Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he solution we need is u(x, t) = 5 e x3 3t . Comment: you can see that the initial condition was chosen carefully so that the method would work. However, for this particular pde, we can make the method work for many other initial conditions! Here’s how: • Suppose we are given u (x, 0) = f (x). • We’ve shown that u(x, t) = C5 x e t is a solution to the PDE for any value of C5 and any value of . By the principle of superposition, then, any sum of these solutions will also be a solution. AMATH 350 Assignment #8 Solutions - Fall 2013 Page 7 • This means that we if we construct an infinite series, by letting take on the integer values 0, 1, 2, . . . , and letting C be an arbitrary constant for each one of these values, the series will also represent a solution, as long 1 X as it converges! So, we may write u (x, t) = Cn xn ent . n=0 • If we take this expression and set t = 0, we find that we can match any initial...
View Full Document

This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

Ask a homework question - tutors are online