3D Plot from 3 Polynomial Equations

Here is my code:

f0[\[Lambda]0_, y0_, y1_, x_, k_] :=-(-y0 + k*x*y1 + (1 - k*x^2)*\[Lambda]0^2)^2 + 4*x^2*(1 - k^2*x)*\[Lambda]0^2*Abs[-(k*y0) + y1 + k*\[Lambda]0^2 - x*\[Lambda]0^2]

f1[\[Lambda]1_, y0_, y1_, x_, k_] := -(x*y0 - y1 + (1 - k*x^2)*\[Lambda]1^2)^2 + 4*x^2*(-k + x)*\[Lambda]1^2*Abs[-(k*y0) + y1 - \[Lambda]1^2 + k^2*x*\[Lambda]1^2]

f2[\[Lambda]0_, \[Lambda]1_, x_, k_] :=-1/27 + \[Lambda]0^3 + 3*x*\[Lambda]0^2*\[Lambda]1 + 3*k*x*\[Lambda]0*\[Lambda]1^2 + \[Lambda]1^3


$k$ is some positive number which can be chosen freely for example $k=1$ or $k=3/2$ or $k=2/3$. $f_0$ is only a function of $\lambda_0$, likewise $f_1$ is only a function of $\lambda_1$.

If I would insert $\lambda_0$ from $f_0$ into $f_2$ and insert $\lambda_1$ from $f_1$ into $f_2$ then $f_2$ will be a function of only $x,k,y_0,y_1$.

In this case I would like to have a 3D plot with x axis =$y_0$, y-axis=$y_1$, and z-axis =$x$ for $7/81<y_0<1/9$ and $7/81<y_1<1/9$, for some known $k$ as mentioned above and $x$ must be eventually in $0<x<1$, by positivity condition. Is it doable? how can I do that?

EDIT:

I obtain the polynomials via equating them to $0$. Namely, $f_0=0$, $f_1=0$ and $f_2=0$ are all given and known.

Thank you very much.

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You have f2 function of 3 variables (x,y0,y1). This is not 3D function but 4D function. You can try ContourPlot3D[f2, {y0, 7/81, 1/9}, {y1, 7/81, 1/9}, {x, 0, 1}]. –  Algohi Jul 12 '14 at 19:26
@Algohi that should be a 3D function. Let me explain. We can write it as $x=f(y_0,y_1)$ and for each $(y_0,y_1)$ we have a value of $x$. I am only interested in the values of $x$ which are in $[0,1]$. OK I think I forgot to mention that I have $f_0=0$, $f_1=0$, $f_2=0$! I must edit. Sorry for that. –  Seyhmus Güngören Jul 12 '14 at 19:31
In this case, ContourPlot3D[f2==0, {y0, 7/81, 1/9}, {y1, 7/81, 1/9}, {x, 0, 1}] will do the job for you. –  Algohi Jul 12 '14 at 19:42
@Algohi didnt work (( I think I need to mention also that $f_0=0$ and $f_1=0$. –  Seyhmus Güngören Jul 12 '14 at 19:47
Please note that in the code you provided, f0 is not a function of lamda0. –  Algohi Jul 12 '14 at 19:51

First of all f0 and f1 are quadratic in $\lambda_0$ and $\lambda_1$. So you would not get a unique value for them. So lets proceed in a general way.

f0[\[Lambda]0_,y0_,y1_,x_,k_]:=-(-y0+k*x*y1+(1-k*x^2)*\[Lambda]0^2)^2+4*x^2*(1-k^2*x)*\[Lambda]0^2*Abs[-(k*y0)+y1+k*\[Lambda]0^2-x*\[Lambda]0^2]

f1[\[Lambda]1_,y0_,y1_,x_,k_]:=-(x*y0-y1+(1-k*x^2)*\[Lambda]1^2)^2+4*x^2*(-k+x)*\[Lambda]1^2*Abs[-(k*y0)+y1-\[Lambda]1^2+k^2*x*\[Lambda]1^2]

f2[\[Lambda]0_,\[Lambda]1_,x_,k_]:=-1/27+\[Lambda]0^3+3*x*\[Lambda]0^2*\[Lambda]1+3*k*x*\[Lambda]0*\[Lambda]1^2+\[Lambda]1^3

\[Lambda]0s=\[Lambda]0/.Solve[f0[\[Lambda]0,y0,y1,x,k]==0,\[Lambda]0];
\[Lambda]1s=\[Lambda]1/.Solve[f0[\[Lambda]1,y0,y1,x,k]==0,\[Lambda]1];
f[k_,x_,y0_,y1_]=f2[\[Lambda]0s[[1]],\[Lambda]1s[[1]],x,k];


see at last step, I specify one of the values of $\lambda_0$ and $\lambda_1$, i.e. the first value. you can choose any combination like (1,1),(1,2),(2,1)or(2,2).

Then comes the second difficulty. You may not get a graphics output with ContourPlot3D because the final function returns a lot of complex values. You can refine your function or see the Abs, Re or Im value.

ContourPlot3D[Abs[f[1, x, y0, y1]], {y0, 7/81, 1/9}, {y1, 7/81, 1/9}, {x, 0, 1},PlotPoints -> 3, AxesLabel -> {"y0", "y1", "x"}]


Since the function is quite heave I use PlotPoints->3. Otherwise it is taking too much time.

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I posted an answer and the only problem is about NSolve. Please let me know if you have any idea to deal with this problem. –  Seyhmus Güngören Jul 13 '14 at 8:07

Based on the answer of Sumit I manipulated the code so that the code should make what I want. But NSolve takes alot of time. I am still waiting and I dont think that the code will end. Is there a faster way? because I am interested in only the 3D plot which can be obtained purely numerically..

f0[\[Lambda]0_,y0_,y1_,x_,k_]:=-(-y0+k*x*y1+(1-k*x^2)*\[Lambda]0^2)^2+4*x^2*(1-k^2*x)*\[Lambda]0^2*Abs[-(k*y0)+y1+k*\[Lambda]0^2-x*\[Lambda]0^2]

f1[\[Lambda]1_,y0_,y1_,x_,k_]:=-(x*y0-y1+(1-k*x^2)*\[Lambda]1^2)^2+4*x^2*(-k+x)*\[Lambda]1^2*Abs[-(k*y0)+y1-\[Lambda]1^2+k^2*x*\[Lambda]1^2]

f2[\[Lambda]0_,\[Lambda]1_,x_,k_]:=-1/27+\[Lambda]0^3+3*x*\[Lambda]0^2*\[Lambda]1+3*k*x*\[Lambda]0*\[Lambda]1^2+\[Lambda]1^3

\[Lambda]0s=\[Lambda]0/.Solve[f0[\[Lambda]0,y0,y1,x,1]==0,\[Lambda]0];
\[Lambda]1s=\[Lambda]1/.Solve[f0[\[Lambda]1,y0,y1,x,1]==0,\[Lambda]1];
f[x_, y0_, y1_] = f2[\[Lambda]0s[[1]], \[Lambda]1s[[1]], x, 1];
xx := x /. NSolve[f[x, y0, y1] == 0, x];

Plot3D[xx, {y0, 7/81, 1/9}, {y1, 7/81, 1/9}, AxesLabel -> {"y0", "y1", "x"}]

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I think the problem is with the complicated initial functions, which makes the final solution quite difficult, both for Solve and NSolve. That's why I suggest to go with Abs, Re or Im. Your method looks quite good to me. As a cross check generate Table[{x,y0,y1,f[x,y0,y1]},{x,..},{y0,..},{y1,..}]//TableForm and you can get an idea about if there exist a solution in the defined range. –  Sumit Jul 13 '14 at 9:23