Solving for a steady-state (passive cable equation)

[rall1959]

TL;DR Summary

Trying to understand the difference between same assumptions pre- and post- ODE solving.

When trying to model a neuronal dendrite or axon, the cylindrical shape is preferred and cable core conductor theory is applied. Considering only passive current, that is, non-voltage-dependent current, such as leakage potassium current, the general form of the cable equation is ∂²V_m/∂X² = ∂V_m/∂T + V_m, where X = x/λ and T = t/τ, λ the space constant and τ the time constant. This equation can then be used to solve for different special cases.

My question is the following: in order to derive a solution for the special case of a steady-state, one may either (a) consider ∂V_m/∂T = 0, and so V_m = ∂²V_m/∂X², and then solve for V_m (which renders a solution of the form V_m = A_1*exp(X) + A_2*exp(-X)) or (b) one may solve the full cable equation to a general form (V_m = (I_0*r_i*λ/4)*(exp(-X)*erfc(X/(2*sqrt(T)) - sqrt(T)) + exp(X)*erfc(X/(2*sqrt(T)) + sqrt(T))), then consider the steady-state as T -> infinity, which renders the solution V_m = (I_0*r_i*λ/2)*exp(-X), with no component A_1*exp(X), when compared to case (a). It is as if we implicitly consider A_1 = 0 in case (b), but I don't understand why.

This means there is a difference between considering the steady-steady before (case (a)) or after (case (b)) solving the differential equation. Can someone explain this to me? Thank you.

 

[renormalize]

Can someone explain this to me?

You've defined the cable partial differential equation:$$\frac{\partial^{2}V_{m}\left(X,T\right)}{\partial X^{2}}=\frac{\partial V_{m}\left(X,T\right)}{\partial T}+V_{m}\left(X,T\right)\tag{1}$$but to solve this PDE you also need to specify one initial condition for some time (call it $T=0$) and boundary values at two positions (call them $X=0,X=L$):$$V_{m}\left(X,0\right)=f\left(X\right),\quad V_{m}\left(0,T\right)=g_{0}\left(T\right),\quad V_{m}\left(L,T\right)=g_{L}\left(T\right)\tag{2a,b,c}$$So to answer your question we need to know: what are the initial/boundary conditions (2) for your problem?
(Also, it's easier for everyone to read if you post equations in LaTeX; there's a handy guide at the lower left of your post.)

 

[rall1959]

Hello @renormalize, thanks for your answer (and tip on using LaTeX; I have tried to edit my previous post but I haven't found a way to do it, sorry).

Eq. (1) has a known general solution: $$V_m (X,T) = \frac {I_0\lambda r_i}{4} (e^{-X}erfc(\frac {X}{2 \sqrt {T}} - \sqrt {T}) - e^{X}erfc(\frac {X}{2 \sqrt {T}} + \sqrt {T}))$$ This equation describes the evolution of membrane voltage over time and space across a neuronal dendrite or axon, given only passive attenuation. If we inject a constant current $I_0$ and let the system reach a steady-state, we have $T = \infty$, and given the properties of the error function we get the special solution $$V_m(X,\infty) = \frac {I_0\lambda r_i}{2}e^{-X}.$$ My question is, had we assumed the steady-state prior to solving the PDE ($\frac {\partial V_m(X,T)} {\partial T} = 0$), we would have gotten the second-order ODE $$ \frac {\partial^2V_m(X)} {\partial X^2} = V_m(X)$$ which has a solution of the form $$V_m(X) = A_1 e^{X} + A_2 e^{-X}.$$ Now, this is the same as the previous solution (obtained after solving the PDE and assuming $T = \infty$), provided $$A_1 = 0$$ and $$A_2 = \frac {I_0\lambda r_i}{2}.$$ I fail to understand why this is so.

 

[renormalize]

I fail to understand why this is so.

You still haven't stated the boundary conditions for your PDE, yet those are essential to answer your question. For example, if your cable is semi-infinite in length ($0\le X\le\infty$) you might specify the values $V_m\left(0,T\right)$ and $V_m\left(\infty,T\right)$, as well the value $V_m\left(X,T_0\right)$ for some particular time $T_0$ (which could be at infinite time). That seems to be the case for the specific solution you wrote:$$V_{m}\left(X,T\right)=k\left[e^{-x}\text{erfc}\left(\frac{X}{2\sqrt{T}}-\sqrt{T}\right)+e^{x}\text{erfc}\left(\frac{X}{2\sqrt{T}}+\sqrt{T}\right)\right]\tag{1}$$for it satisfies the boundary conditions:$$V_{m}\left(0,T\right)=2k,\quad V_{m}\left(\infty,T\right)=0,\quad V_{m}\left(X,\infty\right)=2ke^{-x}\tag{2a,b,c}$$ Now, if you instead assume the steady state first, the PDE becomes the ODE:$$\frac{d^{2}V_{m}\left(X\right)}{dX^{2}}=V_{m}\left(X\right)\tag{3}$$with the solution$$V_{m}\left(X\right)=A_{1}e^{X}+A_{2}e^{-X}\tag{4}$$But the point is, this solution must still satisfy the same boundary conditions (2a,b), but with the time $T$ removed, namely:$$V_{m}\left(0\right)=2k,\quad V_{m}\left(\infty\right)=0\tag{5a,b}$$which implies finally:$$A_{1}=0,\quad A_{2}=2k\tag{6a,b}$$This makes solution (4) the same as the long-time solution (2c), as it must be.
Does this answer your question?

 

[rall1959]

This was almost correct. Solution (1) actually has a minus sign before the second exponential ($... -e^{X} \text{erfc}(\frac {X} {2 \sqrt {T}} + \sqrt{T})$) which implies $V_m (0,T) = 2k\text{erf}(\sqrt{T})$. Naturally, if $T = \infty$, $\text {erf}(\infty) = 1$ and $V_m (0,\infty) = 2k$, but the full boundary condition is the former.
The other two boundary conditions are correct for a semi-infinite cable, and a steady-state, respectively.

Does this answer your question?

Yes, thank you. I now understand the problem much better (and learned a little bit of LaTeX on the way). Basically, the general solution (1) is a door to all situations. Then one uses boundary conditions to derive specific solutions according to physical constraints. For instance, if the cable equation is modelling an axon, which is quite long, we may assume the semi-infinite cable model, which takes on boundary conditions (2a,b). In this case we do not assume the steady-state from start, because we also want to study transient voltage changes at each boundary point (besides the steady-state). However, if the cable equation is to model a dendrite, which is much shorter, we cannot assume the semi-infinite cable model, but we may instead just want to know what the steady-state membrane voltage will be within a short distance of current injection. In that case, we assume $\frac {dV_m (X,T)} {dT} = 0$ from the start.
This was very much enlightening, @renormalize, thanks again for your help and patience.