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kalish1
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I have constructed a system of an arbitrary number of ordinary differential equations that describes my model at steady-state. There are ($i+1$) ODEs, with $i$ arbitrary.
Goal: I want to solve the resulting algebraic system (all equations set to $0$) and obtain an analytical expression that captures the dependence of $S_T = \sum\limits_{j=1}^{i+1} S_j$ on $\alpha$.
______
Here is the system:
$$ Q-\left(\dfrac{\alpha}{X} + r_1\right)S_1+\dfrac{\beta}{Y}\sum\limits_{j=2}^{i+1} S_j = 0 \\
\dfrac{\alpha}{X}(S_1-S_i)-\left(\dfrac{\beta}{Y}+r_1\right)\sum\limits_{j=2}^{i} S_j = 0 \\
\dfrac{\alpha}{X}S_i-\left(\dfrac{\beta}{Y}+\sigma r_1 \right)S_{i+1} = 0
$$
The first equation is the ODE for $S_1$, the last is for $S_{i+1}$, and the middle is for $S_2$ to $S_{i}$. I know beforehand that the parameters $Q, \alpha, X, r_1, \beta, Y$ and all $S_j$'s are positive real numbers and $\sigma > 1.$
____
Work in progress:
Note that $$\frac{d}{dt}S_T = \frac{d}{dt} \left(\sum\limits_{j=1}^{i+1}S_j\right) = Q - r_1 \sum\limits_{j=1}^{i} S_i - \sigma r_1 S_{i+1} = Q - r_1 S_T + r_1 (1-\sigma) S_{i+1},$$ implying that $$S_T = \frac{Q}{r_1} + (1-\sigma)S_{i+1} - \frac{1}{r_1} \dot{S}_T.$$
Now notice that upon solving for $S_1$ in the first and second ODE equations respectively, we obtain:
$$S_1 = \dfrac{X}{Y(X r_1 + \alpha)}\left(QY + \beta \sum\limits_{j=2}^{i+1}S_j\right)$$
and
$$S_1 = \dfrac{X(\beta+Y r_1)}{\alpha Y} \sum\limits_{j=2}^{i} S_j + S_i.$$
Question: How can I find $S_{i+1}$ explicitly so that I can (hopefully) solve the linear ODE for $S_T$?
Thanks in advance for help.
This question has been crossposted here: http://math.stackexchange.com/questions/1537553/solving-a-system-of-an-arbitrary-number-of-odes-at-steady-state
Goal: I want to solve the resulting algebraic system (all equations set to $0$) and obtain an analytical expression that captures the dependence of $S_T = \sum\limits_{j=1}^{i+1} S_j$ on $\alpha$.
______
Here is the system:
$$ Q-\left(\dfrac{\alpha}{X} + r_1\right)S_1+\dfrac{\beta}{Y}\sum\limits_{j=2}^{i+1} S_j = 0 \\
\dfrac{\alpha}{X}(S_1-S_i)-\left(\dfrac{\beta}{Y}+r_1\right)\sum\limits_{j=2}^{i} S_j = 0 \\
\dfrac{\alpha}{X}S_i-\left(\dfrac{\beta}{Y}+\sigma r_1 \right)S_{i+1} = 0
$$
The first equation is the ODE for $S_1$, the last is for $S_{i+1}$, and the middle is for $S_2$ to $S_{i}$. I know beforehand that the parameters $Q, \alpha, X, r_1, \beta, Y$ and all $S_j$'s are positive real numbers and $\sigma > 1.$
____
Work in progress:
Note that $$\frac{d}{dt}S_T = \frac{d}{dt} \left(\sum\limits_{j=1}^{i+1}S_j\right) = Q - r_1 \sum\limits_{j=1}^{i} S_i - \sigma r_1 S_{i+1} = Q - r_1 S_T + r_1 (1-\sigma) S_{i+1},$$ implying that $$S_T = \frac{Q}{r_1} + (1-\sigma)S_{i+1} - \frac{1}{r_1} \dot{S}_T.$$
Now notice that upon solving for $S_1$ in the first and second ODE equations respectively, we obtain:
$$S_1 = \dfrac{X}{Y(X r_1 + \alpha)}\left(QY + \beta \sum\limits_{j=2}^{i+1}S_j\right)$$
and
$$S_1 = \dfrac{X(\beta+Y r_1)}{\alpha Y} \sum\limits_{j=2}^{i} S_j + S_i.$$
Question: How can I find $S_{i+1}$ explicitly so that I can (hopefully) solve the linear ODE for $S_T$?
Thanks in advance for help.
This question has been crossposted here: http://math.stackexchange.com/questions/1537553/solving-a-system-of-an-arbitrary-number-of-odes-at-steady-state