Quantitative Methods: Numerical Solution of DEs

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SUMMARY

The discussion focuses on deriving the numerical solution of differential equations (DEs) using first-degree polynomial interpolation over the interval [xn, xn+1]. It establishes that if f(x) is approximated by a linear polynomial at the endpoints xn and xn+1, the resulting formula for y(xn+1) is y(xn+1) = y(xn) + 1/2[f(xn) + f(xn+1)]h. This derivation utilizes the fundamental relationship y' = f(x) and the equation y(xn+1) = y(xn) + f(xn+1) - f(xn) to arrive at the approximation.

PREREQUISITES
  • Understanding of first-degree polynomial interpolation
  • Familiarity with differential equations and their numerical solutions
  • Knowledge of the concept of finite differences
  • Basic skills in calculus, particularly derivatives
NEXT STEPS
  • Study the method of finite differences for numerical differentiation
  • Explore higher-order polynomial interpolation techniques
  • Learn about the Runge-Kutta methods for solving DEs
  • Investigate error analysis in numerical methods for differential equations
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Students and professionals in mathematics, engineering, and computer science who are involved in numerical analysis and the solution of differential equations.

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Homework Statement



Show that if we fit f(x) over [xn,xn+1], with a first-degree polynomial that interpolates f at xn and xn+1, then f(x)=f(xn)+[f(xn+1)-f(xn)](x-xn)/h. Putting that approximation into the relevant equation 1 which is given at relevant equations part derive the approximation:
y(xn+1)=y(xn)+1/2[f(xn)+f(xn+1)]h

Homework Equations



eq.1 : y(xn+1) = y(xn) + f(xn+1) - f(xn)

Also we know that y'=f(x) and y(xn+1)=y(xn)+f(xn)h
 
Physics news on Phys.org
What have you done? If you haven't done so, draw a picture. Your interpolation polynomial is first-degree, which means it's a straight line.
 

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