# Differential Algebraic Equations

This tutorial will introduce you to the functionality for solving differential algebraic equations (DAEs). Other introductions can be found by checking out DiffEqTutorials.jl.

This tutorial assumes you have read the Ordinary Differential Equations tutorial.

In this example we will solve the implicit ODE equation

\[f(du,u,p,t) = 0\]

where `f`

is a variant of the Roberts equation. This equation is a DAE of the form:

\[\begin{aligned} \frac{du}{dt} &= f(u,p,t) \\ 0 &= g(u,p,t) \\ \end{aligned}\]

which is also known as a constrained differential equation, where `g`

is the constraint equation. The Robertson model can be written in the form:

\[\begin{aligned} \frac{dy_1}{dt} &= -0.04y₁ + 10^4 y_2 y_3 \\ \frac{dy_2}{dt} &= 0.04 y_1 - 10^4 y_2 y_3 - 3*10^7 y_{2}^2 \\ 1 &= y_{1} + y_{2} + y_{3} \\ \end{aligned}\]

with initial conditions $y_1(0) = 1$, $y_2(0) = 0$, $y_3(0) = 0$, $dy_1 = - 0.04$, $dy_2 = 0.04$, and $dy_3 = 0.0$.

The workflow for DAEs is the same as for the other types of equations, where all you need to know is how to define the problem. A `DAEProblem`

is specified by defining an in-place update `f(out,du,u,p,t)`

which uses the values to mutate `out`

as the output. To makes this into a DAE, we move all of the variables to one side. Thus, we can define the function:

```
function f(out,du,u,p,t)
out[1] = - 0.04u[1] + 1e4*u[2]*u[3] - du[1]
out[2] = + 0.04u[1] - 3e7*u[2]^2 - 1e4*u[2]*u[3] - du[2]
out[3] = u[1] + u[2] + u[3] - 1.0
end
```

with initial conditions

```
u₀ = [1.0, 0, 0]
du₀ = [-0.04, 0.04, 0.0]
tspan = (0.0,100000.0)
```

and make the `DAEProblem`

:

```
using DifferentialEquations
differential_vars = [true,true,false]
prob = DAEProblem(f,du₀,u₀,tspan,differential_vars=differential_vars)
```

`differential_vars`

is an option which states which of the variables are differential, i.e. not purely algebraic (which means that their derivative shows up in the residual equations). This is required for the algorithm to be able to find consistent initial conditions. Notice that the first two variables are determined by their changes, but the last is simply determined by the conservation equation. Thus, we use `differential_vars = [true,true,false]`

.

As with the other DifferentialEquations problems, the commands are then to solve and plot. Here we will use the IDA solver from Sundials:

```
using Sundials
sol = solve(prob,IDA())
```

In order to clearly see all the features of this solution, it should be plotted on a logarithmic scale. We'll also plot each on a different subplot, to allow scaling the y-axis appropriately.

```
using Plots
plot(sol, xscale=:log10, tspan=(1e-6, 1e5), layout=(3,1))
```

This gives the following plot: