The cable equation A.K.A. the monodomain model Neurons Electric - PowerPoint PPT Presentation

The cable equation A.K.A. the monodomain model

Neurons

Electric flow in neurons The neuron consists of three parts: Dendrite-tree, the “input stage” of the neuron, converges on the soma. Soma, the cell body, contain the “normal” cellular functions Axon, the output of the neuron, may be branched. Synapses at the ends are connected to neighboring dendrites. The axon has an excitable membrane, gives rise to active conduction. Will first look at conduction in the dentrites, passive conduction.

The cable equation, 4.1 The cell typically has a potential gradient along its length. Radial components will be ignored. Notation: V i and V e are intra- and extra cellular potential I i and I e are intra- and extra cellular (axial) current r i and r e are intra- and extra cellular resistance per unit length r i = R c A i where R c is the cytoplasmic resistivity and A i is the cross sectional area of the cable.

Discrete cable

Ohmic resistance assumed: V i ( x + ∆ x ) − V i ( x ) = − I i ( x ) r i ∆ x V e ( x + ∆ x ) − V e ( x ) = − I e ( x ) r e ∆ x In the limit: I i = − 1 ∂ V i I e = − 1 ∂ V e and r i ∂ x r e ∂ x

Conservation of current yields: I i ( x ) − I i ( x + ∆ x ) = − ( I e ( x ) − I e ( x + ∆ x )) = I t ∆ x (1) where I t is transmembrane current, per unit length. In the limit (1) yields: I t = − ∂ I i ∂ x = ∂ I e ∂ x We would like to express I t in terms of V . ∂ 2 V e ∂ 2 V i ( ∂ 2 V ∂ x 2 + ∂ 2 V e 1 ∂ x 2 = − 1 ∂ x 2 = − 1 ∂ x 2 ) r e r i r i ) ∂ 2 V e ∂ 2 V ( 1 + 1 ∂ x 2 = − 1 ∂ x 2 r e r i r i

cont. ) ∂ 2 V e ∂ 2 V ( 1 + 1 ∂ x 2 = − 1 ∂ x 2 r e r i r i 1 ∂ 2 V e ∂ 2 V ∂ 2 V r e r i ∂ x 2 = − ∂ x 2 = − r e + 1 1 r e + r i ∂ x 2 r i so ∂ 2 V e ∂ 2 V I t = ∂ I e ∂ x = − 1 1 ∂ x 2 = ∂ x 2 r e r e + r i

From the membrane model previously derived we have ∂ V I t = p ( C m ∂ t + I ion ) where p is the circumference of the cable. The total expression will be in Ampere/meter. The total 1D cable model is then: ∂ 2 V ∂ V 1 p ( C m ∂ t + I ion ( V )) = ( ∂ x 2 ) r e + r i

Dimensionless form We can scale the variables to reduce the number of parameters. Defines a membrane resistance: 1 = ∆ I ion ∆ V ( V 0 ) R m where V 0 is the resting potential. Multiplication with R m ∂ 2 V ∂ V R m C m R m ∂ t + R m I ion = p ( r i + r e ) ∂ x 2 Here we have assumed r i and r e constant. Defining f = − R m I ion , τ m = C m R m (time constant) and λ 2 m = R m / ( p ( r i + r e )) (space constant squared) we can write ∂ 2 V ∂ V ∂ t − f = λ 2 τ m (2) m ∂ x 2

Introduces the dimensionless variables: T = t /τ m and X = x /λ m We can then write: ∂ T = f + ∂ 2 V ∂ V (3) ∂ X 2 A solution ˆ V ( T , X ) of (3) will imply that V ( t , x ) = ˆ V ( t /τ m , x /λ m ) will satisfy (2).

The reaction term, 4.2 The form of f depends on the cell type we want to study. For the axon I ion ( m , n , h , V ) of the HH-model is a good candidate. For the dendrite, which is non-excitable, a linear resistance model is good. Shift V so V = 0 is the resting potential: ∂ T = ∂ 2 V ∂ V ∂ X 2 − V Need boundary and initial values. Initially at rest: V ( X , 0) = 0

Boundary conditions Types of boundary conditions: Dirichlet: V ( x b , T ) = V b , voltage clamp. Neumann: ∂ V ∂ X = − r i λ m I , current injection. Justification: ∂ V i ∂ x = − r i I i ⇒ ∂ V ∂ x − ∂ V e ⇒ ∂ V r e =0 ∂ x = − r i I i = ∂ x = − r i I i

Branching structures, 4.2.3

Linear cable equation used in each branch: ∂ T = ∂ 2 V ∂ V ∂ X 2 − V General solution in the steady state: V = Ae − X + Be X Two parameters per branch, six in total to determine. Three taken from boundary conditions: current injection in X=0 and voltage clamp at X = L 21 and X = L 22 . Two more from continuity of voltage: V 1 ( L 1 ) = V 21 ( L 1 ) = V 22 ( L 1 )

The sixth condition is obtained from continuity of current: 1 dV 1 1 dV 21 1 dV 22 dX = + R 1 , in R 21 , in dX R 22 , in dX where the input resistance is � R m R c R in = λ m r i = pr i A i Assuming a circular crossection: R in = 2 � R m R c d − 3 / 2 π If R m and R c is not changing the condition becomes: dV 1 dV 21 dV 22 d 3 / 2 dX = d 3 / 2 + d 3 / 2 1 21 22 dX dX

Equivalent cylinders With certain assumptions the dendrite tree can be modelled with a single cable equation. L 21 = L 22 , and they have the same boundary conditions: This gives V 21 = V 22 and thus: dV 1 22 ) dV 21 d 3 / 2 dX = ( d 3 / 2 21 + d 3 / 2 1 dX The critcal assumption is then: d 3 / 2 = d 3 / 2 21 + d 3 / 2 1 22 If so, then we can use a single equation for the whole system. Similar arguments can be made for more complex branching.