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HW 8 Mathematics 503, Mathematical Modeling, CSUF , July 12, 2007

Nasser M. Abbasi

October 8, 2025

Contents

1 Problem 6 page 204 section 3.6

1 Problem 6 page 204 section 3.6

problem:

Write down the equations that determine the solution of the isoperimetric problem

\[{\displaystyle \int \limits _{a}^{b}} p\left ( x\right ) y^{\prime 2}+q\left ( x\right ) y^{2}\ dx\rightarrow \min \]

Subject to

\[{\displaystyle \int \limits _{a}^{b}} r\left ( x\right ) y^{2}dx=1 \]

where \(p,q,r\,\)are given functions and \(y\left ( a\right ) =y\left ( b\right ) =0\).

Answer

Since \(y\left ( x\right ) \) is fixed at each end, this is not a natural boundary problem. Therefore one can use the auxiliary lagrangian approach, where we write the auxiliary Lagrangian \(L^{\ast }\) as

\[ \fbox {$L^{\ast }=L+\lambda G$}\]

Where \(L\left ( x,y,y^{\prime }\right ) =p\left ( x\right ) y^{\prime 2}+q\left ( x\right ) y^{2}\), and \(G=r\left ( x\right ) y^{2}\) and \(\lambda \) is the Lagrangian multiplier. Hence

\[ \fbox {$L^{\ast }=p\left ( x\right ) y^{\prime 2}+q\left ( x\right ) y^{2}+\lambda r\left ( x\right ) y^{2}$}\]

Hence now we write the solution as the Euler-Lagrange equation, but we use \(L^{\ast }\) instead of \(L\)

\begin{align*} L_{y}^{\ast }-\frac {d}{dx}L_{y^{\prime }}^{\ast } & =0\\ \left ( 2q\left ( x\right ) y+2\lambda r\left ( x\right ) y\right ) -\frac {d}{dx}\left ( 2p\left ( x\right ) y^{\prime }\right ) & =0\\ q\left ( x\right ) y+\lambda r\left ( x\right ) y-(p\left ( x\right ) y^{\prime })^{\prime } & =0 \end{align*}

Therefore the differential equation is

\[ \fbox {$(p\left ( x\right ) y^{\prime })^{\prime }-y\left ( q\left ( x\right ) +\lambda r\left ( x\right ) \right ) =0$}\]

This is a sturm-Liouville eigenvalue problem. The solution \(y\left ( x\right ) \) from the above will contain 3 constants. 2 will be found from boundary conditions, and the third, which is \(\lambda \) is found from plugging in the solution \(y\left ( x\right ) \) into the constraint given: \({\displaystyle \int \limits _{a}^{b}} r\left ( x\right ) y^{2}dx=1\)