} provided the functions are appropriately differentiable.

Example: We look at a small example. \begin{align*} u(t,s)&=t-s\qquad\qquad a(t)=4t\qquad b(t)=3t^2\\ f(u(t,s),x)&=u(t,s) x\\ &=(t-s)x \end{align*}

• The left-hand side of (1) gives \begin{align*} \frac{d}{dt}\left(\int_{a(t)}^{b(t)}f(u(t,s),x)\,ds\right)&=\frac{d}{dt}\int_{s=4t}^{3t^2}(t-s)x\,ds\\ &=x\frac{d}{dt}\left(ts-\frac{1}{2}s^2\bigg|_{s=4t}^{3t^2}\right)\\ &=x\frac{d}{dt}\left(3t^3-\frac{9}{2}t^4-4t^2+8t^2\right)\\ &\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)} \end{align*}
• The right-hand side of (1) gives \begin{align*} &f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)-f(u(t,a(t)),x)\,\frac{d}{dt}a(t) +\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds\\ &\qquad=\left(t-\left(3t^2\right)\right)x\cdot 6t-\left(t-(4t)\right)x\cdot 4 +\int_{4t}^{3t^2}\frac{d}{dt}(t-s)x\,ds\\ &\qquad=\left(6t^2x-18t^3x\right)+12tx+x\int_{4t}^{3t^2}\,ds\\ &\qquad=\left(6t^2-18t^3\right)x+12tx+\left(3t^2-4t\right)x\\ &\qquad\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)} \end{align*} and we see according to (1) the blue marked results coincide as expected.