how can in plot(teta,Xb)

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)

[Walter Roberson]

Your Bi and Da appear to be vectors of values, and those vectors appear to be different lengths. How are you planning to deal with the vectors? Are you wanting to create length(Bi) * length(Da) different plots, selecting one value from each of two vectors?

[Salaheddin Hosseinzadeh]

What's Xb? Have you tried to plot what you equated? If yes what errors you received?

[CHANDRA SHEKHAR BESTA]

yes sir,

I want to create length(Bi) * length(Da) different plots, selecting one value from each of two vectors

[CHANDRA SHEKHAR BESTA]

Already i know teta values from 0:0.2:5, by solving equation; i want to find Xb.

After finding the Xb, I want to plot(teta,Xb)

[Salaheddin Hosseinzadeh]

So the question is how to find Xb? :(

Wait a second, you're confusing me! You already have teta values! not by solving this equation, you actually want to find Xb from this equation you wrote!

You have math prblem not a MATLAB problem!

[CHANDRA SHEKHAR BESTA]

Yes, I have already teta values.

I want to find Xb,

then plot(teta,Xb)

[Salaheddin Hosseinzadeh]

Ok, sounds better now

I'll post you an answer in some minutes ;)

[CHANDRA SHEKHAR BESTA]

Thank you i am waiting for your post.

Answer 1

[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!

Answer 2

[Walter Roberson]

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.