I am confused about linear time-varying system. For a time varying system, the output is given by \begin{align} y(t)=\int x(\tau) h_{\tau}(t) d\tau, \end{align} where $ h_{\tau}(t)$ is the output of the system given the input $\delta(t-\tau)$. By manipulating the integration above, we have \begin{align} y(t)=\int x(t-\tau) h_{t-\tau}(t) d\tau \end{align} I found that in some refeneces, we can write above as $y(t)=\int x(t-\tau) h(t,\tau) d\tau$, where $h(t,\tau)=h_{t-\tau}(t)$. To avoid confusion, here I let $g(t,\tau)=h_{t-\tau}(t)$. Thus, \begin{align} y(t)=\int x(t-\tau) g(t,\tau) d\tau \end{align}
From my understanding, the $g(t,\tau)$ is just obtained by substituting the variables in $ h_{t-\tau}(t)$. However, I am not sure whether we can treat $g(t,\tau)$ as the system output of some input signal.
Based on the definition of $h_{\tau}(t)$, though $ h_{\tau}(t)$ is the output of the system given the input $\delta(t-\tau)$, $h_{\tau}(t)$ in fact defines a function over $(t,\tau), \forall t,\tau$. Therefore, the value of $h_{t-\tau}(t)$ can be obtained by substituting the pair $(t,t-\tau)$ in to $h_{\tau}(t)$.
However, can we treat $h_{t-\tau}(t)$ as the output of $\delta(t-(t-\tau))$, i.e., $\delta(\tau)$?
Here, how can we understand the input $\delta(\tau)$? In my view, I think the input of a system should be a function over $t$, and $\tau$ is just a parameter. Does this mean that the input is a constant?