$c\in(0,1)$ , $0<a<1-c<b$, $f(n)=1-c(n+1)$, $\Pi=\left\{\{\pi_n,n\geq 0\}, t\in[a,b], N^*\in N_+:\pi_n=\left\{ \begin{align} &b^n, &n< N^*,\\ &b^{N^*}t, &n= N^*,\\ &b^{N^*}ta^{n-N*}, &n> N^*. \end{align} \right.\right\}$. Solve $\pi_n\in\Pi$ (actually, the value of $t$ and $N^*$) from $\sum_{n=0}^{\infty}f(n)\pi_n= 0$
My ideas:
$(1)$ $\pi_n$ changes at speed $b$ before $N^*$, at speed $a$ after $N^*$. The solution exists since $0<a<1-c<b$ and $\sum_{n=0}^{\infty}f(n)(1-c)^n$. But there may exist not only one solution.
$(2)$ First, I wanna ignore $t$, let $\pi_n=\left\{ \begin{align} &b^n, &n \leq N^*,\\ &b^{N^*}a^{n-N*}, &n> N^*. \end{align} \right.$ find $N^*= \underset{N\in N_+}{argmax}\{\sum_{n=0}^{\infty}f(n)\pi_n\geq 0\}$. Then find $t$ using $N^*$ and $\sum_{n=0}^{\infty}f(n)\pi_n= 0$. But even in this easier case, it's not easy to calculate $N^*$.
$(3)$ $f(n)$ is a linear decreasing function. $\pi_n$ is like a mutation of geometric series. $\sum_{n=0}^{\infty}f(n)\pi_n= 0$ reminds me of inner product and orthogonality. I'm not sure if this is a kind of problem that has a general method to solve. Can anyone give me some hints or an approach to problem-solving? Your comments, hints, or alternative answers are highly appreciated. Thank you!