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APM3701 EXAM PACK 2023

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APM3701 EXAM PACK 2023

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  • October 21, 2023
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APM3701
EXAM
PACK
2023

, Second Order Linear Partial Differential Equations

Part I

Second linear partial differential equations; Separation of Variables; 2-
point boundary value problems; Eigenvalues and Eigenfunctions



Introduction

We are about to study a simple type of partial differential equations (PDEs):
the second order linear PDEs. Recall that a partial differential equation is
any differential equation that contains two or more independent variables.
Therefore the derivative(s) in the equation are partial derivatives. We will
examine the simplest case of equations with 2 independent variables. A few
examples of second order linear PDEs in 2 variables are:

α2 uxx = ut (one-dimensional heat conduction equation)

a 2 uxx = utt (one-dimensional wave equation)

uxx + uyy = 0 (two-dimensional Laplace/potential equation)


In this class we will develop a method known as the method of Separation of
Variables to solve the above types of equations.




© 2008, 2012 Zachary S Tseng E-1 - 1

,(Optional topic) Classification of Second Order Linear PDEs


Consider the generic form of a second order linear partial differential
equation in 2 variables with constant coefficients:

a uxx + b uxy + c uyy + d ux + e uy + f u = g(x,y).

For the equation to be of second order, a, b, and c cannot all be zero. Define
its discriminant to be b2 – 4ac. The properties and behavior of its solution
are largely dependent of its type, as classified below.


If b2 – 4ac > 0, then the equation is called hyperbolic. The wave
equation is one such example.

If b2 – 4ac = 0, then the equation is called parabolic. The heat
conduction equation is one such example.

If b2 – 4ac < 0, then the equation is called elliptic. The Laplace
equation is one such example.


In general, elliptic equations describe processes in equilibrium. While the
hyperbolic and parabolic equations model processes which evolve over time.



Example: Consider the one-dimensional damped wave equation
9uxx = utt + 6ut.

It can be rewritten as: 9uxx − utt − 6ut = 0. It has coefficients a = 9, b = 0,
and c = −1. Its discriminant is 9 > 0. Therefore, the equation is hyperbolic.




© 2008, 2012 Zachary S Tseng E-1 - 2

, The One-Dimensional Heat Conduction Equation


Consider a thin bar of length L, of uniform cross-section and constructed of
homogeneous material. Suppose that the side of the bar is perfectly
insulated so no heat transfer could occur through it (heat could possibly still
move into or out of the bar through the two ends of the bar). Thus, the
movement of heat inside the bar could occur only in the x-direction. Then,
the amount of heat content at any place inside the bar, 0 < x < L, and at any
time t > 0, is given by the temperature distribution function u(x,t). It
satisfies the homogeneous one-dimensional heat conduction equation:

α2 uxx = ut


Where the constant coefficient α2 is the thermo diffusivity of the bar, given
by α2 = k / ρs. (k = thermal conductivity, ρ = density, s = specific heat, of the
material of the bar.)




Further, let us assume that both ends of the bar are kept constantly at 0
degree temperature (abstractly, by connecting them both to a heat reservoir


© 2008, 2012 Zachary S Tseng E-1 - 3

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