solve ODE with functions that are variable dependent

Question

I have the following set of equations:

$\frac{dv}{dt}=v-v_s(v,u)$

$\frac{du}{dt}=u-u_s(v,u)$

$v_s(v,u)=b \left(\frac{1}{1+a(u,v)}-\frac{e^{-v}}{a(u,v)+e^{-v}}\right)$

$u_s(v,u)=b \left(\frac{e^{u}}{a(u,v)+e^{u}}-\frac{1}{1+a(u,v)}\right)$

$a(v,u)=\frac{1}{u}\left(1-e^{-v}\right)+\frac{1}{v}\left(1-e^u\right)$

with the following initial conditions:

$u(0)=u_0$, $v(0)=v_0$

I need to obtain $v_s$ and $u_s$ first for every step $t$, and then plug them in the differential equations of $\frac{dv}{dt},\frac{du}{dt}$.

Since $v_s$ and $u_s$ have a recursion relation I can't manage to write a script that will solve these equations.

This is what I've obtained so far:

a[v_, u_] := (1/u) (1 - E^(-v)) + (1/v) (1 - E^u);
vs[v_, u_] := b (1/(a[v, u] + 1) - (E^(-v)))/(E^(-v) + a[v, u]);
us[v_, u_] := b (E^u/(E^u + a[v, u]) - 1/(1 + a[v, u]));
b = 1; u0 = 1; v0 = 1;
NDSolve[{v'[t] == v[t] - vs[v[t], u[t]], u'[t] == u[t] - us[v[t], u[t]],
   v[0] == v0, u[0] == u0}, {v[t], u[t]}, {t, 0, 100}];

Full guidance or a scheme that will generalize this problem will be accepted as an answer.

Answer 1

If $u$ and $v$ depend on $t$, this must be set whenever they appears, if you do not want to define functions, unless you need them afterwards.

I have just retyped your code, and plot the solutions. Please, take a look, I am not sure what you are expecting though.

Just an additional note: the range for $t$ was shortened to avoid overflow in the calculations by NDSolve. I assume that you will take care of this in your application.

Hope this helps:

a = (1/u[t]) (1 - E^(-v[t])) + (1/v[t]) (1 - E^u[t]);
vs[t] = b (1/(a + 1) - (E^(-v[t])))/(E^(-v[t]) + a);
us[t] = b (E^u[t]/(E^u[t] + a) - 1/(1 + a));
b = 1;

sol = NDSolve[{v'[t] == v[t] - vs[t], u'[t] == u[t] - us[t], v[0] == 1, u[0] == 1}, {v[t], u[t]}, {t, 0, 40}];

{u[t], v[t]} = {u[t], v[t]} /. sol[[1]];
Plot[{u[t], v[t]}, {t, 0, 40}, PlotRange -> {0, 1000},PlotStyle -> {Blue, Red}, PlotLegends -> "Expressions"]