Solving First order ODEs simultaneously

Hello, needed help figuring out why I cannot obtain a solution. I'm sure this is a solvable solution however I keep getting a warning saying no solution is found. Is there any mistake I'm making in the code?
Everything is a constant except E, Sr(t) & Er(t).
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
Unrecognized function or variable 'Ea'.
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
odes = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
[SrSol(t),ErSol(t)] = dsolve(odes,conds)

4 件のコメント

Walter Roberson
Walter Roberson 2023 年 9 月 28 日
Ea = 123; %just to have SOME value
k1 = 42; %just to have SOME value
k2 = 13; %just to have SOME value
krev1 = 48; %just to have SOME value
Sa = 5; %just to have SOME value
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
odes = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
sol = dsolve(odes,conds)
Warning: Unable to find symbolic solution.
sol = [ empty sym ]
sol2a = dsolve(ode2a)
sol2a = 
so3a = dsolve(ode3a)
so3a = 
I don't know that I would call those "solvable"
Valerie
Valerie 2023 年 9 月 28 日
Is there a way to solve this mathematically?
Walter Roberson
Walter Roberson 2023 年 9 月 28 日
Ea = 123; %just to have SOME value
k1 = 42; %just to have SOME value
k2 = 13; %just to have SOME value
krev1 = 48; %just to have SOME value
Sa = 5; %just to have SOME value
% Rigorous Solution Case #1
syms Sr(t) Er(t) E;
E = Ea - Er(t);
ode2a = diff(Sr(t),t) == -(k1*(Ea - Er(t))*Sr(t)) + krev1*Er(t);
ode3a = diff(Er,t) == (k1*(Ea - Er(t))*Sr(t)) - (krev1+k2)*Er(t);
eqns = [ode2a; ode3a];
cond1 = Sr(0) == Sa;
cond2 = Er(0) == 0;
conds = [cond1; cond2];
[eqs,vars] = reduceDifferentialOrder(eqns, [Sr(t), Er(t)])
eqs = 
vars = 
[M,F] = massMatrixForm(eqs,vars)
M = 
F = 
f = M\F
f = 
odefun = odeFunction(f,vars)
odefun = function_handle with value:
@(t,in2)[in2(2,:).*4.8e+1-in2(1,:).*5.166e+3+in2(2,:).*in2(1,:).*4.2e+1;in2(2,:).*-6.1e+1+in2(1,:).*5.166e+3-in2(2,:).*in2(1,:).*4.2e+1]
InitConditions = double(rhs(conds)) %watch out for order though!
InitConditions = 2×1
5 0
[T, Y] = ode45(odefun, [0 0.01], InitConditions);
subplot(2,1,1); plot(T, Y(:,1)); title(string(vars(1)))
subplot(2,1,2); plot(T, Y(:,2)); title(string(vars(2)))
%that almost looks like the initial conditions are reversed.
%what happens if we try reversing the conditions?
[Tr, Yr] = ode45(odefun, [0 0.01], flipud(InitConditions));
figure
subplot(2,1,1); plot(Tr, Yr(:,1)); title(string(vars(1)))
subplot(2,1,2); plot(Tr, Yr(:,2)); title(string(vars(2)))
Valerie
Valerie 2023 年 9 月 28 日
Thank you so much!

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 採用された回答

Torsten
Torsten 2023 年 9 月 28 日
移動済み: Torsten 2023 年 9 月 28 日

1 投票

I'm quite sure there is no analytical solution for your system of ODEs since the right-hand sides are nonlinear in the unknown functions (term Er(t)*Sr(t)).

7 件のコメント

Valerie
Valerie 2023 年 9 月 28 日
Alright, thank you.
Torsten
Torsten 2023 年 9 月 28 日
A numerical solution using one of the ODE integrators (e.g. ode15s) should be possible.
But you must give values to all the constants involved.
Ea = 123; %just to have SOME value
k1 = 42; %just to have SOME value
k2 = 13; %just to have SOME value
krev1 = 48; %just to have SOME value
Sa = 5; %just to have SOME value
fun = @(t,y)[-(k1*(Ea - y(2))*y(1)) + krev1*y(2);(k1*(Ea - y(2))*y(1)) - (krev1+k2)*y(2)];
y0 = [Sa;0];
tspan = [0 1];
[T,Y] = ode15s(fun,tspan,y0);
plot(T,Y)
Valerie
Valerie 2023 年 9 月 28 日
Thanks so much! Is there a difference between the different ODE integrators?
Torsten
Torsten 2023 年 9 月 28 日
The ode integrators are mainly classified as being suited to solve "nonstiff" (ode45) and "stiff" (ode15s) problems.
In your case with only two equations, I'd take a stiff integrator. It needs a little more memory, but is safe and efficient for both stiff and nonstiff problems.
Valerie
Valerie 2023 年 9 月 28 日
Thank you so much for your patience! Very new to mathematica
Sam Chak
Sam Chak 2023 年 9 月 28 日
Wolfram Mathematica uses DSolve. 😅
Valerie
Valerie 2023 年 9 月 29 日
@Sam Chak I rage quit mathematica this week and is how I eneded up on MATLAB lol but thank you!

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