STEADY STATE TEMPERATURES IN A QUARTER PLANE

The discontinuous boundary value problem of steady state temperatures in a quarter plane gives rise to a pair of dual integral equations which are not of Titchmarch type. These dual integral equations are considered in this paper.


INTRODUCTION.
We consider the problem of steady state temperatures in a quarter plane (see Fig. 1), whose edge x 0 is losing heat to environment at zero temperature according to Newton's Law of cooling while on the edge y 0, temperature is controlled on portion of this edge, while the heat input is known on the remaining part.Typically, this problem is governed by: Find u u(x,y) such that Vu 0u + Ou 0 in x > 0, y > 0; (1.1a) f-m= 0 on x 0 in y > 0; (1.1b) and either (1) u(x,0) f,(x) in 0 < x < 1 (1.2a) and Uy(X,0) -g(x) in x > 1 (1.2b) or (2) uy(x,0) -f,(x) in 0 < x < 1 and u u(x,0) g,(x) in x > 1.   where the subscript denotes differentiation w.r.t, that variable.
Also, in each case we require that ul be bounded at infinity.
An appropriate representation for u u(x,y) in this case is u(x,y) f f(t)(a sin xt + cos xt) e -ty dt in x > 0, y > 0.
We propose to solve such dual integral equations for the function f(t) in this paper.We point out that these equations are not of Titchmarch type (because the kernel k(x,t) a sinxt + cos xt is not a Fourier Kernel) and to our knowledge, have not been considered before.While the kernel k(x,t) has been successfully inverted [1, page 70], dual integral equations involving this kernel have not been considered previously.We shall attempt oly a formal solution of these dual integral equations, and shall assume 'throughout that the functions f(x) and gL(x) are continuous in 0 <_ x <_ 1 and in x _> 1 respectively. 2.
And for both the cases, the equations (2.3) give F(x) e -'ax fXeatf,(t)dr, 0 < x < 1, (2.6a) and G(x) e -ax fXeatg1(t dt,+ Be -ax in x > 1, (2.6b) It remains to determine the constant B. We shall determine this constant by the (physically realistic) condition that the quantity u(x,o) is continuous at x 1.

3.
SOLUTION FOR THE FIRST CASE.
In this case, the dual integral equations (1.5) are reduced to dual equations m x f f(t)sin xt dt F(x) e -ax f e ft(t) dt in 0 < x < 1 and f tf(t)sinxt dt= e -ax fXet gt(t)dt + Be -ax in x > 1.
In deriving equation (3.3a), we have used the fact that F(0) 0.
(3.14)This proves the continuity of u(x,0) at x 1.It can also be seen that if B is given by (3.8), then under suitable restrictions on the data, u(x,0) as given by equation (3.9) is bounded as x 4.
The case of a 0 is completely different, because for bounded u, the representation u(x,y) f(R)f(t)(a sin xt / cosxt)e -ty dt is no more valid.The correct representation now is u(x,y) c, + f tf(t)(cos xt)e -ty dt.
where C is a constant.
And, Case Find C and f(t), such that us consider dual equations (5.2) in Case 1.We shall again determine C1, by the requirement that u(x,0) is continuous at x 1.We have from (5.2) f(R)tf(t)cosxt dt fl(x)-Ct, 0 < x < 1 and f tf(t)sinxt dt f (x)dx g(x), say.
Once again, it can be seen from (5.7) that if g(x) is suitably restricted then u(x,o) is bounded as x (R).
For Case 2, the solution is given by (4.2) in the limit as a 0'.It should be pointed out that the problem posed by equations (5.2) has been considered by Sneddon [4, page 99].Sneddon considers the problem (5.2) with C =0 and gl(x) 0. He then imposes the condition that the heat input on y 0 must remain finite as x 1-and arrives at the conclusion that we must have F(1) 0. All this, however, is a special case of our equation (5.8) wherein if C 0 and GI(1 0, we get F(1) 0. It would appear therefore that this problem ought to be posed as we have done it.
For the particular case of gl(x) 0, the problem posed by equations (5.3) has also been considered by Sneddon  [5, page 26].For this particular case, our solution coincides with his.
We shall now consider some special cases.6.
(6.1b) with the (additional) requirement that the quantity f f(t) (a sinxt + cosxt) dt is continuous at x 1.
In this case f(t) --i--+ 1 f u J0(ut) K0(au du K0(a) The graphs of u(x,0) for several values of p are given in Fig.