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bvpT5

 

 

bvpT5
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{eqnarray*} 	\lambda z'' = t z' + z -( 1 +\lambda \pi^{2}) \cos(\pi t) + \pi t \sin(\pi t), \;\;\;z(-1) = -1, \;\;\; z(1) = -1, 	\end{eqnarray*}

with

    \[ 	z \in \mathbb{R} , \;\;\; t\in [-1,1]. 	\]

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*} 	\left(\begin{array}{c} 	y_1\\ 	y_2 	\end{array}\right)'= 	\left(\begin{array}{c} 	y_2\\ 	\frac{1}{\lambda}f(t,y_1,y_2) 	\end{array}\right), 	\end{equation*}

where

    \begin{equation*} 	f(t,z,z') = t z' + z -( 1 +\lambda \pi^{2}) \cos(\pi t) + \pi t \sin(\pi t), 	\end{equation*}

and

    \[ 	(y_1,y_2)^T \in \mathbb{R}^{2} , \ \ \ t \in [-1,1]. 	\]

The boundary conditions are obtained from

    \begin{equation*} 	\left( 	\begin{array}{cc} 	1 & 0 \\ 	0 & 0 \\ 	\end{array} 	\right) 	\left(\begin{array}{c} 	y_{1}(-1)\\ 	y_{2}(-1) 	\end{array}\right) 	+ 	\left( 	\begin{array}{cc} 	0 & 0 \\ 	1 & 0 \\ 	\end{array} 	\right) 	\left(\begin{array}{c} 	y_{1}(1)\\ 	y_{2}(1) 	\end{array}\right)=\left(\begin{array}{c} 	-1 \\ 	-1 	\end{array}\right). 	\end{equation*}

Exact solution

    \[z(t) = \cos(\pi t).\]

The problem has a turning point at t=0 but the solution is smooth.

 
 

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