how to write to solve this type of system of equations ?

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Risposte (2)

Wan Ji
Wan Ji il 12 Set 2021
Hey friend
Just expand the left items of the two equations, extract u'' and v'', then an ode45 solver is there for you.

Walter Roberson
Walter Roberson il 12 Set 2021
Modificato: Walter Roberson il 12 Set 2021
I was not able to figure out what is being raised to 10/9 . I used squiggle instead.
syms u(r) v(r)
syms N squiggle real
assume(r, 'real')
du = diff(u);
dv = diff(v);
left1 = diff(r^(N-1)*du^3)
left1(r) = 
right1 = r^(N-1) * sqrt(u) * sqrt(v) / (3*r^(2/3) * sqrt(1 + 9*squiggle^(10/9)/(10*(3*N-2)^(1/3))))
right1(r) = 
left2 = diff(r^(N-1)*dv^3)
left2(r) = 
right2 = r^(N-1) * u * v / (3*r^(2/3))
right2(r) = 
eqn1 = left1 == right1
eqn1(r) = 
eqn2 = left2 == right2
eqn2(r) = 
ic = [u(0) == 1, v(0) == 1, du(0) == 0, dv(0) == 0]
ic = 
sol = dsolve([eqn1, eqn2, ic])
Warning: Unable to find symbolic solution.
sol = [ empty sym ]
string(eqn1)
ans = "3*r^(N - 1)*diff(u(r), r)^2*diff(u(r), r, r) + r^(N - 2)*(N - 1)*diff(u(r), r)^3 == (r^(N - 1)*u(r)^(1/2)*v(r)^(1/2))/(3*r^(2/3)*((9*squiggle^(10/9))/(10*(3*N - 2)^(1/3)) + 1)^(1/2))"
string(eqn2)
ans = "3*r^(N - 1)*diff(v(r), r)^2*diff(v(r), r, r) + r^(N - 2)*(N - 1)*diff(v(r), r)^3 == (r^(N - 1)*u(r)*v(r))/(3*r^(2/3))"
string(ic)
ans = 1×4 string array
"u(0) == 1" "v(0) == 1" "subs(diff(u(r), r), r, 0) == 0" "subs(diff(v(r), r), r, 0) == 0"
Lack of a symbolic solution means that you would have to do numeric solutions -- but you cannot do a numeric solution to infinity, and you certainly would not get a formula out of it.
  5 Commenti
NIRUPAM SAHOO
NIRUPAM SAHOO il 12 Set 2021
syms p(t) m(t) t Y
Eqns = [diff((t^(100-1))*(diff(p(t),t))) == (t^(100-1))*(t-1)*exp(t)*p(t)*m(t);
diff((t^(100-1))*(diff(m(t),t))) == (t^(100-1))*(t-1)*exp(t)*p(t)^(1/2)*m(t)^(1/2)]
[DEsys,Subs] = odeToVectorField(Eqns);
DEFcn = matlabFunction(DEsys, 'Vars',{t,Y});
tspan = [0,100];
y0 = [0 0 0 0];
[t,Y] = ode45(DEFcn, tspan, y0);
plot(t,Y)
## i write this but the graph does not shows . please help me

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