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bvpT2

 

 

bvpT2
Contributor:  testset of J.R. Cash
Discipline:   
Accession:  2013

 

 

 

Short description:

A second order differential equations that is reduced to a first order system of 2 equations.

Applicable solvers:

all the solvers supported by the Test Set.

 

Plots of the solution <- click to generate the plots of the solution and the textual output

 

Mathematical description:

 

 

The problem is

    \begin{equation*} 	\begin{matrix} 	\lambda z'' = z', \quad z(0)=1, \quad z(1)=0, 	\end{matrix} 	\end{equation*}

with

    \begin{equation*} 	\begin{matrix} 	z \in \mathbb{R}, \quad t\in [0,1]. 	\end{matrix} 	\end{equation*}

We write this problem in first order form by defining y_1=z, and y_2=z', yielding a system of differential equations of the form

    \begin{equation*} 	\begin{matrix} 	\left(\begin{matrix} 	y_1\\ 	y_2 	\end{matrix}\right)'= 	\left(\begin{matrix} 	y_2\\ 	\frac{1}{\lambda}f(y_2) 	\end{matrix}\right), 	\end{matrix} 	\end{equation*}

where

    \begin{equation*} 	\begin{matrix} 	f(z')= z', 	\end{matrix} 	\end{equation*}

and

    \begin{equation*} 	\begin{matrix} 	(y_1,y_2)^T \in \mathbb{R}^{2}, \quad t \in [0,1]. 	\end{matrix} 	\end{equation*}

The boundary conditions are obtained from

    \begin{equation*} 	\begin{matrix} 	\left( 	\begin{matrix} 	1 & 0 \\ 	0 & 0 \\ 	\end{matrix} 	\right) 	\left(\begin{matrix} 	y_{1}(0)\\ 	y_{2}(0) 	\end{matrix}\right) 	+ 	\left( 	\begin{matrix} 	0 & 0 \\ 	1 & 0 \\ 	\end{matrix} 	\right) 	\left(\begin{matrix} 	y_{1}(1)\\ 	y_{2}(1) 	\end{matrix}\right)=\left(\begin{matrix} 	1 \\ 	0 	\end{matrix}\right). 	\end{matrix} 	\end{equation*}

Exact solution

    \begin{equation*} 	\begin{matrix} 	z(t) = (1 - \exp((t - 1) /\lambda)) / (1 - \exp(-1/\lambda)). 	\end{matrix} 	\end{equation*}

 

 

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