J4James

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9 years, 68 days

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To make the picture more clearer, Here I plotted both the roots, as for as the nature of the problem goes.

 

@Markiyan Hirnyk @Carl Love 

Thanks @Markiyan Hirnyk for your prompt response.

If you see the attached maple sheet, I got exactly the same output as yours. 

But my worry is the lower part (y-axis, from 0..0.5) of the plot, which is missing from my

output compare to the orignal plot.

Thanks 

@Preben Alsholm Thanks dear for the help

@Preben Alsholm How to plot f(x,y) in 3D from eq1 only with bcs using the numerical solution?

@Preben Alsholm Dear thanks once again for your kind help. Please ignore my private msg. 

@Preben Alsholm Once we sub the numerical values of f(y) into P2 then why we need using y=0?

Is taking y=O has any effect on the results?

@Preben Alsholm Im sorry to forgot to mention that all these paramters are positive.

k>0, epsilon>0.

@Preben Alsholm Thanks for your interest. For the following parameters, I'm facing accuracy problem.

h:=evalf(1+epsilon*sin(1)):

params:={k=0.5,B=1,epsilon=0.5};

@Preben Alsholm Thanks dear as always for your helpful and clear response.

As you mentioned about writing a code, can you proivde any hint or any other example where a code (FDM, etc) has been adopted/used in maple to solve elliptic pdes?   

Cheers!

@Preben Alsholm 

Thanks to all,

If I understood correctly, then 

#_C1=0 considering symmetry and all the parameters are positive

convert(diff(eval(%,_C1=0),r), polynom)=0; 

-C1/G3+(1/2)*G5*_C2*r = 0

when we sub r=0 using the bcs, then we left with nothing as far as _C2 is concern.

Am I doing something wrong? 

 

@Alejandro Jakubi Thanks for your kind help.

@mehdi jafari 

Eq1 := (diff(psi(y), y, y, y, y))/qq-M^2*(diff(psi(y), y, y))-b*y = 0;

Eq1 := diff(psi(y), y, y, y, y)-M*(diff(psi(y), y, y))-Gr*b*y = 0;

except new parameters qq, M^2 nothing is new.

@Carl Love 

Here is my_try.mw

@sharena2 see the attached sheet.

hydro.mw

@sharena2 

I think it should work in Maple 16, 17, 18

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