# Tips for solving differential equations in Julia

Contents

Thoughts and tips about solving differential equations in Julia using DifferentialEquations.jl.

## How to solve a system of ODEs

DifferentialEquations.jl | overview

### Define the problem

ODE function could be one of the two forms:

• f(u, p, t), the out of place form returning du.
• f!(du, u, p, t), the in-place form where du is the output.

where

• u is the vector of state variable(s), or called dependent variables. (Actually u can also be a scalar value in the case of a single state variable)
• p are the parameters for the function, often constants thoughout the solution.
• t is the independent variable, usually time in chemical / physical systems.

For an example, to model the exponential growth / decay $\frac{du}{dt} = pu$ , write the right hand side (RHS) as the ODE function.

 1  f(u, p, t) = p*u 

And then you can define the problem

 1  prob = ODEProblem(f, u0, tspan, p) 

### Initial conditions

The initial conditions u0 could be one of the three forms: (when in doubt, use the first one)

• A vector
• A scalar (if the RHS function takes a scalar u)
• The functional form: u(p, t0) returning the initial conditions depending on parameters.

### Time span

The time span ts could be one of the three forms: (when in doubt, use the first one)

• A tuple: (tstart, tend).
• A scalar: tend, equivalent to (0, tend).
• functional: t(p) which returns (tstart, tend), when the time span depends on parameters.

The type of timespan matters: Usually you’ll want to set timespan to a real number (e.g 1.0). If you set it to an integer 1, the Julia solver will try to solve the problem in integers and fractions and exit with not enought precision error.

### Parameters

Parameters (p) in Julia could be tuples, namedtuples, arrays, functions, etc., as long as they could run the RHS function f(u, params, t) you wrote.

### More on the problem object and interface

Problem interface | DiffEq docs

Problem objects are immutable, but you can remake a problem based on an existing one.

 1  probNew = remake(prob, p=newParameters) 

remake() is extensively used in ensemble simulations to vary the fields (e.g. u0, p) in the problem object.

## Dealing with Discontiuities and numerical instabilities

From a post in the Julia forum.

For example, there is a conditional statement in your model like k9*X3/(k10*k11/(k12*(1.-min(1.,k13+k14*X5))) + X3). Calculating derivatives across this discontiuity caused by min() might cause an error in the solver.

### Option 1: Turn off autodiff to use finite differencing instead

Set autodiff=false in the solver

 1 2  sol = solve(prob,Rosenbrock23(autodiff=false)) sol = solve(prob,Rodas4(autodiff=false)) 

Alternativly use CVODE_BDF since it uses finite differencing by default.

However, disable autodiff increases the number of evaluation significantly and makes the solver less efficient.

### Option 2: Replace the discontinuities with smooth functions

Replace step functions / min() / max() with tanh() or exp() for switch-like behavior. This increases problem stiffness but it is efficient for stiff ODE solvers.

 1 2 3  smooth_min(x, y, k=10) = (exp(-k * x) + exp(-k * y)) / (exp(x) + exp(y)) @@ -96,13 +128,21 @@ sol = solve(prob,Rodas4(autodiff=false), callback=cb) sol = solve(prob,CVODE_BDF(), callback=cb) 

If the discontiuities is large in the callback you might want to propose a smaller dt to ensure numerical stability.

 1 2 3 4  function affect!(integrator) # Do things set_proposed_dt!(integrator, 0.01) end 

## Plotting solutions

You can pass more options for the DifferentialEquations.jl solution to the plot() function thanks to Plot recipes.

• vars : Choosing which variable(s) to plot by passing vars to plot() function. Functions dependent on the state variables are also suported.
 1 2 3 4  f(x,y,z) = (sqrt(x^2+y^2+z^2),x) plot(sol,vars=(f,1,2,3)) @@ -127,8 +167,6 @@ plt.plot(sol(ts,idxs=i),sol(ts,idxs=j),sol(ts,idxs=k))