1

Let $h=h(x,t): \mathbb{R}\times [0,+\infty)\longmapsto \mathbb{R}$ and suppose $h,h_x \in C(\mathbb{R}\times [0,+\infty))$.

Consider the function $$F(x,t)=\int_{0}^{t}h(x-c(t-s),s)ds,$$ where $c$ is a constant.

Is it true that $F(x,t)\in C^1(\mathbb{R}\times [0,+\infty))$?

In particular, how can I show that $F$ is continuous on $\mathbb{R}\times [0,+\infty)$? Does there exist a generalization of Leibniz integral rule (in the present case, the integrand depends on two parameters x,t) to calculate $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$?

Thanks in advance!

eleguitar
  • 486
  • 5
    Leibniz integral rule could be of some help. Also, you can look at Proof of Leibniz's formula. – user539887 Jul 11 '18 at 09:31
  • I'm having trouble with Leibniz Integral rule because of "s" parameter. – eleguitar Jul 11 '18 at 18:06
  • $s$ is not a parameter, it is the variable of integration. – user539887 Jul 11 '18 at 18:18
  • I'm sorry, that was a poor word choice. If I well understand Leibniz integral rule then we have $$\frac {\partial }{\partial t}\int_{0}^{t} h(x-c(t-s),s)ds=h(x,t)+\int_{0}^{t} \frac {\partial }{\partial t} h(x-c(t-s),s)ds=\h(x,t)+(-c)\int_{0}^{t}h_x(x-c(t-s),s)ds,$$ and in a similar way $$\frac {\partial }{\partial x}\int_{0}^{t} h(x-c(t-s),s)ds=\int_{0}^{t} \frac {\partial }{\partial x} h(x-c(t-s),s)ds= \int_{0}^{t} h_x(x-c(t-s),s)ds.$$ Is that correct? – eleguitar Jul 12 '18 at 15:57
  • Anyway, to use the chain rule inside the integral to write that $$\frac{\partial}{\partial t}h(x-c(t-s),s)=h_x(x-c(t-s),s)\cdot \frac{\partial (x-c(t-s))}{\partial t}+h_t(x-c(t-s),s) \cdot \frac{\partial (s)}{\partial t}=h_x(x-c(t-s),s) \cdot(-c)$$ I also need that $h=h(x,t)$ is differentiable with respect to the second variable, not only with respect to $x$. – eleguitar Jul 12 '18 at 16:54
  • I probably do not understand Leibniz Integral rule in the case of two parameters (x and t) instead of just one. – eleguitar Jul 12 '18 at 17:57
  • 1
    It appears that your calculations are O.K. However, the notation $h_x$ is rather inopportune. It would be much better to write $D_1h$ (you have the partial derivative in the first variable evaluated at a point where the first variable takes value $x-c(t-s)$). The notations $\frac{\partial}{\partial t}h$ and $\frac{\partial}{\partial x}h$ are O.K. – user539887 Jul 12 '18 at 21:06
  • what is $c?,,$ – zhw. Jul 13 '18 at 17:08
  • @zhw $c$ si a constant. – eleguitar Jul 13 '18 at 18:24

1 Answers1

0

PERSONAL ATTEMPT

I think a generalization of Leibniz integral rule can be something like this

"Let $f=f(x,t,s)$ be a continuous function in some region $[x_0,x_1]\times[t_0,t_1]\times[s_0,s_1]$ of the $(x,t,s)$ space. Let $\alpha=\alpha(x,t)$ and $\beta=\beta(x,t)$ be continuously differentiable functions on $[x_0,x_1]\times[t_0,t_1]$ such that $s_0\leq\alpha(x,t)\leq\beta(x,t)\leq s_1\,\,\forall (x,t) \in [x_0,x_1]\times[t_0,t_1]$.

Then $$F=F(x,t)=\int_{\alpha(x,t)}^{\beta(x,t)}f(x,t,s)ds$$ is continuous on $[x_0,x_1]\times[t_0,t_1]$.

Moreover, if partial derivatives of $f$ with respect to $x$ and $t$, $\frac{\partial f(x,t,s)}{\partial x} \,\text{and}\, \frac{\partial f(x,t,s)}{\partial t}$, exist and are continuous on $[x_0,x_1]\times[t_0,t_1]\times[s_0,s_1]$ then $F=F(x,t)\in C^1([x_0,x_1]\times[t_0,t_1])$ and $$\frac{\partial F(x,t)}{\partial x}=\int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f(x,t,s)}{\partial x}ds+ f(x,t,\beta(x,t))\cdot \frac{\partial \beta(x,t)}{\partial x}- f(x,t,\alpha(x,t))\cdot \frac{\partial \alpha(x,t)}{\partial x},$$ $$\frac{\partial F(x,t)}{\partial t}=\int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f(x,t,s)}{\partial t}ds+ f(x,t,\beta(x,t))\cdot \frac{\partial \beta(x,t)}{\partial t}- f(x,t,\alpha(x,t))\cdot \frac{\partial \alpha(x,t)}{\partial t}$$ for all $(x,t)\in [x_0,x_1]\times[t_0,t_1]$."

eleguitar
  • 486