how can in plot(teta,Xb)

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CHANDRA SHEKHAR BESTA
回答： Walter Roberson 2014-3-14
Sir,
I have a function like;
teta=((2/Bi)*Xb)+(1-3*(1-Xb)^n+2*(1-Xb))+((6/Da)*(1-(1-Xb)^m));
Where
teta=0:0.2:5;
n=2/3;
m=1/3;
Bi=constant's(inf, 100, 10, 1, 0.1)
Da=constant(100, 10, 1, 0.1)
I want to plot(teta,Xb)
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Salaheddin Hosseinzadeh
Salaheddin Hosseinzadeh 2014-3-14
Ok, sounds better now
I'll post you an answer in some minutes ;)
CHANDRA SHEKHAR BESTA
CHANDRA SHEKHAR BESTA 2014-3-14
Thank you i am waiting for your post.

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回答（2 个）

Salaheddin Hosseinzadeh
We have to solve your equation so that we get Xb values for known values of teta
So let's do it as such
teta=0:0.2:5;
n=2/3;
m=1/3;
Bi=(inf, 100, 10, 1, 0.1)
Da=(100, 10, 1, 0.1)
Bi=100;
Da=100;
for i= 1:length(teta)
fxb=@(Xb) teta(i)-((2/Bi)*Xb)+(1-3*(1-Xb)^n+2*(1-Xb))+((6/Da)*(1-(1-Xb)^m))
valXb(i)=fzero(fxb,0); %this gives you Xb for (Bi=100,Da=100,teta=0:.25:5)
end
plot(teta,valXb)
title(['teta VS Xb for Bi=',num2str(Bi),'Da=',num2str(Da)])
Please work out he rest on your own, for different values of Bi and Da, either use for loop or matrix multiplication.
Hope that helped. code is not tested let me know if it had errors, sorry fo that
Good Luck!
Walter Roberson
Walter Roberson 2014-3-14
There are up to three real-valued solutions for each teta. For some Bi, Da combinations, there are no real-valued solutions. For other combinations, there are three real-valued solutions until teta passes a threshold, after which there is only one real-valued solution.
The algebraic solution for Xb in terms of teta is pretty messy.

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