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# Maxima wants to know whether %k1 is positive or negative

Trying to solve a differential equation in Sage:

sage:  de = -diff(y, x, 2) - G*M/y^2
sage: desolve(de, y, ics=[0, 1000])


I get a type error and a question:

 TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(%k1>0)', see assume? for more details)
Is %k1 positive or negative?


But when I use assume(%k1>0) I get a syntax error, saying that there's something wrong with the percent sign.

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## Comments

It's just a syntax. In place of %k1 substitute an actual parameter from your equation. For example, assume(G>0).

It's not just a syntax : Maxima asks for the sign of a constant it has itself introduced. This is, IMHO, a bug.

I also think that it is a bug. SageMath does not send the assumptions to Maxima solver.

SageMath does not send the assumptions to Maxima solver.

Nope. See my answer below. When the assumptions on G and M are omitted, Maxima's question is about of the sign of GM%k1.

Then what is the problem with the following code?

var('G,M')

y=function('y')(x)

assume(G*M>0)

de = -diff(y, x, 2) - G*M/y^2

desolve(de, y, ivar=x, ics=[0, 1000])

## 1 Answer

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As said in comment, the problem is that Maxima introduces a arbitrary constant %k1 and asks the user an hypothesis about it :

sage: G,M=var("G,M")
sage: y=function("y")
sage: with assuming(G>0, M>0): Sol=desolve(de, y(x), ivar=x)


[ Snip.. ]

TypeError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(%k1>0)', see assume? for more details)
Is %k1 positive or negative?


The workaround to this problem is to add this assumption to what Maxima uses. This leads to another problem : there's no way to define a variable whose Maxima name would be %k1. The workaround is to ruin this directly in Sage's default Maxima interpreter :

sage: maxima_calculus("assume(%k1>0);")
[%k1>0]
sage: with assuming(G>0, M>0): Sp=desolve(de, y(x), ivar=x)
sage: maxima_calculus("forget(%k1>0);")
[%k1>0]
sage: maxima_calculus("assume(%k1<0);")
[%k1<0]
sage: with assuming(G>0, M>0): Sn=desolve(de, y(x), ivar=x)


One gets implicit answers, woese getting explicit solutions is ... problematic :

sage: Sp+Sn
[1/2*sqrt(2)*(G*M*arctan(sqrt(-(G*M*_K1*y(x) - G*M)/y(x))/sqrt(G*M*_K1)) + sqrt(G*M*_K1)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x))*y(x))/(sqrt(G*M*_K1)*G*M*_K1) == _K2 + x,
-1/2*sqrt(2)*(G*M*arctan(sqrt(-(G*M*_K1*y(x) - G*M)/y(x))/sqrt(G*M*_K1)) + sqrt(G*M*_K1)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x))*y(x))/(sqrt(G*M*_K1)*G*M*_K1) == _K2 + x,
1/4*sqrt(2)*(G*M*log(-(sqrt(2)*sqrt(-G*M*_K1) - sqrt(2)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x)))/(sqrt(2)*sqrt(-G*M*_K1) + sqrt(2)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x)))) + 2*sqrt(-G*M*_K1)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x))*y(x))/(sqrt(-G*M*_K1)*G*M*_K1) == _K2 + x,
-1/4*sqrt(2)*(G*M*log(-(sqrt(2)*sqrt(-G*M*_K1) - sqrt(2)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x)))/(sqrt(2)*sqrt(-G*M*_K1) + sqrt(2)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x)))) + 2*sqrt(-G*M*_K1)*sqrt(-(G*M*_K1*y(x) - G*M)/y(x))*y(x))/(sqrt(-G*M*_K1)*G*M*_K1) == _K2 + x]


More readably :

$$\frac{\sqrt{2} {\left(G M \arctan\left(\frac{\sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}{\sqrt{G M K_{1}}}\right) + \sqrt{G M K_{1}} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}} y\left(x\right)\right)}}{2 \, \sqrt{G M K_{1}} G M K_{1}} = K_{2} + x$$

$$-\frac{\sqrt{2} {\left(G M \arctan\left(\frac{\sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}{\sqrt{G M K_{1}}}\right) + \sqrt{G M K_{1}} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}} y\left(x\right)\right)}}{2 \, \sqrt{G M K_{1}} G M K_{1}} = K_{2} + x$$

$$\frac{\sqrt{2} {\left(G M \log\left(-\frac{\sqrt{2} \sqrt{-G M K_{1}} - \sqrt{2} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}{\sqrt{2} \sqrt{-G M K_{1}} + \sqrt{2} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}\right) + 2 \, \sqrt{-G M K_{1}} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}} y\left(x\right)\right)}}{4 \, \sqrt{-G M K_{1}} G M K_{1}} = K_{2} + x$$

$$-\frac{\sqrt{2} {\left(G M \log\left(-\frac{\sqrt{2} \sqrt{-G M K_{1}} - \sqrt{2} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}{\sqrt{2} \sqrt{-G M K_{1}} + \sqrt{2} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}}}\right) + 2 \, \sqrt{-G M K_{1}} \sqrt{-\frac{G M K_{1} y\left(x\right) - G M}{y\left(x\right)}} y\left(x\right)\right)}}{4 \, \sqrt{-G M K_{1}} G M K_{1}} = K_{2} + x$$

Neither Sympy nor Fricas can solve thos ODE ; Mathematica also returns an implicit solution :

sage: mathematica("DSolve[-y''[x]-G*M/y[x]^2==0, y[x], x]")
Solve[
(-((G*M*Log[G*M + C*y[x] + Sqrt[C]*Sqrt[C + (2*G*M)/y[x]]*y[x]])/
C^(3/2)) + (Sqrt[C + (2*G*M)/y[x]]*y[x])/C)^2 == (x + C)^2,
y[x]]


i. e., more readibly :

$$\text{Solve}\left[\left(\frac{y(x) \sqrt{c_1+\frac{2 G M}{y(x)}}}{c_1}-\frac{G M \log \left(\sqrt{c_1} y(x) \sqrt{c_1+\frac{2 G M}{y(x)}}+c_1 y(x)+G M\right)}{c_1^{3/2}}\right){}^2=\left(c_2+x\right){}^2,y(x)\right]$$

Note : since your ODE is of the second order, you need two boundary conditions uin order to determine your solution ; your ics gives only one...

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## Comments

So I tried solving this in wxmaxima:  eq: 'diff(y, x, 2)=-G*M/y^2 sol2: ode2(eq, y, x) bc2(sol2, x=0, y=1000, x=pi*sqrt(1000^3/8*G*M), y=5) ` wxmaxima just returns []

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Asked: 2021-04-07 14:49:01 +0200

Seen: 52 times

Last updated: Apr 08