Kyusik,

Can you summarize what you have tried and which errors/problems you got 
with these approaches?
What exactly do you mean with "csc(theta) or sec(theta) can't be used in 
cylindrical coord system"?

Best,
Daniel

Am Freitag, 9. September 2016 06:17:44 UTC+2 schrieb hank...@gmail.com:
>
> Thanks for your answer. But, I still don't know how I can deal with this 
> error.
>
> Anyway, Thank you very much.
>
> Kyusik. 
>
> 2016년 9월 5일 월요일 오후 9시 47분 21초 UTC+9, Wolfgang Bangerth 님의 말:
>>
>> On 09/05/2016 06:04 AM, hank...@gmail.com wrote: 
>> > 
>> > K_Inv[0][0]=2*r/(cos(theta)*cos(theta)), 
>> > K_Inv[1][1]=2*r/(sin(theta)*sin(theta)), K_Inv[2][2]=r 
>> > 
>> > r is calculated by "r=sqrt(x^2+y^2) ", and theta is calculated by 
>> "theta=x/y" 
>> > 
>> > But, as you can expect that ,on the points where |x|<0.0000001 or 
>> > |y|<0.000001,  cos(theta) or sin(theta) is almost zero. So, It seems 
>> that It 
>> > causes the singularity. 
>> > 
>> > So, At first I tried to change the above 2 element in K_Inv like 
>> this... 
>> > 
>> > K_Inv[0][0]=2*r/(cos(theta)*cos(theta)+del) 
>> >  K_Inv[1][1]=2*r/(sin(theta)*sin(theta)+del), 
>>
>> Instead of this approach, you should use the following: if r>delta, use 
>> the 
>> formula above. If r<=delta, use l'Hopital's rule to get an expression 
>> that 
>> avoids the division by zero. 
>>
>> That said, you still have a problem for those values of x,y where theta 
>> is 
>> close to zero or one, but r is not. For example, for x=0, y=1, you have 
>>    r=1 
>>    theta=pi/2 
>>    cos(theta)=0 
>>    K_inv[0,0] = 2/0 
>> You need to think about what you want to do with this situation. 
>>
>> Best 
>>   W. 
>>
>> -- 
>> ------------------------------------------------------------------------ 
>> Wolfgang Bangerth          email:                 bang...@colostate.edu 
>>                             www: http://www.math.colostate.edu/~bangerth/ 
>>
>>

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