I have to find parameter for $C\text{, }R_0\text{, }\alpha\text{, }T_0\text{, }T_A\text{, }T_L\text{ and }\epsilon \sigma A$ based on pairs of $I,t$ with $T\left(t\right)=T_L$. Also I know the pair $I, T$, for which $ \frac{\partial T}{\partial t} = 0 $. All unknown are real and strictly positive. $$ \frac{\partial T}{\partial t} = \frac{\left( R_0 \left( 1 + \alpha \left( T - T_0 \right) \right) \right) I^2 - \epsilon \sigma A \left( T^4 - T_A^4 \right)}{C} \\ T\left(0\right) = T_A $$ Further I can assume that: $$ T_A \le T_0 \\ T_L > T_A \\ R_0 = 200e-6 \\ T_0 \text{ is probably } 23+273.15 \text{ or in the range of } T_A\\ T_A \text{ is probably } 85+273.15 \text{ or } 105+273.15 \\ T_L \ll 1734+273.15 $$

For simplification, I have already substituted complex sub-expressions and factored out $T$: $$ \frac{\partial T}{\partial t} = D T^4 + E T + F \\ \text{ with: } \\ D = \frac{- \epsilon \sigma A}{C} \\ E = \frac{R_0 \alpha I^2}{C} \\ F = \frac{R_0 I^2 - R_0 \alpha T_0 I^2 + \epsilon \sigma A T_A^4}{C} \\ $$ This can easily be decomposed into: $$ \frac{1}{D T^4 + E T + F} \partial T = \partial t $$

Thanks to @mattos I know that this is a Chini equation, for which the Chini invariant is independent of $t$: $$ \frac{\partial y\left(t\right)}{\partial t} = f\left(t\right) y^n\left(t\right) + g\left(t\right) y\left(t\right) + h\left(t\right) \\ f\left(t\right)=D \\ n=4 \\ y\left(t\right)=T \\ g\left(t\right)=E \\ h\left(t\right)=F \\ C = f^{−n−1}\left(t\right) h^{−2n+1}\left(t\right) \left(f\left(t\right)\frac{\partial h\left(t\right)}{\partial t}−h\left(t\right)\frac{\partial f\left(t\right)}{\partial t}+n f\left(t\right) g\left(t\right) h\left(t\right)\right)^n n^{−n} \\ C = D^{−4−1} F^{−2*4+1} \left(D\frac{\partial F}{\partial t}−F\frac{\partial D}{\partial t}+4 D E F\right)^4 4^{−4} \\ C = D^{−5} F^{−7} \left(0 D−0 F+4 D E F\right)^4 4^{−4} \\ C = D^{−5} F^{−7} 4^4 D^4 E^4 F^4 4^{−4} \\ C = D^{−1} F^{−3} E^4 \\ C = \frac{E^4}{D F^3} $$ Obviously $C$ is independent of $t$ and $n=4$.

And from there I have no idea how to continue. Referenced, which do not really fit I found are:

Side information:

  • Finally I need the function $ t\left(I, T_A\right)\text{ for which } T\left(t\right)=T_L $.
  • The solution should be analytical to allow error calculations.
  • The final formula has to be calculated in a low-power embedded system.
  • In a first step, we could assume that $ \alpha = 0 $

P.S.: My math lectures are roughly 20 years ago, please feel free to improve my notation.

  • $\begingroup$ First, you don't need $*$ for multiplication, just write things next to each other to imply a product i.e $\epsilon \sigma A$ instead of $\epsilon * \sigma * A$. Also, use \ge for $\ge$ and \ll for $\ll$. Secondly, this is a Chini differential equation and probably isn't exactly solveable. You may need to resort to numerical solutions. $\endgroup$ – mattos May 19 '19 at 13:18
  • $\begingroup$ I take it back about not being explicitly solvable, apparently it is in your case. This link has more details. In it, you'll see an equation for a Chini invariant, which, if constant, apparently yields a method for a solution to your problem. The Chini invariant in your case is constant. However, the book which describes the method of solution is in German (I don't speak it) and unavailable to me, so perhaps someone else on this site can help you. $\endgroup$ – mattos May 19 '19 at 13:29

You have to solve : $$\frac{d T}{d t} = D\:T^4 + E\:T + F$$ This is an ODE of separable kind. $$t=\int \frac{dT}{D\:T^4+E\:T+F}+C_0$$ $C_0$ is a constant.

Firstly, solve the algebraic equation $\quad D\:X^4+E\:X+F=0$ for X.

This is a quartic equation. The analytic solution is complicated : http://mathworld.wolfram.com/QuarticEquation.html

But if you really want an explicit solution to your problem you cannot avoid to solve it.

Suppose that you does solve it you get the four roots, say $X_1,X_2,X_3,X_4$

More simply, solve numerically the quartic equation and you get the numerical values of $X_1,X_2,X_3,X_4$.

Secondly, write the fraction on the form $$\frac{1}{D\:T^4+E\:T+F}=\frac{C_1}{T-X_1}+\frac{C_2}{T-X_2}+\frac{C_3}{T-X_3}+\frac{C_4}{T-X_4}$$ You have to compute $C_1,C_2,C_3,C_4$ in terms of $D,E,F,X_1,X_2,X_3,X_4$. Very boring job if you want an explicit solution. Much easier with numerical calculus if a numerical solution is sufficient.

At this point $C_1,C_2,C_3,C_4$ are known. $$t=\int \left(\frac{C_1}{T-X_1}+\frac{C_2}{T-X_2}+\frac{C_3}{T-X_3}+\frac{C_4}{T-X_4}\right)dT+C_0$$ $$t=C_0+\ln\bigg( (T-X_1)^{C_1} (T-X_2)^{C_2} (T-X_3)^{C_3}(T-X_4)^{C_4}\bigg) $$ At this point you got $t(T)$, that is $t$ as a function of $T$ with an unknown parameter $C_0$ in it.

Thirdly, one have to determine $C_0$ according to the initial condition.

Fourthly, the major difficulty arises : Inverting the function $t(T)$ in order to obtain the function $T(t)$.

Due to the exponents $C_1,C_2,C_3,C_4$ this cannot be analytically done, except for a few cases of particular values of $C_1,C_2,C_3,C_4$, for example $0$ and/or $1$. This is generally never the cases.

Finally we see that $T(t)$ can only be obtained on numerical form. As a consequence, instead of the arduous above calculus, on a practical viewpoint it is advised to directly solve the ODE $\frac{d T}{d t} = D\:T^4 + E\:T + F$ thanks to numerical method. Such methods are implemented in some math-softwares.

As a conclusion, don't expect an analytical solution for $T(t)$.

An analytical solution is theoretically possible for the inverse function $t(T)$ but the formula would be awfully complicated if we want to write it on a full explicit form (without intermediate variables $X_1,X_2,X_3,X_4,C_1,C_2,C_3,C_4$ which each one is already a complicated formula).

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