# coupled equations with finite difference method

I have these three differential equations in which I need to solve numerically:

$$\frac{dn_0}{dt}= -n_0(t)W_{01}(t) + n_1(t)K_{10}$$

$$\frac{dn_1}{dt}= -n_1(t)W_{12}(t) - n_1(t)K_{10} + n_2(t)K_{21} + n_0(t)W_{01}(t)$$

$$\frac{dn_2}{dt}= n_1(t)W_{12}(t) - n_2(t)K_{21}$$

such that

$$n_0(0)=1$$ $$n_0(N)=0$$ $$n_1(0)=0$$ $$n_1(N)=1$$ $$n_2(0)=0$$ $$n_2(N)=0$$

Using the central finite difference formula:

$$\frac{n_{0}(t + \Delta t) - n_{0}(t - \Delta t)}{2\Delta t}=-n_0(t)W_{01}(t) + n_1(t)K_{10}$$

$$\frac{n_{1}(t + \Delta t) - n_{1}(t - \Delta t)}{2\Delta t}=-n_1(t)W_{12}(t) - n_1(t)K_{10} + n_2(t)K_{21} + n_0(t)W_{01}(t)$$

$$\frac{n_{2}(t + \Delta t) - n_{2}(t - \Delta t)}{2\Delta t}=n_1(t)W_{12}(t) - n_2(t)K_{21}$$

How do I determine the functions $$n0$$, $$n1$$ and $$n2$$ knowing that $$n0 + n1 + n2 =1$$, and that the three equations are coupled?

And I could not understand how to calculate the derivatives, how can I determine their value with the finite difference method without knowing the functions?

Hmm, so I did it now.

$$\frac{n_{0_{i+1}} - n_{0_{i-1}}}{2\Delta t}= -n_0(t_i)W_{01}(t_i) + n_1(t_i)K_{10}$$

with $$t_i= (i-1)\Delta t$$ and $$i= 2,3,... N-1$$ for $$\Delta t=\frac{b-a}{N-1}$$

knowing that this gives a system of the type $$Au=B$$ I can not think of a way to fill the array. I say, on the right side of the equation, what values $$n_0(t_i)$$ and $$n_1(t_i)$$ assume as $$t_i$$ varies?

I tried to implement in scilab, but to no avail

You need to think of your coupled equations as one equation.

Be n=$$\left(\begin{array}{c}n_0\\n_1\\n_2\end{array}\right)$$, you can write it as $$\frac{d}{dt}n= A n$$ with $$A=\left(\begin{array}{ccc}-W_{01}(t)&K_{10} &0\\W_{01}(t)&-W_{12}(t) - K_{10}&K_{21}\\0&W_{12}(t)&-K_{21}\end{array}\right).$$

For the constraint you could either derive a new set of equations by explicitly solving the constraint for $$n_2=1-n_0-n_1$$ and plug it in or if you want to be flexible with linear constraints of such type, you could do the following.

Linear Constraints

We can rewrite the linear constraint equation as $$Bn=l$$ with $$B=\left(\begin{array}{ccc}1&1&1\end{array}\right)$$ and $$l=1.$$

Now be $$I_B$$ a matrix of a basis of the image space of $$B$$ and $$N_B$$ a matrix of a basis of the null space of B, we can do a change of basis by writing $$n$$ as $$n=I_B n_{I}+ N_B n_{N}.$$ The matrices $$I_B$$ and $$N_B$$ can be obtained by using a singular value decomposition on $$B$$. Our new variables are $$n_I$$ and $$n_N$$.

Using this basis in the constraint equation leads to $$B I_B n_I+B N_B n_N=l.$$ By construction $$B N_B n_N= 0$$, we solve the linear system for $$n_I$$ $$B I_B n_I = l.$$

The same basis can be used in the ODE system $$\frac{d}{dt}\left(I_B n_I + N_B n_N \right)= A \left(I_B n_I + N_B n_N \right)$$ $$\frac{d}{dt}I_B n_I=0$$ because it is a constant of motion due to the constraint equation.

Now multiplying both sides with $$N_B^T$$ leads to the reduced final system $$\frac{d}{dt} n_N = N_B^T A I_B n_I + N_B^T A N_B n_N .$$

You can apply your favorite time integrator on the vectorized version of the equation for the variables $$n_N$$, since we haven taken care of the constraints. Remember, the first term on the rhs $$N_B^T A I_B n_I$$ is a known function of $$t$$, since we solved already for $$n_I$$.

Once you have obtained a solution for $$n_N$$, you can use our change of basis to obtain the solution in $$n_0$$ to $$n_2$$.

Time Integration

The simplest method is for sure the Euler method, which is in this case leads to $$n_N^{i+1}=n_N^i+\Delta t \left(N_B^T A(t_i) I_B n_I + N_B^T A(t_i) N_B n_N^i\right).$$

Due to the dependency of the next step from the previous one, you cannot simply build up a system and solve it. Additionally, since $$W_{01}(t)$$ is a function of time, it is possible to a have a stiff or non stiff ODE system. So a recommendation is difficult, a good overview of possible solvers and there application can be found here.

Also you mention too many initial conditions. You only need at one point in time, e.g. the initial conditions at $$t=0$$, $$n_N(0)=N_B^T\left(\left(\begin{array}{c}n_0(0)\\n_1(0)\\n_2(0)\end{array}\right)-I_B n_I\right).$$

I don't know scilab, but it seems that they have time integrators for ODEs, too. So check their documentation how to use them.

• Thank you, and I'm sorry. Due to personal problems I had to leave and I could not thank properly. And I ended up leaving that question aside. I do not quite understand what you did at the end, what does this term $N_B ^ T$ mean you multiply on both sides? Mar 13 '19 at 14:22
• Yes, since $N_B^T N_B=Id$ is the identity matrix, because $N_B$ is a unitary matrix. So you can multiply both sides with this matrix to obtain the equation for $\frac{d}{dt}n_N$. This is useful but not necessary. Because not every ODE solver allows an additional matrix multiplication on the time derivative vector.
– Bort
Mar 14 '19 at 16:42