Solving an implicit equation and plot solution

Question

I am trying to solve numerically a difficult implicit equation and plot the solution. The thing that I want to solve numerically is the following:

$$ Q(h)=\frac{0.13}{1-10^{-4}\log{\frac{Y(h)}{100}}} $$

$$ P(h)=2*10^{4}*\sqrt{\frac{1}{1-10^{-4}\log{\frac{Y(h)}{100}}}}-1/6$$

$$ Y(h)=\sqrt{3*Q(h)*h^2-(P(h)+1/6)*10^{12}} $$

So the point is that I want to solve (numerically is the only way) the implicit equation of Y(h) in terms of h, but note that functions Q(h) and P(h) also depend on Y (so implicitly on h).

I want to get numerically the solution of Y(h) in order to use this to plot Q(h) and P(h) in terms of h.

Someone knows how to do this??? Thanks!

Expressions in format code:

Q[h_] := 0.13/(1 - 10^(-4)*Log[Y[h]/100]);

P[h_] := 2*(10^4)*Sqrt[1/(1 - 10^(-4)*Log[Y[h]/100])] - 1/6;

Y[h_] := Sqrt[3*Q[h]*h^2 - (P[h] + 1/6)*10^12];

Answer 1

You can solve your implicit equation as follows:

First I renamed Y[h]->y

Q[h_] := 0.13/(1 - 10^(-4)*Log[y/100])
P[h_] := 2*(10^4)*Sqrt[1/(1 - 10^(-4)*Log[y/100])] - 1/6

which gives the implicit equation

zero = y - Sqrt[3*Q[h]*h^2 - (P[h] + 1/6)*10^12] /. y -> Exp[logy] //PowerExpand

Thereby I substituted y -> Exp[logy] because of poor scaling. The solution of zero==0 can be evaluated explicitely for h[logy] (not so easy for logy[h])

sol = Solve[zero == 0, h ];
ParametricPlot[ {h, Exp[logy]} /. sol, {logy, 0, 25},AxesLabel -> {h, y}, PlotLabel -> zero == 0,PlotRange -> {{-10^9, 10^9}, {0, 10^9 }}]

Q and P can be plotted

GraphicsRow[{
ParametricPlot[({h, Q[h]} /. y -> Exp[logy] //PowerExpand) /. sol[[2]], {logy, 0, 25}, AspectRatio -> 1,AxesLabel -> {h, "Q[h]"}], 
ParametricPlot[({h, P[h]} /. y -> Exp[logy] // PowerExpand) /.sol[[2]], {logy, 0, 25}, AspectRatio -> 1,AxesLabel -> {h, "P[h]"}]}]