ODEs 3¶
- System of equations
- Higher order derivatives
- Decoupled ODEs
- Adaptive time steps
ODE system¶
- So far we have considered $dy/dt = f(y,t)$, with one equation in one variable.
For a system of ODEs, we have $d\vec{y}/{dt} = \vec{f}(\vec{y},t).$
- For example, for 2 ODEs in 2 variables: $$ \frac{dx}{dt} = g(x,z,t), $$ $$ \frac{dz}{dt} = h(x,z,t).$$
We can write this as $$ \frac{d\vec{y}}{dt} = \vec{f}(\vec{y},t),$$
where $x=y[0]$, $z=y[1]$, $g=f[0]$ and $h=f[1]$.
- Note that each rate depends (in general) on all the variables.
- Note, below, we'll leave off the vector arrow symbols for simplicity.
- Explicit solution is a straightforward extension of the one equation case. Python's array functionality even allows nearly identical codes for systems of equations and for one equation.
- Implicit solutions require solution of a linear system of equations at each step for linear ODE systems. For nonlinear ODE systems, a nonlinear system must be solved at each step.
- Linear:
$$f(y) = Ay + b,$$
$$ y_{k+1} = y_k + \Delta t(Ay_{k+1}+b),$$
$$(I-\Delta tA)y_{k+1} = (y_k + \Delta tb).$$
This last equation has the form $By_{k+1}=c$, which is a linear system solved for $y_{k+1}$ at each step.
- Note, $A$ and $b$ can depend on time, which is not explicitly shown.
- Nonlinear:
$$y_{k+1} = y_k + \Delta t f(y_{k+1},t).$$
- Rearrange and solve the following nonlinear system for the $y_{k+1}$ vector at each step: $$ F(y_{k+1}) = y_{k+1}-y_k - \Delta tf(y_{k+1},t) = 0.$$
- Linear:
$$f(y) = Ay + b,$$
$$ y_{k+1} = y_k + \Delta t(Ay_{k+1}+b),$$
$$(I-\Delta tA)y_{k+1} = (y_k + \Delta tb).$$
This last equation has the form $By_{k+1}=c$, which is a linear system solved for $y_{k+1}$ at each step.
Example¶
The following reactions are given: $$A + B \rightarrow C,$$ $$A + C \rightarrow D.$$
- Reactions have rate constants $k_1 = 1$ and $k_2 = 2$.
- Let $A_0=B_0=1$ and $C_0=D_0=0$.
- Solve to 5 seconds.
- Species concentrations are given by the following rate equations: \begin{align*} \frac{dA}{dt} &= -k_1AB - k_2AC, \\ \frac{dB}{dt} &= -k_1AB, \\ \frac{dC}{dt} &= k_1AB - k_2AC, \\ \frac{dD}{dt} &= k_2AC. \end{align*}
Solve this system using the fourth order RK method: \begin{align*} y_{k+1} &= y_k + h\left(\frac{1}{6}S_1 + \frac{2}{6}S_2 + \frac{2}{6}S_3 + \frac{1}{6}S_4\right),\\ & S_1 = f(y_k), \\ & S_2 = f(y_k+\frac{h}{2}S_1), \\ & S_3 = f(y_k+\frac{h}{2}S_2), \\ & S_4 = f(y_k+hS_3). \end{align*}
import Plots as plt
function odeRK4(f, y0, t)
ns = length(t)-1
y = zeros(length(t), length(y0))
y[1,:] = y0
for k in 1:ns
h = t[k+1] - t[k]
S1 = f(y[k,:], t[k])
S2 = f(y[k,:]+0.5*h*S1, t[k]+0.5*h)
S3 = f(y[k,:]+0.5*h*S2, t[k]+0.5*h)
S4 = f(y[k,:]+ h*S3, t[k]+ h)
y[k+1,:] = y[k,:] + h/6*(S1 + 2*S2 + 2*S3 + S4)
end
return y
end;
function rhsf(ABCD, t)
A = ABCD[1]
B = ABCD[2]
C = ABCD[3]
D = ABCD[4]
k1 = 1
k2 = 2
dAdt = -k1*A*B - k2*A*C
dBdt = -k1*A*B
dCdt = k1*A*B - k2*A*C
dDdt = k2*A*C
return [dAdt, dBdt, dCdt, dDdt]
end;
ABCD_initial = [1,1,0,0]
tend = 5
t = LinRange(0,tend,100)
ABCD = odeRK4(rhsf, ABCD_initial, t);
plt.resetfontsizes(); plt.scalefontsizes(1.5)
plt.plot(t,ABCD, lw=2, label=["A" "B" "C" "D"])
plt.plot!(foreground_color_legend=nothing)
plt.plot!(xlabel="t", ylabel="Concentration")
Higher order derivatives¶
- A single higher order ODE results in a system of first order ODEs. $$y^{\prime\prime\prime}=\frac{d^3y}{dt^3} = f(y,y^{\prime},y^{\prime\prime},t).$$
Question: can we solve this using the techniques we have used for first order ODE's? If so, how?
Can you convert this one third-order equation to a system of three first-order equations?
$$y^{\prime\prime\prime}=\frac{d^3y}{dt^3} = f(y,y^{\prime},y^{\prime\prime},t).$$- let $x=y^{\prime\prime}$, then $y^{\prime\prime\prime}=x^{\prime}$.
- let $z=y^{\prime}$, then $y^{\prime\prime}=z^{\prime}$.
- Then the resulting system of ODEs is:
- This is a system of three first order ODEs in three variables.
Decoupled equations¶
Consider a linear system of ODEs. $$y^{\prime} = Ay+b.$$
In general, each rate equation depends on all the variables.
In our linear algebra review, we discussed how to decouple a system of linear equations.
- How did we do that?
- To decouple the system, change to an eigenvector basis so that each component equation depends only on its own variable (in the new basis)
- Let $V$ be a matrix whose columns are the eigenvectors of matrix $A$.
- Let $\Lambda$ be a diagonal matrix whose diagonal elements are the eigenvalues of $A$.
- Then,
- Insert this into the ODE: $$y^{\prime} = V\Lambda V^{-1}y + b.$$
- Multiply through by $V^{-1}$: $$(V^{-1}y^{\prime}) = \Lambda(V^{-1}y) + (V^{-1}b).$$
- Now, let $\hat{y}=V^{-1}y$, and $\hat{b}=V^{-1}b$: $$\hat{y}^{\prime} = \Lambda \hat{y} + \hat{b}.$$
- Because $\Lambda$ is diagonal, this system is decoupled. That is component $i$ is given by $$\hat{y}^{\prime}_i = \lambda_i\hat{y}_i + \hat{b}_i.$$
- This equation has a simple analytic solution.
- When solved, all the $\hat{y}_i(t)$ are known.
- Then $y(t)$ are given by $$y(t) = V\hat{y}(t).$$
This analysis can be useful for solving ODEs analytically, but also for analyzing (and modifying) stability properties of ODEs.
ODE Step Size¶
- See Numerical Recipes section 16.2 for more detailed explanations.
- When solving an ODE, we needed a step size $\Delta t$ or $h$.
- How should we select the step size?
Consider integrating $dy/dt = \tanh(t)$
- What part of the solution will dictate the step size?
- Do we have to use this step size everywhere?
- If not, how can we make the computer choose the stepsize in an "intelligent" way?
Suppose we know the error $\Delta$ for a given step size $\Delta t=h_1$.
- We'll show how to get $\Delta$ below.
Suppose we set some desired error that we are okay with on a given step, like $|\Delta|\le\epsilon = atol + |y|rtol.$
- A large $y\rightarrow$ rtol controls;
- A small $y\rightarrow$ atol controls.
Given a known error for a known step, how can we change our step to get the desired error? Assume we are using the RK4 method.
Adjust $h$ to get the desired error.
- For a globally $4^{th}$ order method $\rightarrow$ $\Delta = \mathcal{O}(h^5)$ $\rightarrow$ $\Delta\sim h^5$.
- Then $$\frac{\Delta_2}{\Delta_1} = \frac{\epsilon}{\Delta_1} = \left(\frac{h_2}{h_1}\right)^5,$$ $$\rightarrow h_2 = h_1\left(\frac{\epsilon}{\Delta_1}\right)^{1/5}.$$
- So, guess an initial $h_1$.
- If the error $\Delta_1$ is too big, then redo the step using a smaller $h$ as computed using the above equation.
- If the error is too small, take the step, but do the next step with a larger $h$ computed using the equation above.
How to compute $\Delta$¶
- See N.R.
- Two approaches:
- Step doubling
- Felberg.
Step doubling¶
- Consider two grids where we can either take two size $h$ steps, or one size $2h$ step:
(A) <----h---->|<----h----> (B) <---------2h----------> Now, let $\Delta$ = $y_B-y_A$.
Recall, if the error of the method (per step) is $\mathcal{O}(h^5)$, then
\begin{align} y_{exact} &= y_A + 2\mathcal{O}(h^5), \\ y_{exact} &= y_B + \mathcal{O}((2h)^5) = y_B + 32\mathcal{O}(h^5). \end{align}
- The error in $y_{B}$ is 16 times larger than the error in $y_{A}$, so we consider $y_A$ to be exact (compared to $y_B$), and evaluate $\Delta = y_B-y_A$.
- Cost
- Each RK step requires 4 function evaluations.
- 3 total steps (two for grid (A) and one for grid (B)) $\rightarrow$ 12 function evaluations. Actually 11, since the grids share a starting point.
- We compare 11 required using step doubling to 8 required without step doubling.
- 37.5%=(11-8)/8 is the cost increase.
- This cost increase pays for itself in terms of allowing (ideally) at least 37.5% fewer overall steps due to the adaptive stepsize control.
- For a given error, without adaptive stepsize, the whole time domain would have to use the most stringent step size.
Felberg¶
- You can write an RK method where one linear combination of slopes results in an $\mathcal{O}(h^6)$ method, and another combination of the same slopes gives an $\mathcal{O}(h^5)$ method.
- Then let $\Delta=y_{k+1}-\hat{y}_{k+1}$.