# Help with nondimensionalization

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1. Feb 23, 2016

### Wiseguy

• Member warned about posting without the HW template
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.

and I need to convert it to the form

by rescaling and shifting the given variables in the equation as

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!

2. Feb 23, 2016

### Samy_A

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 $\tau_m>0$?)

3. Feb 23, 2016

### Wiseguy

Yes, I should have mentioned earlier. $\tau_m > 0, a_L > 0$ and $V_C > V_L$

4. Feb 23, 2016

### Samy_A

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.

5. Feb 23, 2016

### Wiseguy

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.

6. Feb 23, 2016

### Samy_A

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.

7. Feb 23, 2016

### Wiseguy

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?

8. Feb 23, 2016

### Samy_A

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?