From f73eaf79e845f0548b169a46783c99a106df3fc9 Mon Sep 17 00:00:00 2001
From: luccul <luccul@gmail.com>
Date: Tue, 6 Jul 2010 06:00:21 +0000
Subject: formatting doh

---
 ClassJuly5.page | 19 ++++++++++++++++++-
 1 file changed, 18 insertions(+), 1 deletion(-)

(limited to 'ClassJuly5.page')

diff --git a/ClassJuly5.page b/ClassJuly5.page
index 632ecec..19b3d47 100644
--- a/ClassJuly5.page
+++ b/ClassJuly5.page
@@ -7,28 +7,45 @@ Below are answers to some questions that came up during the lecture.  If anyone
 Suppose instead of having a metal ring we had a metal rod of length $L$, and we kept the ends of the rod at constant temperatures $u_0$ and $u_L$.  How might we solve the heat equation in this context?
 
 There is one obvious solution that satisfies these boundary conditions, namely the time-independent or steady state solution
+
 $$ g(x,t) = u_0 + \frac{u_L- u_0}{L}x $$
+
 This satisfies the heat equation for trivial reasons since it is time-independent and its second spatial derivative is zero, hence both sides of the heat equation are zero independently of one another.
 
 Now suppose that $h(x,t)$ is another solution satisfying the same boundary conditions.  Then the function
+
 $$ u(x,t) = g(x,t) - h(x,t)$$
-also satisfies the heat equation, but it is zero at both endpoints.
+
+also satisfies the heat equation, but it is zero at both endpoints:
+
 $$ u(x,0) = u(x,L) = 0 $$
+
+
 To solve for $h$, it is clear that we only need to solve for $u$.  First we'll use what's called the ``separation of variables'' trick to generate a lot of nice solutions, then hope and pray that any other solution can be expressed as a linear combination of these.
 
 Here's how the separation of variables trick works.  We seek a solution of the form
+
 $$u(x,t) = a(t)b(x)$$
+
 for some functions $a$ and $b$.  Then the heat equation tells us:
+
 $$ \frac{da}{dt}b = a \frac{d^2b}{dx^2} $$
+
 Rearranging, we get:
+
 $$ \frac{da}{dt}/a = \frac{d^2b}{dx^2}/b $$
+
 The left hand side only depends on $t$ and the right hand side depends only on $x$.  Therefore, neither side depends on either $x$ or $t$.  We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$.  
 
 First we solve for $a$.  It satisfies the equation
+
 $$ \frac{da}{dt} = - \lambda^2 a$$
+
 and therefore (up to a constant multiple) it is given by:
+
 $$ a(t) = e^{-\lambda^2 t} $$
 
+
 Next we solve for $b$.  It satisfies the equation
 $$ \frac{d^2b}{dx^2} = -\lambda b $$
 However we have to be a bit more careful in picking our solutions because $b$ is supposed to satisfy the boundary conditions
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