Help with nondimensionalization

Wiseguy
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I am working on an assignment for my neuroscience course, and I'm running into a problem with one question which requires me to rewrite an equation into its nondimensionalized form. The equation is given below.

Screenshot_from_2016_02_23_02_51_56.png


and I need to convert it to the form
Screenshot_from_2016_02_23_02_54_11.png


by rescaling and shifting the given variables in the equation as
Screenshot_from_2016_02_23_02_55_40.png


I have already attempted to follow the nondimensionalization procedure given in the Wikipedia article on it, but however I do it, there are some variables on the R.H.S that simply refuse to go away to give me the form required. I'd be grateful for any help. Thanks!
 
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Wiseguy said:
I am working on an assignment for my neuroscience course, and I'm running into a problem with one question which requires me to rewrite an equation into its nondimensionalized form. The equation is given below.

Screenshot_from_2016_02_23_02_51_56.png


and I need to convert it to the form
Screenshot_from_2016_02_23_02_54_11.png


by rescaling and shifting the given variables in the equation as
Screenshot_from_2016_02_23_02_55_40.png


I have already attempted to follow the nondimensionalization procedure given in the Wikipedia article on it, but however I do it, there are some variables on the R.H.S that simply refuse to go away to give me the form required. I'd be grateful for any help. Thanks!
When you replace ##V## by ##\alpha x+ V_0## and ##I_e## by ##\gamma I + I_0## in the equation, what do you get?
Can't you then choose ##\alpha## in such a way that the term in ##x## vanishes?
(Is ##\tau_m>0##?)
 
Samy_A said:
When you replace ##V## by ##\alpha x+ V_0## and ##I_e## by ##\gamma I + I_0## in the equation, what do you get?
Can't you then choose ##\alpha## in such a way that the term in ##x## vanishes?
(Is ##\tau_m>0##?)
Yes, I should have mentioned earlier. ##\tau_m > 0, a_L > 0## and ##V_C > V_L##
 
Wiseguy said:
Yes, I should have mentioned earlier. ##\tau_m > 0, a_L > 0## and ##V_C > V_L##
In that case, why don't you simply do the suggested computation? You will get a term in ##x²## and a term in ##x##. By choosing appropriate values for ##\alpha## and ##V_0## you can make the coefficient of ##x²## equal to 1, and make the term in ##x## vanish.
 
Samy_A said:
In that case, why don't you simply do the suggested computation? You will get a term in ##x²## and a term in ##x##. By choosing appropriate values for ##\alpha## and ##V_0## you can make the coefficient of ##x²## equal to 1, and make the term in ##x## vanish.

I have tried this. Even if the ##x## term vanishes, how about constant terms like ##V_L##, ##V_C##? Also, no value I choose for ##\alpha## is helping me make the current term ##\frac{R_mI_0}{\alpha\tau_m}## vanish simultaneously.
 
Wiseguy said:
I have tried this. Even if the ##x## term vanishes, how about constant terms like ##V_L##, ##V_C##? Also, no value I choose for ##\alpha## is helping me make the current term ##\frac{R_mI_0}{\alpha\tau_m}## vanish simultaneously.
Can you show your calculation, because I think I managed to do it (of course maybe I have an error in my calculation). Don't forget that after having "fixed" ##x## by choosing appropriate ##\alpha,\ V_0##, you can choose ##\gamma## and ##I_0## to get the desired result.
 
Samy_A said:
Can you show your calculation, because I think I managed to do it (of course maybe I have an error in my calculation). Don't forget that after having "fixed" ##x## by choosing appropriate ##\alpha,\ V_0##, you can choose ##\gamma## and ##I_0## to get the desired result.
I can describe the steps I've tried taking so far.

1. Substitute for both ##V## and ##I_e## as given on both sides of the equation.
2. L.H.S then becomes ##\alpha\tau_m\frac{dx}{dt}##, since ##\frac{dV_0}{dt}## term equals 0.
3. Now divide throughout by ##\alpha\tau_m## to get just ##\frac{dx}{dt}## as needed.
4. Substitute for ##\alpha## with ##\frac{\tau_m}{a_L}## to get coefficient of ##x^2## term as 1.

I'm unclear after this. What can I substitute for ##V_0## to make all those terms disappear?
 
Wiseguy said:
I can describe the steps I've tried taking so far.

1. Substitute for both ##V## and ##I_e## as given on both sides of the equation.
2. L.H.S then becomes ##\alpha\tau_m\frac{dx}{dt}##, since ##\frac{dV_0}{dt}## term equals 0.
3. Now divide throughout by ##\alpha\tau_m## to get just ##\frac{dx}{dt}## as needed.
4. Substitute for ##\alpha## with ##\frac{\tau_m}{a_L}## to get coefficient of ##x^2## term as 1.

I'm unclear after this. What can I substitute for ##V_0## to make all those terms disappear?
After setting ##\alpha =\frac{\tau_m}{a_L}##, what is the coefficient of ##x## is your RHS? Can you chose ##V_0## so that the term in ##x## disappears?
 
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