Note on a Passage in Fourier's Heat (original) (raw)

In finding the motion of heat in a sphere, Fourier expands a function F x {\displaystyle Fx} , arbitrary between the limits x = 0 {\displaystyle x=0} and x = X {\displaystyle x=X} , in a series of the form

a 1 sin ⁡ n 1 x + a 2 sin ⁡ n 2 x + {\displaystyle a_{1}\sin n_{1}x+a_{2}\sin n_{2}x+} & c . {\displaystyle c.}

where n 1 {\displaystyle n_{1}} , n 2 {\displaystyle n_{2}} , & c {\displaystyle c} . are the successive roots of the equation

tan ⁡ n X n X = 1 − h X {\displaystyle {{\tan nX} \over {nX}}=1-hX}

Now Fourier gives no demonstration of the possibility of this expansion, but he merely determines what the coefficients a 1 {\displaystyle a_{1}} , a 1 {\displaystyle a_{1}} , & c {\displaystyle c} would be, if the function were represented by a series of this form. Poisson arrives, by another method, at the same conclusion as Fourier, and then states this objection to Fourier’s solution; but, as is remarked by Mr Kelland (Theory of Heat, p. 81, note), he “does not appear, as far as I can see, to get over the difficulty.” The writer of the following article hopes that the demonstration in it will be considered as satisfactory, and consequently as removing the difficulty.

Let n i X = ε i {\displaystyle n_{i}X=\varepsilon _{i}} , π x X = x ′ {\displaystyle {{\pi x} \over X}=x'} , and F x = f x ′ {\displaystyle Fx=fx'}

Then the preceding series will take the form

a 1 sin ⁡ ε 1 x π + a 2 sin ⁡ ε 2 x π + {\displaystyle a_{1}\sin {{\varepsilon _{1}x} \over \pi }+a_{2}\sin {{\varepsilon _{2}x} \over \pi }+} & c . , {\displaystyle c.,}

the accents being omitted above x {\displaystyle x} .

Now it is shewn by Fourier, that

ε i = ( 2 i − 1 2 − c i ) π , {\displaystyle \varepsilon _{i}=\left({{{2i-1} \over 2}-c_{i}}\right)\pi ,}