A computational approach for the inverse problem of neuronal conductances determination

Abstract

The derivation by Alan Hodgkin and Andrew Huxley of their famous neuronal conductance model relied on experimental data gathered using the squid giant axon. However, the experimental determination of conductances of neurons is difficult, in particular under the presence of spatial and temporal heterogeneities, and it is also reasonable to expect variations between species or even between different types of neurons of the same species.

We tackle the inverse problem of determining, given voltage data, conductances with non-uniform distribution in the simpler setting of a passive cable equation, both in a single or branched neurons. To do so, we consider the minimal error iteration, a computational technique used to solve inverse problems. We provide several numerical results showing that the method is able to provide reasonable approximations for the conductances, given enough information on the voltages, even for noisy data.

Appendix A: Abstract Formulation

In practice, V |_Γ is the data and given such information and under the assumption that Eq. (3) holds, the inverse problem under consideration is to recover or approximate the conductances. The lack of stability, characteristic of ill-posed problems can be tamed by regularization methods (Engl et al. 1996; Kaltenbacher et al. 2008; Kirsch 2011), in particular by the minimal error method.

Consider for simplicity T > 0. Let Ω = {(t, x) : 0 ≤ t ≤ T, 0 ≤ x ≤ L}, and

$$ \begin{array}{@{}rcl@{}} H(F)=\bigl(L^{2}({\Omega})\bigr)^{N_{{\text{ion}}}} =\biggl\{f:{\Omega}\rightarrow\mathbb{R}^{N_{{\text{ion}}}}:{\int}_{\Omega}|f(\xi)|^{2}d\xi<\infty \biggr\}, \\ R(F)=L^{2}({\Gamma})=\biggl\{f:{\Gamma} \rightarrow \mathbb{R}:{\int}_{\Gamma}|f(\xi)|^{2}d\xi<\infty\biggr\}. \end{array} $$

It is well-known that H(F) and R(F) become Hilbert spaces under the inner products

$$ \begin{array}{@{}rcl@{}} {\langle f,h\rangle_{H(F)}={\int}_{\Omega}f(\xi)h(\xi)d\xi,}\\ {\langle f,h\rangle_{R(F)}={\int}_{\Gamma}f(\xi)h(\xi)d\xi,}. \end{array} $$

and the associated norms $\|f\|_{H(F)}=\langle f,f\rangle _{H(F)}^{1/2}$, $\|f\|_{R(F)}=\langle f,f\rangle _{R(F)}^{1/2}$. Note that the inner product on R(F) depends on Γ, see Eq. (11), as follows:

$$ \begin{array}{@{}rcl@{}} \langle f,h\rangle_{R(F)} &=& \alpha_{1}{{\int}_{0}^{L}}{{\int}_{0}^{T}}f(t,x)h(t,x)dtdx\\ &&+\alpha_{2}{{\int}_{0}^{T}}f(t,0)h(t,0)dt +\alpha_{2}{{\int}_{0}^{T}}f(t,L)g(t,L)dt,\\ \end{array} $$

(26)

where α₁, α₂ are as in Eq. (11).

The set $D(F)=\left (L^{\infty }({\Omega })\right )^{N_{{\text {ion}}}}\subset H(F)$ is the Banach space of “essentially” bounded functions (see Kreyszig (1978) for precise definitions). Consider the operator $F:D(F)\subset H(F)\rightarrow R(F)$ defined by F(G) = V |_Γ. Our goal is to find an approximation for G using the minimal error iteration defined by Eq. (6).

In the next Theorem we show how to obtain Eq. (9) from Eq. (6).

Theorem 1

Consider the iteration in Eq. (6). Then Eq. (9) holds

Proof

Given G^{k, δ} ∈ D(F) and $\boldsymbol {\theta }=(\theta _{1},\dots ,\theta _{N_{{\text {ion}}}})\in H(F)$, the Gâteux derivative of F at G^{k, δ} in the direction 𝜃 is given by

$$ F^{\prime}(\boldsymbol{G}^{k,\delta})(\boldsymbol{\theta})=\lim_{\lambda\to0}\frac{F(\boldsymbol{G}^{k,\delta}+\lambda\boldsymbol{\theta})-F(\boldsymbol{G}^{k,\delta})}{\lambda}=W^{k}|_{\Gamma}, $$

(27)

where W^k solves

$$ \left \{\begin{array}{ll} \displaystyle\frac{r_{a}}{2R}W_{xx}^{k}(t,x)-C_{M}{W_{t}^{k}}(t,x)-G_{L}W^{k}(t,x)\\ \displaystyle\qquad\qquad\qquad\qquad\qquad-\underset{i\in{\text{Ion}}}{\sum}G_{i}^{k,\delta}(t,x)W^{k}(t,x)=\underset{i\in{\text{Ion}}}{\sum}\theta_{i}(V^{k,\delta}(t,x)-E_{i})\quad\text{ in }{\Omega},\\ W^{k}(0,x)=0\quad\text{for }0<x<L,\qquad\\ {W^{k}_{x}}(t,0)={W^{k}_{x}}(t,L)=0\quad\text{for }0<t<T, \end{array} \right. $$

(28)

and V^{k, δ} solves Eq. (3) with G_i replaced by $G_{i}^{k,\delta }$. To obtain Eq. (28) from Eq. (27), it is enough to consider the difference between problem in Eq. (3) with coefficients G^{k, δ} + λ𝜃 and G^{k, δ}, divide by λ and take the limit λ → 0.

Let $V^{k,\delta }|_{\Gamma }=F(\boldsymbol {G}^{k,\delta })$. From the minimal error iteration in Eq. (6), we gather that

$$ \begin{array}{@{}rcl@{}} {\langle}\boldsymbol{G}^{k+1,\delta}-\boldsymbol{G}^{k,\delta},\boldsymbol{\theta}{\rangle}_{H(F)}&=&w^{k,\delta}{\langle} F^{\prime}(\boldsymbol{G}^{k,\delta})^{*}(V^{\delta}|_{\Gamma}-F(\boldsymbol{G}^{k,\delta})),\boldsymbol{\theta}{\rangle}_{H(F)} \\ &=&w^{k,\delta}{\langle} F^{\prime}(\boldsymbol{G}^{k,\delta})^{*}(V^{\delta}|_{\Gamma}-V^{k,\delta}|_{\Gamma}),\boldsymbol{\theta}{\rangle}_{H(F)}. \end{array} $$

By definition of adjoint operator,

$$ \begin{array}{@{}rcl@{}} {\langle}\boldsymbol{G}^{k+1,\delta}-\boldsymbol{G}^{k,\delta},\boldsymbol{\theta}{\rangle}_{H(F)}=w^{k,\delta}{\langle} V^{\delta}|_{\Gamma}-V^{k,\delta}|_{\Gamma},F^{\prime}(\boldsymbol{G}^{k,\delta})(\boldsymbol{\theta})\rangle_{R(F)} \\ =w^{k,\delta}{\langle} V^{\delta}|_{\Gamma}-V^{k,\delta}|_{\Gamma},W^{k}|_{\Gamma} {\rangle}_{R(F)}, \end{array} $$

(29)

from Eq. (27).

Although Eq. (29) yields an interesting relation, it carries an impeding dependence on 𝜃 through W^k. It is possible to avoid that by performing some “trick” manipulations.

Multiplying the first equation from (10) by − W^k, and integrating in the intervals [0, T] and [0, L] we gather that

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\frac{r_{a}}{2R} U^{k}_{xx}(t,x)W^{k}(t,x)dtdx\\ &&+{{\int}_{0}^{L}}{{\int}_{0}^{T}}C_{M}{U^{k}_{t}}(t,x)W^{k}(t,x)dtdx\\ &&-{{\int}_{0}^{L}}{{\int}_{0}^{T}}G_{L}U^{k}(t,x)W^{k}(t,x)dtdx\\ &&-{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i \in \text{ion}}{\sum}G_{i}^{k,\delta}(t,x)U^{k}(t,x)W^{k}(t,x)dtdx= \\ &-&\alpha_{1}{{\int}_{0}^{L}}{{\int}_{0}^{T}} \left( V^{\delta}(t,x)-V^{k,\delta}(t,x)\right) W^{k}(t,x)dtdx. \end{array} $$

(30)

Integrating by parts twice the first term from Eq. (30) with respect to the space variable, and using the boundary conditions for W^k we have

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\frac{r_{a}}{2R} U^{k}_{xx}(t,x)W^{k}(t,x)dtdx\\ &&={{\int}_{0}^{L}}{{\int}_{0}^{T}}\frac{r_{a}}{2R}U^{k}(t,x)W_{xx}^{k}(t,x)dtdx\\ &&+\frac{r_{a}}{2R}{{\int}_{0}^{T}} {U^{k}_{x}}(t,x)W^{k}(t,x)|_{0}^{L}dt, \end{array} $$

(31)

where we denote ${U^{k}_{x}}(t,x)W^{k}(t,x)|_{0}^{L}=U^{k}(t,L)W^{k}(t,L)$ − U^k(t, 0)W^k(t, 0). Similarly, integrating by parts the second term from Eq. (30) with respect to time and using the initial condition of W^k and the final condition of U^k, we gather that

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}C_{M}{U^{k}_{t}}(t,x)W^{k}(t,x)dtdx\\ &&=-{{\int}_{0}^{L}}{{\int}_{0}^{T}}C_{M}U^{k}(t,x){W_{t}^{k}}(t,x)dtdx. \end{array} $$

(32)

Substituting Eqs. (31) and (32) in Eq. (30), it follows that

$$ \begin{array}{@{}rcl@{}} {{\int}_{0}^{L}}{{\int}_{0}^{T}}\Biggl(\frac{r_{a}}{2R} W_{xx}^{k}(t,x)-C_{M}{W_{t}^{k}}(t,x)-G_{L}W^{k}(t,x)-\underset{i \in ion}{\sum}G_{i}^{k,\delta}(t,x)W^{k}(t,x) \Biggr)U^{k}(t,x)dt dx \\ =-\alpha_{1}{{\int}_{0}^{L}}{{\int}_{0}^{T}} \bigl(V^{\delta}(t,x)-V^{k,\delta}(t,x)\bigr)W^{k}(t,x)dtdx-\frac{r_{a}}{2R}{{\int}_{0}^{T}}{U^{k}_{x}}(t,x)W^{k}(t,x)|_{0}^{L}dt. \end{array} $$

Substituting the first equation from Eq. (28) in the previous equation, we obtain

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i \in {\text{Ion}}}{\sum}\theta_{i}(V^{k,\delta}(t,x)-E_{i}) U^{k}(t,x) dt dx \\ &=&-\alpha_{1}{{\int}_{0}^{L}}{{\int}_{0}^{T}} \left( V^{\delta}(t,x)-V^{k,\delta}(t,x)\right) W^{k}(t,x)dt dx\\ &&-\frac{r_{a}}{2R}{{\int}_{0}^{T}}{U^{k}_{x}}(t,x)W^{k}(t,x)|_{0}^{L}dt. \end{array} $$

From the boundary conditions from Eq. (10), the following expression holds:

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i \in {\text{Ion}}}{\sum}\theta_{i}(V^{k,\delta}(t,x)-E_{i}) U^{k}(t,x) dt dx= \\ &&-\alpha_{1}{{\int}_{0}^{L}}{{\int}_{0}^{T}} \left( V^{\delta}(t,x)-V^{k,\delta}(t,x)\right) W^{k}(t,x)dt dx\\ &&-\alpha_{2}{{\int}_{0}^{T}}\left( V^{\delta}(t,0)-V^{k,\delta}(t,0)\right)W^{k}(t,0) \\ &&-\alpha_{2}{{\int}_{0}^{T}}\left( V^{\delta}(t,L)-V^{k,\delta}(t,L)\right)W^{k}(t,L)dt. \end{array} $$

From the previous equation and the definition of the inner product in Eq. (26), we have

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i \in {\text{Ion}}}{\sum}\theta_{i}(V^{k,\delta}(t,x)-E_{i}) U^{k}(t,x) dt dx\\ &&\qquad=-{\langle} V^{\delta}|_{\Gamma}-V^{k,\delta}|_{\Gamma},W^{k}|_{\Gamma} {\rangle}_{R(F)}. \end{array} $$

(33)

From Eqs. (29) and (33) we have

$$ \begin{array}{@{}rcl@{}} &&{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i\in{\text{Ion}}}{\sum}\theta_{i}\left( G_{i}^{k+1,\delta}(t,x)-G_{i}^{k,\delta}(t,x)\right)dt dx \\ &&=-w^{k,\delta}{{\int}_{0}^{L}}{{\int}_{0}^{T}}\underset{i\in{\text{Ion}}}{\sum}\theta_{i}(V^{k,\delta}(t,x)-E_{i}) U^{k}(t,x) dt dx. \end{array} $$

Since 𝜃 ∈ H(F) is arbitrary and $L^{\infty }({\Omega })$ is dense in L²(Ω), we gather that the following iteration holds:

$$ \begin{array}{@{}rcl@{}} G_{i}^{k+1,\delta}(t,x)&=&G_{i}^{k,\delta}(t,x)-w^{k,\delta}(V^{k,\delta}(t,x)-E_{i})U^{k}(t,x) \end{array} $$

for all i ∈Ion.

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