Problem 3a: Hyperbolic PDE
This problem holds an elastic string of length X=1 at both ends and deflects it at
its midpoint by a max of u=.1. The string is then released and allowed to perform
undamped oscillations. Since the hyperbolic PDE has 2 second order derivatives with
respect to time and 2 second order derivatives with respect to position there needs to be
an additional boundary condition. The 2 with respect to position are where
u(0,t)=u(1,t)=0. The boundary conditions with respect to time includes the initial
deflection f(X)=u(X,0) which is the piecewise function defined in the problem
description and the initial speed which is defined as function g(X)=0. These 4 boundary
conditions are used in the Hyperbolic fortran file and produced a text file 86 which was
plotted in GNUPLOT. This shows how the function behaves along the length of the
string. It increases to a max and then decreases back to where it started. Text files 84 and
85 shows that the potential starts at approximately .0988 at time zero. It decreases into a
negative maximum at -.988 at t=1s and increases back to a positive maximum at the same
point it started, .0988 at time t=2.05s.
Hyperbolic PDE Plot (File 86)
Problem 3a: Hyperbolic PDE
Hyperbolic PDE Plot (File 85)