 Passive spatial and conductance integration

The spatial panel uses two compartments with membrane potentials $V_s,V_d$, one representing the soma and one a dendrite. Set $v_s=V_s-E_L$, $v_d=V_d-E_L$, with leak reversal potential $E_L=-65\mathrm{mV}$, each membrane capacitance $C=100\mathrm{pF}$, each leak electrical conductance $g_L=5\mathrm{nS}$ and axial electrical conductance $g_c=10\mathrm{nS}$. Electric charge conservation givesThis chosen passive model has two decay time constants, $C/g_L=20\mathrm{ms}$ for the equal-voltage mode and $C/(g_L+2g_c)=4\mathrm{ms}$ for the voltage-difference mode. Starting from rest, a $100\mathrm{pA}$ step into the soma gives, with $u$ the time since onset,while the same step into the dendrite gives$u$ is expressed in $\mathrm{ms}$. Their long-time rises are $12$ and $8\mathrm{mV}$ respectively. Applying both inputs gives $20(1-e^{-u/20})\mathrm{mV}$ exactly: spatial attenuation coexists with linear summation of prescribed electric currents. After the steps end, both modes decay with their respective time constants. This deliberately coarse model is not a reconstructed dendrite or a fit to the experiment above.

The conductance panels use the existing one-compartment synaptic potential equation, with $C=200\mathrm{pF}$, $g_L=10\mathrm{nS}$, $E_L=-65\mathrm{mV}$ and excitatory reversal potential $E_e=0\mathrm{mV}$. For a constant excitatory synaptic conductance $g_e$, the steady state voltage rise and membrane time constant areEach isolated $2\mathrm{nS}$ input has a limiting rise $130/12\approx 10.833\mathrm{mV}$. Adding those isolated rises predicts $21.667\mathrm{mV}$, but opening both together gives $260/14\approx 18.571\mathrm{mV}$. The two conductances share a changing driving force. This is sublinear input summation even though the voltage ordinary differential equation is linear for each prescribed conductance waveform.

Add $g_i=20\mathrm{nS}$ with reversal potential $E_i=E_L$ to the joint $4\mathrm{nS}$ excitation. The limiting rise becomes $260/34\approx 7.647\mathrm{mV}$ and the membrane time constant becomes $200/34\approx 5.882\mathrm{ms}$. The shunt alone leaves rest unchanged. At the joint-excitation-only steady state $V\approx -46.429\mathrm{mV}$, however, its inward electric current would be $20(-65+46.429)\approx -371.429\mathrm{pA}$: it opposes that excursion. The finite $50\mathrm{ms}$ traces in the diagram approach, but need not reach, these limiting voltages. See shunting inhibition for the constant-electric current comparison.