B2 DETEST

Equations:

$$\begin{align}\dot{x}_{0}&=- x_{0} + x_{1}\\\dot{x}_{1}&=- x_{0} - 2.0 \cdot x_{1} + x_{2}\\\dot{x}_{2}&=x_{1} - x_{2}\end{align}$$

Description:

A linear chemical reaction.
Link to DETEST description

Keywords:

min test set

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Plots:

C5 WPD

Initial value problems:

10 (PILBNU)
11 (PILBU)

IVP 10

Class: PILBNU

Initial conditions:

$$\begin{align}x_{0}(0)&=2.0\\x_{1}(0)&=0.0\\x_{2}(0)&=1.0\end{align}$$

Exact Solution^*:

$$\begin{align}x_{0}(t)&=- 1.0 \cdot \left(t - 1.0\right) \cdot e^{- t} - 1.0 \cdot e^{- 2.0 \cdot t} + 2.0 \cdot e^{- t}\\x_{1}(t)&=1.0 \cdot e^{- 2.0 \cdot t} - 1.0 \cdot e^{- t}\\x_{2}(t)&=- 1.0 \cdot t \cdot e^{- t} - 1.0 \cdot e^{- 2.0 \cdot t} + 2.0 \cdot e^{- t}\end{align}$$

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=0.74730
c5: t_us=0.74570 , e_us=9.365952458040283e-12
c6: t_bd=10.000000 , e_bd=2.035712743666228e-10

DynIbex

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

c4: t_c=2.36920
c5: t_us=2.36570 , e_us=9.311384996379957e-12
c6: t_bd=10.000000 , e_bd=1.998524318583793e-10

DynIbex

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

c4: t_c=2.86390
c5: t_us=2.85910 , e_us=6.117217843382150e-12
c6: t_bd=10.000000 , e_bd=1.743849497855103e-10

DynIbex

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

c4: t_c=3.48390
c5: t_us=3.47710 , e_us=4.990619029143772e-12
c6: t_bd=10.000000 , e_bd=1.313581305902901e-10

VNODE-LP

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

c4: t_c=0.00600
c5: t_us=0.00400 , e_us=1.269817584415023e-14
c6: t_bd=10.000000 , e_bd=4.130584763117895e-13

VNODE-LP

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

c4: t_c=0.01010
c5: t_us=0.00540 , e_us=1.269817584415023e-14
c6: t_bd=10.000000 , e_bd=4.130584763117895e-13

VNODE-LP

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

c4: t_c=0.01410
c5: t_us=0.00830 , e_us=1.269817584415023e-14
c6: t_bd=10.000000 , e_bd=4.130584763117895e-13

VNODE-LP

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

c4: t_c=0.05760
c5: t_us=0.04990 , e_us=1.269817584415023e-14
c6: t_bd=10.000000 , e_bd=4.130584763117895e-13

Generate graphics:

DynIbex VNODE-LP

WPD for Problem 10

IVP 11

Class: PILBU

Initial conditions:

$$\begin{align}x_{0}(0)&\subset~\left[ 1.5, 2.5\right]\\x_{1}(0)&=0.0\\x_{2}(0)&=1.0\end{align}$$

Exact Solution^*:

$$\begin{align}x_{0}(t)&\subset~\left(t - 1.0\right) \cdot e^{- t} \cdot \left[ -1.5, -0.5\right] + e^{- 2.0 \cdot t} \cdot \left[ -1.5, -0.5\right] + e^{- t} \cdot \left[ 1.5, 2.5\right]\\x_{1}(t)&\subset~e^{- 2.0 \cdot t} \cdot \left[ 0.5, 1.5\right] + e^{- t} \cdot \left[ -1.5, -0.5\right]\\x_{2}(t)&\subset~t \cdot e^{- t} \cdot \left[ -1.5, -0.5\right] + e^{- 2.0 \cdot t} \cdot \left[ -1.5, -0.5\right] + e^{- t} \cdot \left[ 1.5, 2.5\right]\end{align}$$