The name is absent



70 Lectures on Scientific Subjects

so that

Further, we may differentiate once under the integral sign
so that (ʃ2-
ti)2 is replaced by the derivative of this ex-
pression. If we write ʃ
=tu we see, however, that if и =s∕t > 1,
ds         du

Hence we obtain the final solution for m = 1 :

(4)      A1W= ⅛


(w=→l).
2>


Thus our general conclusion in the case of a harmonic
distribution of the first order is as follows:
In the case of a
given harmonic density function of the first order
= 1)

Fi(r, θ) ≈=Fι(r) cos 0÷Gι(r) sin θ,

where ZZ1W = Ffr) +iGfr) remains finite near r = 0,3 there
exists a corresponding continuous distribution function,

/i(ʃ, v) =√ιW cos φ+i gι(j) sin φ,

likewise bounded near j=0 and harmonic of the first order, if
and only if the integral in
(4) represents a function with a
continuous derivative. In this event the unique solution hfs)
is exhibited in
(4), where of course,

AiW =∕ιW+⅛ιW∙

9∙ THE SOLUTION IN HIGHER HARMONIC CASES m > 1

The method of solution used above applies essentially to
any order m>l. Here we start by writing the identity

•We are assuming that F{r, θ} is continuous for r>0 only.



More intriguing information

1. The name is absent
2. Benchmarking Regional Innovation: A Comparison of Bavaria, Northern Ireland and the Republic of Ireland
3. LOCAL PROGRAMS AND ACTIVITIES TO HELP FARM PEOPLE ADJUST
4. The name is absent
5. The name is absent
6. Empirical Calibration of a Least-Cost Conservation Reserve Program
7. Personal Income Tax Elasticity in Turkey: 1975-2005
8. Towards Teaching a Robot to Count Objects
9. PACKAGING: A KEY ELEMENT IN ADDED VALUE
10. MULTIMODAL SEMIOTICS OF SPIRITUAL EXPERIENCES: REPRESENTING BELIEFS, METAPHORS, AND ACTIONS