Two Compartment Model

Equations:

$$\begin{align}\dot{x}_{0}&=par_{1} \cdot x_{1} + x_{0} \cdot \left(- par_{0} - par_{2}\right)\\\dot{x}_{1}&=- par_{1} \cdot x_{1} + par_{2} \cdot x_{0}\end{align}$$

Description:

A standard example in parameter optimization; describes the way materials or energies are transmitted among the compartments. Please note that the supplied exact solution is correct only for certain initial conditions (e.g., those from the IVPs).

Keywords:

cooperative; min test set

Download file

Plots:

C5 WPD

IVP 118

Class: PILBNU

Initial conditions:

$$\begin{align}x_{0}(0)&=1.0\\x_{1}(0)&=0.0\end{align}$$

Parameters:

$$\begin{align}par_{0}&=1.925403\\par_{1}&=0.232718\\par_{2}&=0.145076\end{align}$$

Exact Solution^*:

$$\begin{align}x_{0}(t)&=0.990293648402508 \cdot e^{- 2.08867009270157 \cdot t} + 0.00970635159749178 \cdot e^{- 0.214526907298434 \cdot t}\\x_{1}(t)&=- 0.0774092401956971 \cdot e^{- 2.08867009270157 \cdot t} + 0.0774092401956971 \cdot e^{- 0.214526907298434 \cdot t}\end{align}$$

View test data:

Solver
Settings
Results

DynIbex

arithmetic: Interval
method: 7
abs. tolerance: 1e-12
initial stepsize: 1e-3
min. stepsize: 1e-5
monotonicity: 0

c4: t_c=0.64770
c5: t_us=0.64700 , e_us=2.520136876960066e-13
c6: t_bd=10.000000 , e_bd=1.608066804714703e-11

DynIbex

arithmetic: Interval
method: 7
abs. tolerance: 1e-12
initial stepsize: 1e-3
min. stepsize: 1e-5
monotonicity: 1

c4: t_c=1.30630
c5: t_us=1.30410 , e_us=2.520136876960066e-13
c6: t_bd=10.000000 , e_bd=1.608066804714703e-11

DynIbex

arithmetic: Interval
method: 9
abs. tolerance: 1e-12
initial stepsize: 1e-3
min. stepsize: 1e-5
monotonicity: 0

c4: t_c=2.14500
c5: t_us=2.14160 , e_us=1e-308
c6: t_bd=10.000000 , e_bd=5.177185069538126e-12

DynIbex

arithmetic: Interval
method: 11
abs. tolerance: 1e-12
initial stepsize: 1e-3
min. stepsize: 1e-5
monotonicity: 0

c4: t_c=4.29370
c5: t_us=4.28890 , e_us=1.354819034737886e-13
c6: t_bd=10.000000 , e_bd=2.955597572240620e-12

VNODE-LP

method: HOE
order: 10
abs. tolerance: 1e-12
rel. tolerance: 1e-12
min. stepsize: 1e-05

c4: t_c=0.00510
c5: t_us=0.00480 , e_us=5.342948306008566e-16
c6: t_bd=10.000000 , e_bd=6.149594722337781e-15

VNODE-LP

method: HOE
order: 15
abs. tolerance: 1e-12
rel. tolerance: 1e-12
min. stepsize: 1e-05

c4: t_c=0.00990
c5: t_us=0.00910 , e_us=9.714451465470120e-17
c6: t_bd=10.000000 , e_bd=8.673617379884035e-17

VNODE-LP

method: HOE
order: 20
abs. tolerance: 1e-12
rel. tolerance: 1e-12
min. stepsize: 1e-05

c4: t_c=0.01490
c5: t_us=0.01360 , e_us=1.457167719820518e-16
c6: t_bd=10.000000 , e_bd=6.418476861114186e-17

VNODE-LP

method: HOE
order: 5
abs. tolerance: 1e-12
rel. tolerance: 1e-12
min. stepsize: 1e-05

c4: t_c=0.04610
c5: t_us=0.04240 , e_us=3.456263053536190e-14
c6: t_bd=10.000000 , e_bd=1.339969801783525e-13

Generate graphics:

DynIbex VNODE-LP

WPD for Problem 118

IVP 119

Class: PILBU

Initial conditions:

$$\begin{align}x_{0}(0)&=1.0\\x_{1}(0)&=0.0\end{align}$$

Parameters:

$$\begin{align}par_{0}&\subset~\left[ 1.9, 2.0\right]\\par_{1}&\subset~\left[ 0.2, 0.3\right]\\par_{2}&\subset~\left[ 0.1, 0.2\right]\end{align}$$

Exact Solution^*:

$$\begin{align}x_{0}(t)&\subset~\left(e^{t \cdot \left[ -2.2795630140987, -1.9616843969807\right]} \cdot \left[ 1.7116843969807, 2.0295630140987\right] + e^{t \cdot \left[ -0.388315603019296, -0.0704369859013001\right]} \cdot \left[ -0.138315603019296, 0.1795630140987\right]\right) \cdot \left[ 0.485642931178632, 0.58025885318566\right]\\x_{1}(t)&\subset~\left(- e^{t \cdot \left[ -2.2795630140987, -1.9616843969807\right]} + e^{t \cdot \left[ -0.388315603019296, -0.0704369859013001\right]}\right) \cdot \left[ 0.0485642931178632, 0.116051770637132\right]\end{align}$$