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bvpT1

 

 

bvpT1
Contributor:  testset of J.R. Cash
Discipline:  academic test
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_1) \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) = (\exp(-t/\sqrt \lambda) - \exp((t-2)/\sqrt \lambda)) / (1 - \exp(-2/\sqrt \lambda)). \end{matrix} \end{equation*}

 

 

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