Is there a fluctuation effect between heat, resistance and current?

This could be analyzed in the same way as a control circuit with feedback. From a practical sense, the heating will be much slower than the other effects, so that will dominate the loop equations. As such, it will exponentially approach equilibrium, unless there's other elements of the system that limit its response (ridiculously enormous inductors, state machines introducing delays, etc).

This is something like a PTC thermistor. which will reach an equilibrium temperature.

To get oscillation you'd have to have a phase shift or delay of some kind. You could probably make an oscillator with a mass transport delay having a a heater heat water flowing in a tube which warms a thermistor downstream and increases the heat to the upstream heater.

I believe it's possible to build a simple physical model with the ideas you provided.

In a simple DC circuit, under a constant voltage V and ohmic resistance R it is possible to use the power equation: $$P = V i = \frac{V^2}{R}$$

If we suppose the system is made of a wire with constant lenght L and cross section area A, the resistance R can be: $$R = \rho \frac{L}{A}, \: where \:\: \rho = resistivity$$

For small temperature T oscillations, the resistivity can be aproximated to: $$\rho = \rho_0(1 + \alpha(T - T_0) ) = \rho_0(1 + \alpha \Delta T)$$

And since there's only solid material heating, the power recieved by the wire is: $$P = \frac{dQ}{dt} = \frac{d}{dt}(mcT) = mc \dot{T} = mc \Delta \dot{T}, \: where \:\: \Delta \dot{T} = \frac{d\Delta T}{dt} = \frac{dT}{dt}$$ Finally, all of this togheter becomes: $$mc \Delta \dot{T} = \frac{V^2 A}{\rho_0 L}\frac{1}{1 + \alpha \Delta T} \Rightarrow \frac{mc \rho_0 L}{V^2 A}\Delta \dot{T} = \frac{1}{1 + \alpha \Delta T}$$ I don't know how to solve this analitically, but there's a valid approximation since I am working with small temperature fluctuations: $$ \frac{1}{1 + \alpha \Delta T} \approx 1 - \alpha \Delta T$$ Now, we can solve it: $$\frac{mc \rho_0 L}{V^2 A}\Delta \dot{T} + \alpha \Delta T - 1 = 0$$

And the solution is: $$\Delta T = C e^{ - t/\tau } + \frac{1}{\alpha} \, , \: where \: \tau = \frac{m c L \rho_0}{\alpha A V^2} \: \: and \: \: C = cte $$

In this model, we see a transient solution followed by a constant one. But remember this is valid just for small temperature fluctuations.