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Consider the following plot:

Plot[(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4 /. T -> 96.43650760992945`, {ϕ, -200, 20}]

I would like to find the value of $\Phi<0$ and $T>0$ to obtain the first minimum of the above function.

These two values can be obtained with 

FindRoot[
  {(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4 == 0 , 
   D[(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4, {ϕ, 1}] == 0}, 
  {{ϕ, -200}, {T, 1}}]

which yields

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.
{ϕ -> -160., T -> 96.4365}

However, I would like to obtain it with NSolve but it does not work:

NSolve[
  {(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4 == 0, 
   D[(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4, {ϕ, 1}] == 0}, 
  {ϕ, T}]

which gives

NSolve::infsolns: Infinite solution set has dimension at least 1. Returning intersection of solutions with -((92291 T)/87992)-(121001 [Phi])/175984 == 1.
{{ϕ -> 0., T -> -0.953419}}

as a solution, which is obviously wrong.

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2 Answers 2

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eqns = {(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4 == 0,
    D[(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4, ϕ] == 0,
    ϕ < 0, T > 0} // Simplify;

Use Solve rather than NSolve

sol = Solve[eqns, {ϕ, T}, Reals][[1]]

(* {ϕ -> -160, T -> 10 Sqrt[93]} *)

Verifying that sol satisfies eqns

And @@ (eqns /. sol)

(* True *)

The approximate values are

sol // N

(* {ϕ -> -160., T -> 96.4365} *)

EDIT: To use NSolve, use the option Method->{"UseSlicingHyperplanes" -> False}

soln = NSolve[eqns, {ϕ, T}, Reals, 
   Method -> {"UseSlicingHyperplanes" -> False}][[1]]

(* {ϕ -> -160., T -> 96.4365} *)
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2
  • $\begingroup$ Thanks but why is NSolve not working? $\endgroup$ Commented Sep 13, 2019 at 8:01
  • $\begingroup$ @ketherok There is a dimensional component to the solution set (when phi is 0). It can be removed to get the finite solution set as follows: In[119]:= poly = (22500 + t^2/3)*phi^2 + 320 phi^3 + phi^4; NSolve[{poly, D[poly, phi], phi*recip - 1} == 0] Out[120]= {{phi -> -160., recip -> -0.00625, t -> 96.4365076099}, {phi -> -160., recip -> -0.00625, t -> -96.4365076099}} $\endgroup$ Commented Sep 13, 2019 at 20:04
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When we use ContourPlot we see that min is at roughly T=0 and $\Phi=-175$

    ContourPlot[(22500 + T^2/3) ϕ^2 + 
  320 ϕ^3 + ϕ^4, {T, -100, 100}, {ϕ, -250, 0}, 
 PlotLegends -> Automatic]

enter image description here

We then use NMinimize

NMinimize[{(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4,   0 < T < 100, -250 < ϕ < 0}, {ϕ, T}]

{-8.80964*10^7, {$\Phi$ -> -176.125, T -> 2.97721*10^-11}}

Plot3D[(22500 + T^2/3) ϕ^2 + 320 ϕ^3 + ϕ^4, {T, -100, 
  100}, {ϕ, -200, 20}]

enter image description here

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