625 
The general solution of the problem must be composed of special 
solutions in the following way : 
6,= SAMS, 
k=1 
SEO eer Tc. setae Ne! 
The constants A,“ can only be determined in connection with the 
value of 0,. The expression given by (18) satisfies the boundary condition 
(10), which holds at the boundary surface of the solid substance and 
the wall of the tube, and is also in agreement with (9). 
The value of the temperature @, prevailing in the liquid is found 
in an analogous way. It is: 
2 yr & (hk) En MGR 
Ee Add lr HEEE 7 OET 
ll a 
The quantities $,%) are the roots of the equation: 
Ya se J, (5) — J, (54) ° : : : y } 2 (20) 
in which: 
1,8 (ai 
Un Za e . . e je . ° . . ) 
From §,) follows p,): 
‘J. Us | Ban a 
en a iaucastilriagedns 82) 
In conclusion the constants A,“ and A,® occurring in (18) and 
(19), must be determined from the conditions (8) at the boundary 
surface of the solid and the liquid phase. By the aid of (18) and 
(19) these conditions become: 
» & (k) Ee „& (k) 
Su ey ieee (23) 
—] ( a =i a 
ao r 8%) r Ek 
= A, (*) A, p,* Je + A ut) Ay i (k) J == 0, v,. (24) 
il a 
Both equations must hold for all values of r. 
The difficulty to find the constants A,’ and A,“ from (23) and 
(24), consists in this that in these equations tore oceur two series 
»< (%) 
Rees 
iis 5 s 
i These series 
of normal functions, viz. J, 
are, indeed, each in themselves orthogonal; but Ee functions of one 
series are not orthogonal to those of the other. The most symmetrical 
way would be to try and find normal functions belonging to the 
whole space, and not, as had been done up to now, either to space 
1 (solid substance) or to space 2 (liquid). There exists, however, a 
simple — though asymmetrical — method, which leads to the pur- 
