Solve 0=u''+u*e^x | Poincare Return Map

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Thread starter Nusc
Start date Nov 1, 2009
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SUMMARY

The discussion focuses on transforming the second-order differential equation \( u'' + u e^x = 0 \) into a planar system. The transformation involves defining \( z = u' \), leading to the system \( \{ u' = z, z' = -e^x u \} \). This can be expressed in matrix form as \( U' = A U \), where \( U \) is the column vector \( (u, z) \) and \( A \) is the matrix \( \begin{pmatrix} 0 & 1 \\ -e^x & 0 \end{pmatrix} \). The discussion also touches on the implications of the Poincare return map for the case when \( u > 0 \).

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Understanding of second-order differential equations
Familiarity with planar systems and state-space representation
Knowledge of matrix algebra, specifically 2x2 matrices
Concept of Poincare return maps in dynamical systems

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Study the derivation of planar systems from higher-order differential equations
Learn about the stability analysis of linear systems using eigenvalues
Explore the properties and applications of Poincare return maps
Investigate numerical methods for solving differential equations, such as Runge-Kutta methods

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Nov 1, 2009

Nusc

How do I write u''+u*e^x = 0 as a planar system?

Last edited: Nov 1, 2009

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Nov 2, 2009

Nusc

Okay its:

Let z=u', then z'=u''=-exp(x)*u.

So the planar system is {u'=z,z'=-exp(x)*u}, or in matrix form, U'=A*U, where U is the column vector (u,z) and A is the 2x2 matrix (0,1,-exp(x),0).

What can be said about the Poincare return map on the transversal u>0?