DEsolution.md

Differential Equations Solution

Differential Equations Solution uses DifferentialEquations.jl to solve the Problems. Build the problem and call generateDEcodes, work done!

generateDEcodes

Overview of generateDEcodes and generateJuMPcodes

generateDEcodes and generateJuMPcodes using the same way to build problem, and they both can solve Fixed Value(Example1) and Free End(Example2). But generateDEcodes cannot deal with End constraint and Add variable limit. Meanwhile, generateDEcodeshave lower accuracy than generateJuMPcodes. Maybe accuracy can be improved by pass different solver parameters. You can have a try.

Example	generateJuMPcodes	generateDEcodes
1. Fixed Value	✅	✅
2. Free Value	✅	✅
3. Constraint	✅	❌
4. Multiple x and u	✅	✅
5. variable limit	✅	❌

✅ = supported

❌ = not supported

Example 1: Fixed value

To solve:

$$min \int_{0}^{2} u^2dt \newline s.t. ~~~~~ \dot{\boldsymbol{x}} =\begin{bmatrix}0&1 \newline 0&0\end{bmatrix}\boldsymbol{x}+ \begin{bmatrix}0 \newline 1 \end{bmatrix}u \newline \boldsymbol{x}(0) = \begin{bmatrix} 1 \newline 1 \end{bmatrix}, \boldsymbol{x}(2)=\begin{bmatrix} 0 \newline 0 \end{bmatrix}$$

Just define variables and build functions. Call generateDEcodes and get the results.

The analytical solution of $x_1$ is

$$x_1(t) = 0.5t^3-1.75t^2+t+1$$

and we can campare the difference between them by using Mean Square Error(MSE).

using OptControl, Statistics, ModelingToolkit
@variables t u x[1:2]
f = [0 1; 0 0] * x + [0, 1] * u
L = 0.5 * u^2
t0 = [1.0, 1.0]
tf = [0.0, 0.0]
tspan = (0.0, 2.0)
sol = generateDEcodes(L, f, x, u, tspan, t0, tf)
nu = [sol.u[i][1] for i in 1:length(sol.u)]
xs = collect(range(tspan[1], tspan[2], length=length(sol.u)))
an = @.(0.5 * xs^3 - 1.75 * xs^2 + xs + 1)
mean((an - nu).^2)

Example 2: Free End

To solve:

$$min \int_{0}^{2} u^2dt \newline s.t. ~~~~~ \dot{\boldsymbol{x}} =\begin{bmatrix}0&1 \newline 0&0\end{bmatrix}\boldsymbol{x}+ \begin{bmatrix}0 \newline 1 \end{bmatrix}u \newline \boldsymbol{x}(0) = \begin{bmatrix} 1 \newline 1 \end{bmatrix}, \boldsymbol{x}(1)=\begin{bmatrix} 0 \newline free \end{bmatrix}$$

If variable is free, use nothing.

The analytical solution of $x_1$ is

$$x_1(t) = t^3-3.0*t^2+t+1$$

and get MSE.

using OptControl, Statistics, ModelingToolkit
@variables t u x[1:2]
f = [0 1; 0 0] * x + [0, 1] * u
L = 0.5 * u^2
t0 = [1.0, 1.0]
tf = [0.0, nothing]
tspan = (0.0, 1.0)
N = 100
sol = generateDEcodes(L, f, x, u, tspan, t0, tf)
nu = [sol.u[i][1] for i in 1:length(sol.u)]
xs = collect(range(tspan[1], tspan[2], length=length(sol.u)))
an = @.(xs^3 - 3.0 * xs^2 + xs + 1)
mean((an - nu).^2)

Example 3: Multiple x and u

To solve:

$$min \int_{0}^{2} u^2dt \newline s.t. ~~~~~ \dot{\boldsymbol{x}} =\begin{bmatrix}0&1 \newline 0&0\end{bmatrix}\boldsymbol{x}+ \begin{bmatrix}1&0 \newline 0&1 \end{bmatrix}\boldsymbol{u} \newline \boldsymbol{x}(0) = \begin{bmatrix} 1 \newline 1 \end{bmatrix}, \boldsymbol{x}(2)=\begin{bmatrix} 0.0 \newline free \end{bmatrix}$$

The analytical solution of $u_1,u_2$ is

$$\begin{matrix} u_1=-\frac{9}{14}\\u_2=\frac{9}{14}*t-\frac{9}{7} \end{matrix}$$

and get MSE.

using OptControl, Statistics, ModelingToolkit
@variables t u[1:2] x[1:2]
f = [0 1; 0 0] * x + [1 0; 0 1] * u
L = 0.5 * (u[1]^2 + u[2]^2)
t0 = [1.0, 1.0]
tf = [0.0, nothing]
tspan = (0.0, 2.0)
sol = generateDEcodes(L, f, x, u, tspan, t0, tf)
nu1 = [sol.u[i][5] for i in 1:length(sol.u)]
nu2 = [sol.u[i][6] for i in 1:length(sol.u)]
xs = collect(range(tspan[1], tspan[2], length=length(sol.u)))
an_u2 = @.(9 / 14 * xs - 9 / 7)
res1 = mean((-9 / 14 .- nu1).^2)
res2 = mean((an_u2 - nu2).^2)
(res1,res2)