Where is the fixed point? -- Matlab is cluless too

Question

Consider the differential equation $$\dot{k}(t)=f(k)-(r+t_1)k-f(1-k)+(r+t_2)(1-k)$$ where $k,t_1,t_1\in[0,1]$, $r\in\mathbb{R}$ and $f:[0,1]\to\mathbb{R}_+$. I'd like to solve for the fixed point $k_\infty$ which verifies $\dot{k}(t)=f(k_\infty)-(r+t_1)k_\infty-f(1-k_\infty)+(r+t_2)(1-k_\infty)=0$. To find an explicit solution let $f(k)=k^\alpha$ with $\alpha\in(0,1)$. I tried to solve it with matlab. The code reads

syms k a r t1 t2 sol = solve(k^a-(r+t1)*k-(1-k)^a+(r+t2)*(1-k) == 0, k)

which gave Warning: Explicit solution could not be found. > In solve at 179 In Untitled2 at 2

Any idea?

Answer 1

Thanks for the response. I have no reason whatsoever to believe that an analytic solution exists. Actually I wrote up a model and tried to solve it. Since I don't know how to isolate $k$ here I thought maybe I could solve it numerically. However I am not that familiar with numerical methods and it's more like troubleshooting. I took care of the bounds now

syms k a r t1 t2 assume(1 >= k & k >= 0) assumeAlso(1 > a & a > 0) assumeAlso(r > 0) assumeAlso(1 >= t1 & t1 >= 0) assumeAlso(1 >= t2 & t2 >= 0) sol = solve(k^a-(r+t1)*k-(1-k)^a+(r+t2)*(1-k) == 0, k)

which gives Warning: 1 equations in 3 variables. New variables might be introduced. Maybe fsolve is more suitable?! Since $r$, $t_1$ and $t_2$ are endogenous a universal solution like $k_\infty=g(r,t_1,t_2)$ is preferred, however I'm not certain if it actually exists.