I want to check if something algebraically simplifies to 0 when i sub in values

clc
clear
syms q1 q2 q3 g Y u a rho e
E=[q2,((q2^2)/q1)*(1-(g-1)*.5)+(g-1)*q3,((q3*q2)/q1)+(g-1)*((q3*q2)/q1)-.5*(g-1)*(q2^3/q1^2)]
Q=[q1,q2,q3]
J=jacobian(E,Q)
pretty((2*q2^3*(g/2 - 1/2))/q1^3 - (q2*q3)/q1^2 - (q2*q3*(g - 1))/q1^2)
pretty(q3/q1 - (3*q2^2*(g/2 - 1/2))/q1^2 + (q3*(g - 1))/q1)
pretty(q2/q1 + (q2*(g - 1))/q1)


C=det(J-Y*eye(3))
C2=simplify(C)

so i have Y= u, u-a and u+a

q1= rho, q2=rho*u, and q3=e

all of these are symbolic and have no values but i want to sub in all the q's and 1 by 1 the Y's to see if they simplify to 0.

how would i do this?

Answer 1

[John D'Errico]

syms q1 q2 q3 g Y u a rho e
E=[q2,((q2^2)/q1)*(1-(g-1)*.5)+(g-1)*q3,((q3*q2)/q1)+(g-1)*((q3*q2)/q1)-.5*(g-1)*(q2^3/q1^2)];
Q=[q1,q2,q3];
J=jacobian(E,Q);
pretty((2*q2^3*(g/2 - 1/2))/q1^3 - (q2*q3)/q1^2 - (q2*q3*(g - 1))/q1^2)

  3 / g   1 \
q2  | - - - | 2
    \ 2   2 /     q2 q3   q2 q3 (g - 1)
--------------- - ----- - -------------
        3            2           2
      q1           q1          q1

pretty(q3/q1 - (3*q2^2*(g/2 - 1/2))/q1^2 + (q3*(g - 1))/q1)

       2 / g   1 \
     q2  | - - - | 3
q3       \ 2   2 /     q3 (g - 1)
-- - --------------- + ----------
q1           2             q1
           q1

pretty(q2/q1 + (q2*(g - 1))/q1)

q2   q2 (g - 1)
-- + ----------
q1       q1

C=det(J-Y*eye(3))

C = 

C2=simplify(C)

C2 = 

Now, if you have q1,q2,q3

C2 = subs(C2,[q1,q2,q3],[rho,rho*u,e])

C2 = 

expand(C2)

ans = 

So nothing special happening so far. Now you want t osub in those possible values for Y. Just do it.

simplify(expand(subs(C2,Y,[u,u+a,u-a])))

ans = 

Simple enough. When Y == u, everything goes away, but not so in the other cases.

Another way of doing this is to see for which values of Y, C2 would be zero.

Ysol = solve(C2 == 0,Y)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

Ysol = 

It finds three solutions, one of which is the case you wanted. The other two are alternatives. I could check under which conditions they apply, but most likely it would just tell me that rho cannot be zero.

Answer 2

[Torsten]

Euler equations of gas dynamics ?

Eigenvalues and Eigenvectors of the Jacobian are well-studied. Why reinvent the wheel ?

syms q1 q2 q3 g Y u a rho e
E=[q2,((q2^2)/q1)*(1-(g-1)*.5)+(g-1)*q3,((q3*q2)/q1)+(g-1)*((q3*q2)/q1)-.5*(g-1)*(q2^3/q1^2)];
Q=[q1,q2,q3];
J=jacobian(E,Q);
[S,V] = eig(J)

S = 

V = 

S = simplify(subs(S,[q1 q2 q3],[rho rho*u e]))

S = 

V = simplify(subs(V,[q1 q2 q3],[rho rho*u e]))

V = 

[joshua payne]

yes youre correct, but i was simply curious if i could do all the algebra in matlab and show it simplifies to 0