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Mathematical Methods for Physicists: A concise introduction - Site Map

Mathematical Methods for Physicists: A concise introduction - Site Map

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SOLUTIONS IN POWER SERIES<br />

Solutions in power series<br />

In many problems in physics and engineering, the di€erential equations are of<br />

such a <strong>for</strong>m that it is not possible to express the solution in terms of elementary<br />

functions such as exponential, sine, cosine, etc.; but solutions can be obtained as<br />

convergent in®nite series. What is the basis of this method? To see it, let us<br />

consider the following simple second-order linear di€erential equation<br />

d 2 y<br />

dx 2 ‡ y ˆ 0:<br />

Now assuming the solution is given by y ˆ a 0 ‡ a 1 x ‡ a 2 x 2 ‡, we further<br />

assume the series is convergent and di€erentiable term by term <strong>for</strong> suciently<br />

small x. Then<br />

and<br />

dy=dx ˆ a 1 ‡ 2a 2 x ‡ 3a 3 x 2 ‡<br />

d 2 y=dx 2 ˆ 2a 2 ‡ 2 3a 3 x ‡ 3 4a 4 x 2 ‡:<br />

Substituting the series <strong>for</strong> y and d 2 y=dx 2 in the given di€erential equation and<br />

collecting like powers of x yields the identity<br />

…2a 2 ‡ a 0 †‡…2 3a 3 ‡ a 1 †x ‡…3 4a 4 ‡ a 2 †x 2 ‡ˆ0:<br />

Since if a power series is identically zero all of its coecients are zero, equating to<br />

zero the term independent of x and coecients of x, x 2 ; ...; gives<br />

and it follows that<br />

The required solution is<br />

y ˆ a 0<br />

2a 2 ‡ a 0 ˆ 0; 4 5a 5 ‡ a 3 ˆ 0;<br />

2 3a 3 ‡ a 1 ˆ 0; 5 6a 6 ‡ a 4 ˆ 0;<br />

3 4a 4 ‡ a 2 ˆ 0; <br />

a 2 ˆ a 0<br />

2 ; a 3 ˆ a 1<br />

2 3 ˆa 1<br />

3! ; a 4 ˆ a 2<br />

3 4 a 0<br />

4!<br />

a 5 ˆ a 3<br />

4 5 ˆ a1<br />

5! ; a 6 ˆ a 4<br />

5 6 ˆa 0<br />

6! ; ...:<br />

1 x2<br />

2! ‡ x4<br />

4! x6<br />

6! ‡ !<br />

‡ a 1 x x3<br />

3! ‡ x5<br />

5! ‡ !<br />

;<br />

you should recognize this as equivalent to the usual solution<br />

y ˆ a 0 cos x ‡ a 1 sin x, a 0 and a 1 being arbitrary constants.<br />

85

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