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An Introduction to Auction Theory$

Flavio M. Menezes and Paulo K. Monteiro

Print publication date: 2004

Print ISBN-13: 9780199275984

Published to Oxford Scholarship Online: April 2005

DOI: 10.1093/019927598X.001.0001

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(p.159) Appendix B Differential Equations

(p.159) Appendix B Differential Equations

Source:
An Introduction to Auction Theory
Publisher:
Oxford University Press

We collect here a few results on ordinary differential equations that are used in the text.

B.1 The Simplest Differential Equation

Suppose f:ℝ → ℝ is a continuous function. Let x 0 and A be given real numbers. If we want to find a differentiable function y:ℝ → ℝ such that

(B.1)
y ( x ) = f ( x ) , x ∈ℝ; y ( x 0 ) = A ,
we say that we want to solve the differential equation y′ = f with initial condition y(x 0) = A. This equation is solved using the Fundamental Theorem of Calculus.

Theorem 27 Fundamental Theorem of Calculus If two differentiable functions g(x) and h(x) have the same derivative then they differ by a constant.

To solve (B.1) consider ψ ( x ) = A + x 0 x f ( y ) d y . Then note that ψ(x 0) = A and ψ′(x) = f(x) for every x since f(·) is continuous. Suppose now that y(·) solves (B.1). Then since y′(x) = f(x) = ψ′(x) we have by the fundamental theorem of calculus that ψ(x) = y(x) + C for some constant C. Choosing x = x 0 we see that C = 0.

B.2 Integrating Factor

The integrating factor is used to solve linear differential equations. It is used several times in this book. It can be applied to equations of the following form:

b ( x ) + b ( x ) Q ( x ) = R ( x ) .
(p.160) To understand how it works note that the left-hand side of the above equation has a slight similarity to the derivative of a product: (bp)′ = bp + bp′. Suppose we multiply the left-hand side by a function P(x). It becomes
b ( x ) P ( x ) + b ( x ) P ( x ) Q ( x ) .
If we can choose P(x) such that P′(x) = P(x)Q(x) (this P is called an integrating factor) we transform the left-hand side into the derivative of the product b(x)P(x):
( b ( x ) P ( x ) = b ( x ) P ( x ) + b ( x ) P ( x ) = ( b ( x ) + b ( x ) Q ( x ) ) P ( x ) = R ( x ) P ( x ) .
Now to find P such that P′ = PQ is easy. Note that
( log P ) = P P = Q .
Thus P(x) = eQ(x) dx is an integrating factor. And our differential equation has a solution:
b ( x ) = R ( x ) P ( x ) P ( x ) .
The reader may wonder why we did not mention the initial condition. This will, however, be taken care of in the text each time we use the integrating factor.