PSI - Issue 40

A.I. Chanyshev et al. / Procedia Structural Integrity 40 (2022) 97–104 Chanyshev A.I. at al. / Structural Integrity Procedia 00 (2022) 000 – 000

99 3

( ) ( ) u f z g z   , where f and g are arbitrary functions.

and the solution is of the form Function u must be real, so

(5)

( ) ( ) u f z f z   ,

that is, it must be a doubled real part of the function ( ) Re ( ) Im ( ) f z f z i f z   . To find the function ( ) f z , we use the boundary conditions (2), (3). Applying them, we find

(6)

( i i f a e f a e     ) (

) 2 ( )   

,

2 ( )   

( i i e f a e e f a e         ) ( i i

)



(7)



.

a

We differentiate (6) by the polar angle  . As a result, instead of (6), (7), we have the system

  

( i i a e f a e i a e f a e a e f a e a e f a e                    ) ( ( ) ( i i i i i i i

) 2 ( ),    

(8)

) 2 ( ).    

From (8) we find

( i i i a e f a e  

( )       ( ) i

(9)

)



.

 

Hence, given that

( ) d d 

( ) d d 

d f

d f

 



  

  

( f a e 

i

)



,

,

,





i

i i a e d

( d a e

)



 



we obtain that

( i d f a e d  )

( )       . ( ) i d

Integrating it, we find

( i f a e

(10)

) ( )       ( ) i



(owing to (3), the constant of integration can be placed in a function ( )   ). In this formula

1 ln z i a

i z e а 



 

i z a e   ,

,

.

Hence it follows that

1

1

z

z

  

  

  

  

( )

ln

ln

f z

i









(11)

.

i a

i a

Owing to (10), this formula is valid at r a  . Solution (11) is continued inside the contour r a  . It is obvious, that the boundary conditions (2), (3) are satisfied. Moreover, at each point inside the circle, the Laplace equation (1)