Does this system of non-linear differential equations have an analytical solution?

Question

I have this system of linear differential equations: \begin{align*} \frac{dA}{dt} & = k_1 - k_2AB - k_3A\\ \frac{dB}{dt} & = -k_2AB - k_4B\\ \frac{dC}{dt} & = k_2AB - k_5C\\ \frac{dD}{dt} & = k_6C \end{align*} Where $k_{i}$ are constants for $i\in\{1,2,3,4,5,6\}$.

Will an analytic solution to these equations exist? Is there an easy way to tell?

Thank you in advance

Answer 1

PARTIAL SOLUTION (case $k_3=k_4=k_5$)

Upon eliminating the $k_2AB$ terms, the system in $A,B,C$ becomes

\begin{align*} \dot{A}-\dot{B}&=k_1-k_3(A-B)\\ \dot{B}+\dot{C}&=-k_3(B+C)\\ \dot{A}+\dot{C}&=k_1-k_3(A+C) \end{align*}

which may be integrated directly to yield

\begin{align*} A-B&=\frac{k_1-\bigg(k_1-k_3(A_0-B_0)\bigg)\exp(-k_3t)}{k_3}&=f_1\\ B+C&=(B_0+C_0)\exp(-k_3t)&=f_2\\ A+C&=\frac{k_1-\bigg(k_1-k_3(A_0+C_0)\bigg)\exp(-k_3t)}{k_3}&=f_3 \end{align*}

where $A_0=A(0)$ and so on. It follows that $B_0+C_0=0$ and hence $B(t)+C(t)=0$.

In matrix form,

$\begin{pmatrix} A\\ B\\ C\\ \end{pmatrix}= \begin{pmatrix} 1&-1&0\\ 0&1&1\\ 1&0&1\\ \end{pmatrix}^{-1} \begin{pmatrix} f_1\\ 0\\ f_3\\ \end{pmatrix}$

and it is clear that there exists no unique solution because the $3\times 3$ matrix is singular.

Answer 2

\begin{align*} A'=\frac{dA}{dt} & = k_1 - k_2AB - k_3A \tag 1\\ B'=\frac{dB}{dt} & = -k_2AB - k_4B \tag 2\\ C'=\frac{dC}{dt} & = k_2AB - k_5C \tag 3\\ D'=\frac{dD}{dt} & = k_6C \tag 4 \end{align*}

The equations $(1)$ and $(2)$ are independent from $(3)$ and $(4)$.

From $(2)$ :
$$A=-\frac{1}{k_2} \left(\frac{B'}{B}+k_4 \right) \tag 5$$ $$A'=-\frac{1}{k_2} \left(\frac{B''}{B}-\frac{(B')^2}{B^2} \right)$$

Putting them into $(1)$ : $$-\frac{1}{k_2} \left(\frac{B''}{B}-\frac{(B')^2}{B^2} \right)= k_1 +(B'+k_4B) +\frac{k_3}{k_2} \left(\frac{B'}{B}+k_3k_4 \right)$$

$$BB'' -B'^2 +(k_2B +k_3)B B' +(k_1k_2+k_3k_4)B^2+k_2k_4B^3=0$$ This is a second order non-linear ODE, with only one unknown function. Of course, they are an infinity of solutions $B(t)$. But most likely in the general case the solutions cannot be expressed on a closed form.

It is certain that closed form exist in case of some particular values of the coefficients. In those particular cases, $B(t)$ which is obtained can be put into equation (5) leading to $A(t)$. Then, putting $A$ and $B$ into (3) leads to a first order linear ODE which theoretically can be solved. May be not in practice, if the functions involved become too complicated. So, in some cases, $C(t)$ can be obtained. Then, $D(t)=k_6\int C(t)dt$.$

Answer 3

Suppose $k_3=k_4$ and $k_2=-k_3^2/k_1$. Then one solution to the first two equations is \begin{align} A(t)&=\frac{k_1}{k_3}\left(1-\frac{1}{k_3(t-T)}\right),\\ B(t)&=-\frac{k_1}{k_3^2(t-T)}, \end{align} where $T$ is a constant. The other two are then straightforward to integrate.