The following rate equations describe a chemical reaction

$$\frac{d{y}_1}{dt}=-0.04y_1+{10}^4{y}_2{y}_3$$ $$\frac{d{y}_2}{dt}=0.04y_1-{10}^4{y}_2{y}_3-{3*10}^7{y}_2^2$$ $$\frac{d{y}_3}{dt}={3*10}^7{y}_2^2$$

This system time domain is [0,1000] with initial conditions y₁ = 1, y₂ = y₃ = 0

Solution

We pick T1 to represent the time variable and Y1, Y2, Y3 to represent the differential (state) variables, and define the system right-hand-side formulas in terms of our selected variables in range A1:A3 as shown in Table 1. We also assign the initial conditions for the state variables as shown in Table 1.

Next, we evaluate the array formula =IVSOLVE(A1:A3, (T1,Y1:Y3), {0,1000}) in range G4:J25 to obtain the solution shown in Table 2. The first argument to IVSOLVE is the reference to the system RHS formulas defined in A1:A3. The second parameter is the reference to the selected system variables. Note that we have used Excel union operator, (), to combine the time variable T1 and the state variables Y1:Y3 into one reference. The third parameter is the time domain for the problem which is passed using Excel constant array syntax, {}, for convenience. Since we have used default format for output, IVSOLVE reports the solution at a uniform time subdivision of the interval [0,1000] based on the number of allocated rows in the allocated range for output. You can control the output by allocating a larger range or using other optional formats for parameter 3 to control the subdivison or the exact outputed time points.

Consider the following differential algebraic system of equations: $$\frac{d{y}_1}{dt}=-0.04y_1+{10}^4{y}_2{y}_3$$ $$\frac{d{y}_2}{dt}=0.04y_1-{10}^4{y}_2{y}_3-{3*10}^7{y}_2^2$$ $$0=y_1+y_2+y_3-1$$

This is a variation of Example 1 where y₃ is treated as an algebraic variable. This system time domain is [0,1000] with initial conditions y₁ = 1, y₂ = y₃ = 0

Solution

The solution steps are similar to Example 1. The RHS formulas are defined in A4:A6 as shown in Table 1 in terms of the selected variables T1 and Y1, Y2, Y3. Note that Y3 represents now an algebraic variable rather than a differential variable.

Next, we evaluate the array formula =IVSOLVE(A4:A6, (T1,Y1:Y3), {0,1000}, 1) in an allocated range to obtain the solution. Note that we pass 1 in the 4rth parameter to tell the IVSOLVE that we have one algebraic equation in the system. Algebraic equations and variables must be ordered last. Here the solver will treat the last formula as an algebraic equation along with variable Y3.

The new results, are only slightly different from Example 1, however, we extract below a sample to show that the solution satisfies the algebraic constraint Y1+Y2+Y3-1=0 as shown in Table 2.

Next, we demonstrate how to use additional optional arguments.

Passing System Jacobian, and Custom Tolerances

We define the system Jacobian in M3:O5 and custom tolerances for variables Y1, Y2 and Y3 in the vector Q3:Q5 as shown in Table 3. We name these two ranges in Excel as jacob and rtol respectively.

We evaluate the updated array formula =IVSOLVE(A4:A6, (T1,Y1:Y3), {0,1000,16}, 1, rtol, jacob) in an allocated array M7:P25 and obtain the new solution shown in Table 4. Notice that we requested in argument 3 to use exactly 16 subdivisions for the output time interval. This fills rows 8 through 24. Any extra rows in the allocated array are filled with zeros.

The following system with 12 rate equations arises from chemical kinetics.

We solve this system in the time interval [0,100] starting from initial conditions: y₁ = 1 and 0 for the rest.

Solution

The steps to solve this system are similar to Example 1. We define the RHS system formulas in A26:A37 using variables T1, and Y1:Y12 as shown in Table 1. Variable Y1 is assigned the value 1 for the initial condition whereas the remaining variables Y2:Y12 are left blank consistent with the initial condition of 0.

To obtain the solution, we evaluate the array formula =IVSOLVE(A26:A37, (T1,Y1:Y12), {0,100}) in allocated array J4:V26. The results are plotted in the figure below.