Solve symbolic equation system

Question

Lu Zhao 2021-2-26

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评论： Lu Zhao 2021-3-2

There is a symbolic equation system like this and wonder if any one could help me out by solving this equation system symbolically:

4 equations with 4 unknown variables [u ̂ , v ̂, w ̂, p ̂ ]. (u^, v^, w^, p^ are function of "r", namely, u(r), v(r), w(r), p(r))
All other symbolic parameters are known variables, i.e., A, r, m, alpha, U(r), V(r), W(r) ).

Here are the equations:

Objective :

Combine the 4 equations into 1 single equation in terms of u^ only, namely, cancel out v ̂, w ̂, p ̂ .
Find expressions of v ̂, w ̂, p ̂ in terms of u^ : i.e., find expression of v^ in termed of u^, v^=f (u^). Same for expressions of w^ and p^, in terms of u^.

Here is a short program I made and I know it is not even close to the answer:

clc
clear
syms U(r) V(r) W(r) u(r) v(r) w(r) p(r) A r m alpha
                                      % f(u, v, w, p)
eqn1=A*u+diff(U,r)*v+alpha*p==0;      % f(u, v,    p)=0
eqn2=-A*v-2*W/r*w+diff(p,r)==0;       % f(   v, w, p)=0
eqn3=(diff(W,r)+W/r)*v+A*w+m/r*p==0;  % f(   v, w, p)=0
eqn4=alpha*u+v/r+diff(v,r)+m/r*w==0;  % f(u, v, w   )=0
solve([eqn1,eqn2,eqn3,eqn4],[v w p])

The above program doesn solve for "v" w" "p" in terms of "u". I wondered is there anyway to do it? I really appreciate your time and help.

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Answer 1

Walter Roberson 2021-2-26

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That is a series of ODE. You will not be able to solve it with solve(). You could try it with

dsolve([eqn1, eqn2, eqn3, eqn4])

but the reality is that you will get back emptiness.

There is a trivial solution with u, v, r, and p all being constant 0s. Other than that there does not appear to be any closed form solutions.

Any time you have an unknown function involved U(r), V(r), W(r) then you are not going to get a closed form solution except possibly trivial solutions (like the constant zeros.)

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Walter Roberson 2021-3-2

在 MATLAB Online 中打开

Why would du/dr appear in the final equation? You do not have du/dr anywhere in the original equations.

syms U(r) V(r) W(r) u(r) v(r) w(r) p(r) A r m alpha

eqn1=A*u+diff(U,r)*v+alpha*p==0;

eqn2=-A*v-2*W/r*w+diff(p,r)==0;

eqn3=(diff(W,r)+W/r)*v+A*w+m/r*p==0;

eqn4=alpha*u+v/r+diff(v,r)+m/r*w==0;

syms pr vr wr

temp = subs([eqn1(r); eqn2(r); eqn3(r); eqn4(r)], [p(r), v(r), w(r)],[pr,vr,wr])

temp =

PR = solve(temp(1),pr)

PR =

temp2 = subs(temp(2:end), pr, PR)

temp2 =

VR = solve(temp2(1),vr)

VR =

temp3 = subs(temp2(2:end), vr, VR)

temp3 =

WR = solve(temp3(1), wr)

WR =

single_ode = subs(temp3(2:end), wr, WR)

single_ode =

None of p, v, w are differentiated, so I do not see any reason why following the steps would introduce du/dr ?

Lu Zhao 2021-3-2

I really appreciate your patience and explaning to me step by step. I think it's because the missing dpdr in eqn(b). Here is how I derived the final ODE by hand:

(Background information: U(r), V(r), W(r) are actually three component of velocity, to make the equations easier to solve at begining, we assume U(r)=constant, namely, d(U(r))/dr=0. )

This reduced eqn(a) to:

Actually, from eqn(b) and (c), we could get expression of v, w in terms of p:

Finally, substitute v, w, p in terms of u into the eqn(d), the final single ODE was obtained:

(My guess to the missing dudr happened when substitute v(r) w(r) p(r) with vr wr pr. It's a great way to simplify the equation, but somehow it makes the third term "dpdr=0" for eqn(b), and thus missing dudr and du^2dr^2.