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How can I determine the order of a symbolic differential equation?

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I'm writing a function that takes a differential equation in symbolic form as an argument and I want to determine the order of the equation in terms of certain variables.
Example 1 A first-order system:
syms t s y(t) u(t) R L
diff_eqn = R*y(t) + L*diff(y(t), t) == u(t); % differential equation
The order (w.r.t. y(t)) is 1.
Example 2 A second-order system:
syms t s y(t) u(t) omega_n z K
diff_eqn = 1/omega_n^2*diff(y(t), t, 2) + 2*z/omega_n*diff(y(t), t) + y(t) == K*u(t);
The order is 2.
I would also like to know the order w.r.t. u(t) if possible as well which in general might not be 0.

Accepted Answer

Ayush Gupta
Ayush Gupta on 10 Sep 2020
There doesn’t exist a direct function to determine the order of a differential equation. However, there is a workaround, and we can use the reduceDifferentialOrder and get newvars from where we can get the last element and see the occurrence of t and this is one plus than the order of equation. Refer to the following code:
syms x(t) y(t) f(t)
eqs = [diff(x(t),t,t) == diff(f(t),t,t,t), diff(y(t),t,t,t) == diff(f(t),t,t)];
vars = [x(t), y(t)];
[newEqs, newVars, R] = reduceDifferentialOrder(eqs, vars)
l = length(newVars);
s = string(newVars(l,1));
order_of_equation = count(s, 't') -1;
  1 Comment
Paul
Paul on 2 Apr 2023
It's a bit easier if dealing wth ODEs as in the Question, at least to find the order of the ODE
syms t s y(t) u(t) R L
diff_eqn = R*y(t) + L*diff(y(t), t) == u(t) % differential equation
diff_eqn = 
%The order (w.r.t. y(t)) is 1.
numel(odeToVectorField(diff_eqn))
ans = 1
% Example 2 A second-order system:
syms t s y(t) u(t) omega_n z K
diff_eqn = 1/omega_n^2*diff(y(t), t, 2) + 2*z/omega_n*diff(y(t), t) + y(t) == K*u(t)
diff_eqn = 
numel(odeToVectorField(diff_eqn))
ans = 2
Finding the order wrt u(t) would take more work.

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More Answers (2)

Ganesh
Ganesh on 1 Apr 2023
Edited: Walter Roberson on 1 Apr 2023
function order = order_polynomial(poly1,x)
count=0;
temp=1;
while(temp~=0)
poly1=diff(poly1,x);
if poly1==0
temp=0;
else
count=count+1;
end
end
disp(count);
end

Walter Roberson
Walter Roberson on 1 Apr 2023
Moved: Walter Roberson on 1 Apr 2023
p = poly2sym(randi([-9,9],1,randi(15)))
p = 
poly_degree = length(coeffs(p,'all'))-1
poly_degree = 6
  3 Comments
Paul
Paul on 1 Apr 2023
If the coefficient vector is double there's no need to go the Symbolic route.
rng('default')
c = [0 0 0 randi([-9,9],1,randi(15))] % add some leading zeros
c = 1×16
0 0 0 8 -7 8 3 -8 -4 1 9 9 -7 9 9 0
tic
for ii = 1:1e3
p = poly2sym(c);
poly_degree_s = length(coeffs(p,'all'))-1;
end
toc
Elapsed time is 2.962911 seconds.
tic
for ii = 1:1e3
poly_degree_d = numel(c) - find(c,1,'first');
end
toc
Elapsed time is 0.004062 seconds.
[poly_degree_s poly_degree_d]
ans = 1×2
12 12
Walter Roberson
Walter Roberson on 2 Apr 2023
@Paul is of course correct that if you have a vector of coefficients then the difference between the length of the vector and the position of the first non-zero tells you about the degree.
However... the original question deals with symbolic polynomials. My creation of p with intermediate numeric form was just to have some polynomial to work with, and to demonstrate that I my code worked with polynomials of different degrees, not just something that "happened" to work with a particular length.

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