Solve Assignment 4 - Introduction to Partial Differential Equations | MATH 442, Assignments of Differential Equations

Material Type: Assignment; Class: Intro Partial Diff Equations; Subject: Mathematics; University: University of Illinois - Urbana-Champaign; Term: Spring 2010;

Typology: Assignments

Pre 2010

Uploaded on 03/11/2009

koofers-user-i2m
koofers-user-i2m 🇺🇸

4

(1)

8 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 442 Homework 4 Feedback
Questions 1 and 2. Conservation of energy for the wave equation in a region
Din 3-dimensions:
E(t) = 1
2ZD
[u2
t+c2u· u]dV
E0(t) = ZD
[ututt +c2u· ut]dV by the product rule
=ZD
[ut(c2∇·∇u) + c2u· ut]dV
by the Wave Equation utt =c2u=c2∇·∇u
=c2ZD
· (utu)dV by the product rule for divergence
=c2Zbdy D
(utu)·~n dS by the Divergence Theorem
= 0
because:
(i) under Dirichlet BC, u(~x, t) = 0 for all ~x bdy Dand all timplies
ut(~x, t) = 0 for all ~x bdy Dand all t;
(ii) under Neumann BC, u·~n =∂u
∂n = 0 for all ~x b dy Dand all t.
pf2

Partial preview of the text

Download Solve Assignment 4 - Introduction to Partial Differential Equations | MATH 442 and more Assignments Differential Equations in PDF only on Docsity!

Math 442 — Homework 4 Feedback

Questions 1 and 2. Conservation of energy for the wave equation in a region D in 3-dimensions:

E(t) =^12

D^ [u

(^2) t + c (^2) ∇u · ∇u] dV E′(t) =

D^ [ututt^ +^ c

(^2) ∇u · ∇ut] dV by the product rule

D^ [ut(c

(^2) ∇ · ∇u) + c (^2) ∇u · ∇ut] dV by the Wave Equation utt = c^2 ∆u = c^2 ∇ · ∇u = c^2

D^ ∇ ·^ (ut∇u)^ dV^ by the product rule for divergence = c^2

bdy D^ (ut∇u)^ ·^ ~n dS^ by the Divergence Theorem = 0 because:(i) under Dirichlet BC, u(~x, t) = 0 for all ~x ∈ bdy D and all t implies ut(~x, t(ii) under Neumann BC,) = 0 for all ~x ∈ bdy D ∇and allu · ~n = t ;∂u ∂n = 0 for all^ ~x^ ∈^ bdy^ D^ and all^ t.

Question 3. Dissipation of energy for the diffusion equation in a region3-dimensions: assuming u D in t =^ k∆u^ in the region^ D, we have E(t) =^12

D^ u

(^2) dV

E′(t) =

D^ uut^ dV = k

D^ u(∇ · ∇u)^ dV^ by the Diffusion Equation^ ut^ =^ k∆u^ =^ k∇ · ∇u = k

D^ [∇ ·^ (u∇u)^ − ∇u^ · ∇u]^ dV^ by the product rule for divergence = k

bdy D^ (u∇u)^ ·^ ~n dS^ −^ k

D^ |∇u|

(^2) dV by the Divergence Theorem

= −k

D^ |∇u|

(^2) dV using the Dirichlet BC u = 0 or the Neumann BC ∇u · ~n = 0 ≤ 0. Thus the energy is dissipated: E′(t) ≤ 0.

Section 2.3 #6. Letsatisfies w ≤ 0 when w t == 0 and when u − v, so that x w= 0 satisfies the diffusion equation, and, `. Then w ≤ 0 for all x, t, by the Maximum Principle. That is, u ≤ v for all x, t.