KatePirs

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8 years, 174 days

MaplePrimes Activity


These are replies submitted by KatePirs

Thank you very much!

Thank you very much!

I'm sorry, how do I get the value of x separately, in order to assign a value to the new variable.

I'm sorry, how do I get the value of x separately, in order to assign a value to the new variable.

I'm sorry, I understand. Thank you again!

I'm sorry, I understand. Thank you again!

Thank you so much! I'm sorry, but I'm new to working with this program.
Could you explain me please, what means " U1(.3)"   

0.3 Which variable it corresponds to  ?

Thank you so much! I'm sorry, but I'm new to working with this program.
Could you explain me please, what means " U1(.3)"   

0.3 Which variable it corresponds to  ?

@Preben Alsholm  Thank you for your decision. But I have already solved this problem by adding a spacestep = 1/5. Now everything is resolved as it should, without changing the boundary conditions, but I'm not sure that 1/5 is the correct value. Do not fully understand what value should be taken.

@Preben Alsholm  Thank you for your decision. But I have already solved this problem by adding a spacestep = 1/5. Now everything is resolved as it should, without changing the boundary conditions, but I'm not sure that 1/5 is the correct value. Do not fully understand what value should be taken.


unfortunately it's not right,
because when I use t=0, like in intial conditions I don't get the right plot which should be for ur = 1/cosh(x), but I get smth else(


unfortunately it's not right,
because when I use t=0, like in intial conditions I don't get the right plot which should be for ur = 1/cosh(x), but I get smth else(

These boundary conditions are equivalent to the fact, that this function at infinity tend to zero. But however I changed 500 to 100 (-500 to -100) and get the result!

I do not know if I can change this, but still thank you very much for your help. I am very grateful to you! 

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