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Partial differential equations

Contents
Contents

iii

Chapter 1
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1
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2
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3
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4
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1
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The following are some
standard tests for determining the convergence or divergence of such a series
...

(2) Ratio test:
n+1
If an = 0 for all n, and if limn→∞ aan = L, then
• converges (absolutely) if L < 1,
• diverges if L > 1
...


n

an

(3) Root test:
1
If limn→∞ |an | n = L, then n an
• converges (absolutely) if L < 1,
• diverges if L > 1
...

(4) Refined root test:
1
If lim supn→∞ |an | n = L, then n an
• converges (absolutely) if L < 1,
• diverges if L > 1
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Remark 1
...
In the ratio or root test, the existence of L is a pre-requisite
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(See wikipedia for an
explanation
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1
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Power series
A power series in x is a series of the form
∞

an (x − x0 )n ,

(1
...
(Note that there is one series for each value of x
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1

2

1
...
1) converges for some (real or complex) number
x1 = x0
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It can be shown by applying the refined root test that
the series is (absolutely) convergent for all x such that |x − x0 | < r, that is, in the
open interval (or disc) of radius r centred at x0
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We do not consider them
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Definition 1
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The largest positive number R, including ∞, such that a given
power series converges in the open interval (or disc) {|x − x0 | < R} is called the
radius of convergence of the power series
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gence of the power series is
(2) Root test:
1
If limn→∞ |an | n = L, then the radius of convergence of the power series
1
is L
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For example, the geometric series n xn has radius of convergence R = 1
...
3
...
3
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Since power series can be differentiated term by term as often as we please, real
analytic functions are infinitely differentiable, and the above power series representation is necessarily the Taylor series of f around x0 , that is,
f (n) (x0 )

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A power series is real analytic in its interval of convergence
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The Taylor
series
x5
x3
+
+
...

2!
4!
x2
x3
ex = 1 + x +
+
+
...
4
...
However this may not always be the case
...
)
f (x) =
1 + x2
The series on the right has a radius of convergence 1 and the identity only holds
for |x| < 1
...

The function x1/3 is not differentiable at 0 and hence not real analytic in any
interval containing 0
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For instance,
1
1 1
(x − 1)2
x1/3 = (1 + (x − 1))1/3 = 1 + (x − 1) + ( − 1)
+
...
(This is the binomial theorem which we will prove below
...
But it is not real analytic at 0 since f (n) (0) = 0 for all n
and the Taylor series around 0 is identically 0
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If a function of a complex variable is differentiable in a region, then it is complex
analytic in that region
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)
1
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Solving a linear ODE by power series
Consider the initial value problem
(1
...

Theorem 1
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Let r > 0 be less than the minimum of the radii of convergence
of the functions p, q, r and g expanded in power series around a
...
Then there is a unique solution to the initial
value problem (1
...
3)

an (x − x0 )n

y(x) =
n≥0

whose radius of convergence is at least r
...
See [1, Chapter 5, Appendix A]
...
In particular, it applies to the first order linear ODE for which
the initial condition is simply the value at x0
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3)
into (1
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In most examples, x0 = 0 and the functions p, q, r and g are
polynomials
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POWER SERIES

Example 1
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Consider the first order linear ODE
y ′ − y = 0,

y(0) = 1
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4 applies and it will yield a solution
n
which is real analytic on all of R
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The initial condition
n
y(0) = 1 implies a0 = 1
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This is a 2-step recursion
...
Thus y(x) =
1 n
x
n n! x (which we know is e )
...
6
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We proceed as in the previous example
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This time we get a 3-step recursion:
(n + 1)an+1 = 2an−1 for n ≥ 1,
and a1 = 0
...
For the even coefficients, note that
2
1
na2n = a2n−2 , so a2n = 1/n!
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Example 1
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Consider the function f (x) = (1 + x)p where |x| < 1 and p is any
real number
...

(1 + x)y ′ = py,

y(0) = 1
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Since 1 + x is zero
at x = −1, we are guaranteed a solution only for |x| < 1
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We get a 2-step recursion:
p−n
an+1 =
an for n ≥ 0
...

(1 + x)p = 1 + px +
2
This is the binomial theorem
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Also check by the ratio test that the power series has radius of convergence 1
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8
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(The initial conditions are left unspecified
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There is a general method to solve recursions of
this kind
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Consider the quadratic
λ2 + λ − 2 = 0 (which can be written directly from the constant-coefficient ODE)
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The general solution to the recursion is then
an = α

1
(−2)n
+β

...
4
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So the general solution to the ODE is
y(x) = αex + βe−2x
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But this may not always be possible
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F
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, New York, 1972, International Series in Pure and Applied Mathematics

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