Joel French

2022-07-01

I have a simple, two object thermodynamic model with radiation and advection. This model consists of two first order quadratic differential equations, what I would like to solve analytically. The equations can be simplified to

$\frac{d}{dt}x={a}_{0}{x}^{4}+{a}_{1}{y}^{4}+{a}_{2}x+{a}_{3}y+{a}_{4}$

$\frac{d}{dt}y={a}_{5}{x}^{4}+{a}_{6}{y}^{4}+{a}_{7}x+{a}_{8}y+{a}_{9}$

I'm trying to find the analytic solution because I would like to use the model in an embedded system with limited resources, where solving it numerically is not feasible. So far I have tried to solve this with Maxima without any success. I have also consulted with my college textbooks, but they cover only linear differential equation systems.

$\frac{d}{dt}x={a}_{0}{x}^{4}+{a}_{1}{y}^{4}+{a}_{2}x+{a}_{3}y+{a}_{4}$

$\frac{d}{dt}y={a}_{5}{x}^{4}+{a}_{6}{y}^{4}+{a}_{7}x+{a}_{8}y+{a}_{9}$

I'm trying to find the analytic solution because I would like to use the model in an embedded system with limited resources, where solving it numerically is not feasible. So far I have tried to solve this with Maxima without any success. I have also consulted with my college textbooks, but they cover only linear differential equation systems.

Jordin Church

Beginner2022-07-02Added 11 answers

I am afraid you are out of luck. Even finding the equilibrium is a very complicated endeavor. With general coefficients what you have for the equilibrium is a fourth order polynomial.

My suggestion is you try numerically, there is no general simple analytic solution. You don't actually need all that much resources to solve the problem numerically, since you can always try Euler:

Take $h>0$ a time step. The approximations for ${x}_{i}=x(ih),{y}_{i}=y(ih)$ are

${x}_{i}=x(ih),{y}_{i}=y(ih)$

The computational cost should be minimal and the approximation for small h is fairly accurate. If you find this method insufficiently accurate you can always try better numerical methods.

My suggestion is you try numerically, there is no general simple analytic solution. You don't actually need all that much resources to solve the problem numerically, since you can always try Euler:

Take $h>0$ a time step. The approximations for ${x}_{i}=x(ih),{y}_{i}=y(ih)$ are

${x}_{i}=x(ih),{y}_{i}=y(ih)$

The computational cost should be minimal and the approximation for small h is fairly accurate. If you find this method insufficiently accurate you can always try better numerical methods.

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B) $\frac{5}{9}$ degree on fahrenheit scale

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I Completed the square on the bottom but what do you do now?

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

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$K=0$

$A=F(s)\ast (s+a)$