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 | 1 | initial version |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
HTH,
| | 2 | No.2 Revision |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
HTH,
| | 3 | No.3 Revision |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
EDIT : Nontwithstanding my laziness, here's the abstract of my solution :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
D2={(K:=var("k", n=3))[u[1]]:u[0].simplify_full().factor()
for u in DE1.coefficients(f0)}
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
from _operator import lt, eq, gt
Sol=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2).full_simplify().factor())]
Sol2=H(t=t)==cases(Sol2)
which is way yooo wide to LaTeX. Asking Mathematica a bit of help :
Sol2=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol2+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2)._mathematica_().FullSimplify().sage(locals={("Tan", 1):tan, ("Coth", 1):coth}))]
Sol2=H==cases(Sol2)
$$ \left[\left(-k_{1}^{2} + 4 \, k_{0} k_{2} > 0, -\frac{1}{6} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} \tan\left(\frac{1}{2} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} {\left(C + t\right)}\right) - \frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}}\right), \left(-k_{1}^{2} + 4 \, k_{0} k_{2} = 0, -\frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}} + \frac{1}{3 \, {\left(C + t\right)}}\right), \left(-k_{1}^{2} + 4 \, k_{0} k_{2} < 0, -\frac{C_{2} n - \frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} \coth\left(\frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} {\left(C + t\right)}}{2 \, {\left(C_{2} t + C_{3}\right)} \sqrt{2 \, w + 3}}\right)}{\sqrt{2 \, w + 3}}}{6 \, {\left(C_{2} t + C_{3}\right)}}\right)\right] $$
HTH,
| | 4 | No.4 Revision |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
EDIT : Nontwithstanding my laziness, here's the abstract of my solution :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
D2={(K:=var("k", n=3))[u[1]]:u[0].simplify_full().factor()
for u in DE1.coefficients(f0)}
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
from _operator import lt, eq, gt
Sol=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2).full_simplify().factor())]
Sol2=H(t=t)==cases(Sol2)
which is way yooo wide to LaTeX. Asking Mathematica a bit of help :
Sol2=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol2+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2)._mathematica_().FullSimplify().sage(locals={("Tan", 1):tan, ("Coth", 1):coth}))]
Sol2=H==cases(Sol2)
$$ \left[\left(-k_{1}^{2} H\left(t\right) = \begin{cases}{-\frac{1}{6} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} \tan\left(\frac{1}{2} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} {\left(C + t\right)}\right) - \frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}}} & {-k_{1}^{2} + 4 \, k_{0} k_{2} > 0, -\frac{1}{6} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} \tan\left(\frac{1}{2} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} 0}\{-\frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}} + \frac{1}{3 \, {\left(C + t\right)}\right) - \frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}}\right), \left(-k_{1}^{2} t\right)}}} & {-k_{1}^{2} + 4 \, k_{0} k_{2} = 0, -\frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}} + \frac{1}{3 \, {\left(C + t\right)}}\right), \left(-k_{1}^{2} + 4 \, k_{0} k_{2} < 0, -\frac{C_{2} 0}\{-\frac{C_{2} n - \frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} \coth\left(\frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} {\left(C + t\right)}}{2 \, {\left(C_{2} t + C_{3}\right)} \sqrt{2 \, w + 3}}\right)}{\sqrt{2 \, w + 3}}}{6 \, {\left(C_{2} t + C_{3}\right)}}\right)\right] C_{3}\right)}}} & {-k_{1}^{2} + 4 \, k_{0} k_{2} < 0}\end{cases} $$
HTH,
| | 5 | No.5 Revision |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
EDIT : Nontwithstanding my laziness, here's the abstract of my solution :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
D2={(K:=var("k", n=3))[u[1]]:u[0].simplify_full().factor()
for u in DE1.coefficients(f0)}
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
from _operator import lt, eq, gt
Sol=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2).full_simplify().factor())]
Sol2=H(t=t)==cases(Sol2)
which is way yooo wide to LaTeX. Asking Mathematica a bit of help :
Sol2=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol2+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2)._mathematica_().FullSimplify().sage(locals={("Tan", 1):tan, ("Coth", 1):coth}))]
Sol2=H==cases(Sol2)
$$ H\left(t\right) \begin{cases}{a} & {a > 0}\{0} & {a = \begin{cases}{-\frac{1}{6} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} \tan\left(\frac{1}{2} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} {\left(C + t\right)}\right) - \frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}}} 0}\{-a} & {-k_{1}^{2} + 4 \, k_{0} k_{2} > 0}\{-\frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}} + \frac{1}{3 \, {\left(C + t\right)}}} & {-k_{1}^{2} + 4 \, k_{0} k_{2} = 0}\{-\frac{C_{2} n - \frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} \coth\left(\frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} {\left(C + t\right)}}{2 \, {\left(C_{2} t + C_{3}\right)} \sqrt{2 \, w + 3}}\right)}{\sqrt{2 \, w + 3}}}{6 \, {\left(C_{2} t + C_{3}\right)}}} & {-k_{1}^{2} + 4 \, k_{0} k_{2} {a < 0}\end{cases} $$
HTH,
| | 6 | No.6 Revision |
After running your code, typing your solution in text mode gives :
sage: sol
[H(t) == ilt(1/2*(2*(2*w*(H(0) - 3*laplace(H(t)^2, t, g1626) - laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626)) + 3*H(0) - 9*laplace(H(t)^2, t, g1626) - 3*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(2*n + 1) + (2*C*C2^(2*n)*g1626*n*gamma(2*n, C3*g1626/C2) + (2*C*C2^n*w*gamma(n + 1, C3*g1626/C2) + C*C2^n*gamma(n + 1, C3*g1626/C2))*g1626^n)*e^(C3*g1626/C2))*g1626^(-2*n - 2)/(2*w + 3), g1626, t),
sigma(t) == ilt(-1/4*(C*C2^n*e^(C3*g1626/C2)*gamma(n + 1, C3*g1626/C2) + (w*laplace(P(t)^2/(C2*t + C3)^(2*n), t, g1626) - 6*laplace(H(t)^2, t, g1626) - 6*laplace(H(t)*P(t)/(C2*t + C3)^n, t, g1626))*g1626^(n + 1))*g1626^(-n - 1), g1626, t),
P(t) == ilt(C2^n*n*e^(C3*g1626/C2)*gamma(n, C3*g1626/C2)/g1626^n, g1626, t)]
which should lead you to :
sage: sol[0].operands()[1].operator()
ilt
sage: ilt
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
Cell In[23], line 1
----> 1 ilt
NameError: name 'ilt' is not defined
This is a (long-standing) implementation defect. ilt is Maxima's inverse Laplace transform operator ; from maxima_calculus.ilt? :
-- Function: ilt (<expr>, <s>, <t>)
Computes the inverse Laplace transform of <expr> with respect
to <s> and parameter <t>. <expr> must be a ratio of
polynomials whose denominator has only linear and quadratic
factors; there is an extension of ‘ilt’, called ‘pwilt’
(Piece-Wise Inverse Laplace Transform) that handles several
other cases where ‘ilt’ fails.
The gxxxx symbols, generated by Maxima's solver, denote integration constants, to be solved to satisfy boundary conditions.
By the way, the first and third equations of your system are ordinary equations ; the only differential equation of your system is the second :
$$ H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{6 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{H\left(t\right) P\left(t\right)}{{\left(C_{2} t + C_{3}\right)}^{n}} - \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{2}{3} \, \Sigma\left(t\right) = 0 $$
$$ 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{w P\left(t\right)^{2}}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{2}{3} \, \Sigma\left(t\right) + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
$$ {\left(C_{2} t + C_{3}\right)}^{n - 1} C_{2} n = P\left(t\right) $$
We can solve the ordinary equations for $P(t)$ an $\Sigma(t)$ (by temporarily substituting variable for these expressions) and obtain a form f the differential equation where they no longer appear :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
$$ \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + 2 \, H\left(t\right)^{2} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{6 \, C_{2}^{2} t^{2} H\left(t\right)^{2} - C_{2}^{2} n^{2} w + 6 \, C_{2} C_{3} n H\left(t\right) + 6 \, C_{3}^{2} H\left(t\right)^{2} - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(C_{2}^{2} n H\left(t\right) + 2 \, C_{2} C_{3} H\left(t\right)^{2}\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} + \frac{\partial}{\partial t}H\left(t\right) = 0 $$
We can rewrite this differential equation by isolating the differential term, which, after resubstituting to a variable is equal to :
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
$$ -\frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} - 2 \, f_{0}^{2} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} - \frac{6 \, C_{2}^{2} f_{0}^{2} t^{2} - C_{2}^{2} n^{2} w + 6 \, C_{3}^{2} f_{0}^{2} + 6 \, C_{2} C_{3} f_{0} n - {\left(C C_{2}^{2} t^{2} + 2 \, C C_{2} C_{3} t + C C_{3}^{2}\right)} {\left(C_{2} t + C_{3}\right)}^{n} + 6 \, {\left(2 \, C_{2} C_{3} f_{0}^{2} + C_{2}^{2} f_{0} n\right)} t}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
This equation is polynomial in $f0$ ; we can rewrite it as :
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
$$ \frac{\partial}{\partial t}H\left(t\right) = \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2}^{2} t^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{2} C_{3} t}{3 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} + \frac{C_{2}^{2} n^{2} w}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 2} C_{2}^{2} n^{2} w}{6 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{n} C C_{3}^{2}}{6 \, {\left(C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}\right)}} - {\left(\frac{C_{2}^{2} t^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{2 \, C_{2} C_{3} t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{3}^{2}}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + 2\right)} H\left(t\right)^{2} - {\left(\frac{C_{2}^{2} n t}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}} + \frac{C_{2} C_{3} n}{C_{2}^{2} t^{2} + 2 \, C_{2} C_{3} t + C_{3}^{2}}\right)} H\left(t\right) + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{3 \, {\left(C_{2} t + C_{3}\right)}^{n}} + \frac{{\left(C_{2} t + C_{3}\right)}^{2 \, n - 1} {\left(C_{2} t + C_{3}\right)}^{n} C C_{2} n - {\left(C_{2} t + C_{3}\right)}^{2 \, n} C}{{\left(C_{2} t + C_{3}\right)}^{n} {\left(2 \, w + 3\right)}} $$
which is of the general form :
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
$$ \frac{\partial}{\partial t}H\left(t\right) = k_{2} H\left(t\right)^{2} + k_{1} H\left(t\right) + k_{0} $$
which can be (awkwardly) solved by Sage :
sage: desolve(DEg, H(t=t), ivar=t)
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
[Snip... ]
TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*k0*k2-k1^2>0)', see `assume?` for more details)
Is 4*k0*k2-k1^2 positive or negative?
sage: with assuming (4*k0*k2-k1^2>0): Solp=desolve(DEg, H(t=t), ivar=t) ; Solp
2*arctan((2*k2*H(t) + k1)/sqrt(-k1^2 + 4*k0*k2))/sqrt(-k1^2 + 4*k0*k2) == _C + t
An implicit solution, which can be made explicit :
sage: Solp.solve(H(t=t))
[H(t) == 1/2*(sqrt(-k1^2 + 4*k0*k2)*tan(1/2*sqrt(-k1^2 + 4*k0*k2)*_C + 1/2*sqrt(-k1^2 + 4*k0*k2)*t) - k1)/k2]
Similarly :
sage: with assuming (4*k0*k2-k1^2<0): Soln=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Soln
[H(t) == -1/2*((k1 + sqrt(k1^2 - 4*k0*k2))*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k1 + sqrt(k1^2 - 4*k0*k2))/(k2*e^(sqrt(k1^2 - 4*k0*k2)*_C + sqrt(k1^2 - 4*k0*k2)*t) - k2)]
sage: with assuming (4*k0*k2-k1^2==0): Solz=desolve(DEg, H(t=t), ivar=t).solve(H(t=t)) ; Solz
[H(t) == -1/2*(_C*k1 + k1*t + 2)/(_C*k2 + k2*t)]
There you have a solution for the general form of the equation. All that remains to be done is to replace the coefficients by their values (which I am too lazy to do...) and to frame the three solution cases in a cases expression.
EDIT : Nontwithstanding my laziness, here's the abstract of my solution :
D1=dict(zip((H(t=t), sigma(t=t), P(t=t)), var("f", n=3)))
Sol1=solve([des[u].subs(D1) for u in (0, 2)], (f1, f2))
ID1={D1[u]:u for u in D1.keys()}
DE=des[1].subs(D1).subs(Sol1[0]).subs(ID1)
DE1=((-DE)+diff(H(t=t), t)).lhs().subs(D1)
D2={(K:=var("k", n=3))[u[1]]:u[0].simplify_full().factor()
for u in DE1.coefficients(f0)}
DE2=diff(H(t=t), t)==sum([u[0]*H(t=t)^u[1] for u in DE1.coefficients(f0)])
DEg=diff(H(t=t), t)==sum([(K:=var("k", n=3))[u]*H(t=t)^u for u in range(3)])
from _operator import lt, eq, gt
Sol=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol+=[(cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2).full_simplify().factor())]
Sol2=H(t=t)==cases(Sol2)
which is way yooo wide to LaTeX. Asking Mathematica a bit of help :
Sol2=[]
for cond in [f(4*k0*k2-k1^2,0) for f in (gt, eq, lt)]:
with assuming(cond): Sol2+=[(cond, Sol2+=[[cond, desolve(DEg, H, ivar=t).solve(H)[0].rhs().subs(D2)._mathematica_().FullSimplify().sage(locals={("Tan", 1):tan, ("Coth", 1):coth}))]
1):coth})]]
# Refining display
Sol2[2][1]=Sol2[2][1]._mathematica_().Factor().sage(locals={("Tan", 1):tan, ("Coth", 1):coth})
Sol2=H==cases(Sol2)
Since this site's Mathjax desn't seem to support cases, here is its argument :
$$ \begin{cases}{a} & {a \left[-k_{1}^{2} + 4 \, k_{0} k_{2} > 0}\{0} & {a 0, -\frac{1}{6} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} \tan\left(\frac{1}{2} \, \sqrt{-\frac{C_{2}^{2} n^{2}}{{\left(C_{2} t + C_{3}\right)}^{2}} - \frac{6 \, {\left(2 \, {\left(C_{2} t + C_{3}\right)}^{n} C_{2} n + {\left(C_{2} t + C_{3}\right)} {\left(2 \, w + 1\right)}\right)} {\left(C_{2} t + C_{3}\right)}^{n - 1} C}{2 \, w + 3}} {\left(C + t\right)}\right) - \frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}}\right] $$
$$ \left[-k_{1}^{2} + 4 \, k_{0} k_{2} = 0}\{-a} & {a 0, -\frac{C_{2} n}{6 \, {\left(C_{2} t + C_{3}\right)}} + \frac{1}{3 \, {\left(C + t\right)}}\right] $$
$$ \left[-k_{1}^{2} + 4 \, k_{0} k_{2} < 0}\end{cases} 0, -\frac{C_{2} n \sqrt{2 \, w + 3} - \sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} \coth\left(\frac{\sqrt{C_{2}^{2} n^{2} {\left(2 \, w + 3\right)} + 12 \, {\left(C_{2} t + C_{3}\right)}^{2 \, n + 1} C C_{2} n + 6 \, {\left(C_{2} t + C_{3}\right)}^{n + 2} C {\left(2 \, w + 1\right)}} {\left(C + t\right)}}{2 \, {\left(C_{2} t + C_{3}\right)} \sqrt{2 \, w + 3}}\right)}{6 \, {\left(C_{2} t + C_{3}\right)} \sqrt{2 \, w + 3}}\right] $$
HTH,
