## DSolve unable to solve the known second-order differential equation

I am trying to solve a time dependent second order differential equation with a known solution. The equations I want to solve are

$$x & # 39; & # 39; (t) = frac {-x (t)} {(x (t) ^ 2 + y (t) ^ 2) ^ {1.5}}$$

and

$$y & # 39; & # 39; (t) = frac {-y (t)} {(x (t) ^ 2 + y (t) ^ 2) ^ {1.5}}$$

This clearly has the solution x (t) = Cos (t + a), y (t) = Sin (t + b), where a and b can be defined according to the boundary conditions. However, when I try to solve this problem in Mathematica, it is impossible to return the solution, even if I include the boundary conditions.

eqns1 = {(x(t)^2 + y(t)^2)^-1.5*x''(t) == -
x(t), (x(t)^2 + y(t)^2)^-1.5*y''(t) == - y(t), x(0) == 0,
y(0) == 1};
DSolve(eqns1, {x, y}, t)


But it comes back


DSolve({(x^(Prime)(Prime))(t)/(x(t)^2 + y(t)^2)^1.5 == -x(t), (
y^(Prime)(Prime))(t)/(x(t)^2 + y(t)^2)^1.5 == -y(t), x(0) == 0,
y(0) == 1}, {x, y}, t)


that is, it just tells me that Mathematica cannot solve this problem. Does anyone know why i can't find a solution or what is the problem? Is there a known way around this problem?

Thank you

Posted on

## fa.functional analysis – Trace and second-order inverse trace on the space with a Gibbs measure

Consider $$(t, x) in (0, T) times ( mathbb {R} ^ d, d mu)$$, where the measure $$d mu (x) = K ^ {- 1} exp (-U (x)) dx$$ is a reasonable Gibbs measure (it responds to an inequality of Poincaré or log-Sobolev, we can, for simplicity, start with Gaussian $$d mu (x) = (2 pi) ^ {- frac {d} {2}} exp (- dfrac {| x | ^ 2} {2}) dx$$). We define the spaces of Sobolev $$L ^ 2 (I times mu) = {f: int _ { mathbb {R} ^ d} int_0 ^ T f (t, x) ^ 2 dtd mu (x) < infty }$$ and $$H ^ 1 (I times mu) = {f: f, partial_t f, nabla_x f in L ^ 2 (I times mu) }$$ and $$H ^ 2 (I times mu)$$ in a similar way. Now, considering everything $$f in H ^ 1 (I times mu)$$, can we find $$u in H ^ 2 (I times mu)$$, such as $$u (t = 0, cdot) = u (t = T, cdot) = 0$$ and $$partial_t u (t = 0, cdot) = f (0, cdot), partial_t u (t = T, cdot) = f (t = T, cdot),$$ such as $$| u | _ {H ^ 2 (I times mu)} the C | f | _ {H ^ 1 (I times mu)}?$$

Posted on

## set theory – Why is Skolem's theorem essentially second-order?

The (philosophical) intuition beyond this question is that, as far as the skolemization of a $$exists forall$$ The first-order formula introduces a Skolem constant that refers to an arbitrary individual (ie, an arbitrarily chosen individual), so that the skolemization of a $$forall exists$$ The first-order formula introduces a Skolem function that (should) refer to an arbitrary function.

Nevertheless, it is common knowledge that, if the equisatisfiability of a $$forall exists$$ The first order formula and its normal Skolem form (SNF) can be proven in FOL as: $$| forall x exists x varphi (x, y) | iff | forall x varphi (x, f (x)) |$$
The logical equivalence of a first-order formula and its SNF can only be proven in SOL (I'll call it Skolem's theorem), assuming the axiom of choice (AC) is: $$forall x exists x varphi (x, y) templates exists f forall x varphi (x, f (x))$$ My question is: What is the ZFC hypothesis (or functionality) that makes the Skolem theorem essentially second-order? It seems to me that the metalanguage of FOL – namely ZFC – does not make it possible to prove the logical equivalence between a first-order formula and its SNF within FOL.

I will now propose a solution – but any other suggestion is welcome. I think this problem is closely related to the fact that in ZFC, although each definable set can be proved (see my question Why in ZFC, there is also an expressible set?), All existing sets can not be defined. Let me explain: Skolem's functions in an extended model are by no means unique. However, if the first order formula is like:
$$forall x exists! x varphi (x, y)$$
Then the logical equivalence with its SNF can be proved in FOL without AC as: $$forall x exists! x varphi (x, y) templates forall x varphi (x, f (x))$$ Moreover, this last equivalence defines Skolem's function for the antecedent formula. This is because in ZFC you can define a selection set t (ie a Skolem function) for another set s Yes Yes s contains a definable element (Fraenkel 1973, Foundations of Set Theory). In the example above, s contains a definable element because $$exists! x$$ is a singleton. However, the ZFC proof of Skolem's theorem also requires considering a model without a definable element – such as the set of all the good orders of the set of real numbers. At this point, AC must guarantee the existence of a function (Skolem) that can not be defined. Is it correct?

If this argument is valid, there is a tension with the initial (philosophical) intuition. Let me explain: to prove the logical equivalence of $$forall exists$$ For first order formulas and their SNF, we need AC – and therefore SOL – because there are models without definable elements in which the values ​​of Skolem functions can not be defined. However, to prove the logical equivalence of $$exists forall$$ first order formulas and their SNF neither AC nor SOL are required. But how is it possible that the added Skolem constant denotes a non-definable element? To return to the initial (philosophical) intuition, it seems that in the metalanguage of FOL – namely ZFC – we can refer to arbitrary objects but not to arbitrary functions. Why This last question is delicate: although the first proof requires AC – and so it is done in ZFC – the last proof can be given in a (pure) theoretical model framework (as for Henkin's theory).

Posted on

## lo.logic – Is there a conservative or non-conservative extension of the finitely axiomatized theory of second-order sets (like MK and TG)?

(1) For MK and TGIs there a conservative or nonconservative extension finely axiomatisable?

(2) Does "each" second-order theory have a finitely axiomatized conservative / non-conservative extension?

(3) We need the theory of third order sets to hold CH actually complete (as Ultimate-L).

Does a "useful" second-order theory have a finitely axiomatized conservative / non-conservative extension?

Posted on

## NDSolve function problem and interpolated when solving a second-order differential equation

I'm getting an interpolated NDSolve function by solving a second-order differential equation. If I generate a list of interpolated function than create a new interpolated function.Is the two functions are the same or different. Because I get two different results using these two functions.

Posted on

## differential equations – Asymptotic solution of a second-order ODE containing InverseFunction

I have basically a second order differential equation given by ode below. In order to solve it, I have to get an asymptotic solution where $$g (x)$$ must go to the & # 39; infinity that will be used after initial conditions (c & # 39; is to say, $$g ( infty), g ( infty)$$) to evolve the differential equation with the help of NDSolve.
For the asymptotic solution, I come with the code below that uses the Frobenius method. The difficulty comes from the fact that the differential equation contains a function TP which is a inverse function of the rho (r, q). I guess my code does not work with a function not defined by the user. My questions are:

(1) Is there a way to correct my code?
(2) Instead of using InverseFunction to invert rho (r, q), is there a better way to invert functions so that it works with the way code is created?

asymp(p_) := {ode = (2/TP(x, q) - p/TP(x, q)^(p + 1)) TP(x, q)^2 dTP g'(x) + TP(x, q)^2 g''(x) - l (l + 1) g(x);
b = 1;
q = -1;
(Alpha) = l + 1;
rho(r_, q_) := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1(1/2, 1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q));
TP = InverseFunction(Function({r, q}, rho(r, q)), 1, 2);
dTP = (Sqrt(1 - (b/TP(x, q))^(1 - q)) ((b/TP(x, q))^(1 - q))^(1/(1 - q)) TP(x, q))/b;
g(r_) := Sum(a(i)/r^(i + (Alpha)), {i, 0, ORDINF + 5});
ORDINF = 3;
ss = FullSimplify(Series(ode, {x, (Infinity), ORDINF}));
eqsINF = Table(SeriesCoefficient(ss, i) == 0, {i, 2, ORDINF});
yinf = Table(a(i), {i, 1, ORDINF - 1});
seriesINF = Simplify(Solve(eqsINF, yinf))((1));
Rasymp = Collect(Simplify(Sum(a(i)/x^(i + (Alpha)), {i, 0, ORDINF - 1}) /. seriesINF /. a(0) -> 1), x);
dRasymp = Collect(Simplify(D(Rasymp, x)), x)}


By the way, $$p> 0$$ and $$q < 1$$. Any help or hint would help. Thank you!

Posted on

## Second-order cybernetics and human improvement

What does second-order cybernetics have to do with human enhancement?

Posted on

## Differential Equations – Finding the General Solution for Second-Order EDOs

Thank you for contributing an answer to MathOverflow!

• Please make sure to respond to the question. Provide details and share your research!

But to avoid

• Make statements based on the opinion; save them with references or personal experience.

Use MathJax to format equations. MathJax reference.

Posted on

## Tracing – Resolution of a second-order ODE with ParametricNDSolveValue

I'm trying to numerically solve a non-linear ODE on the 0-to-0 interval R, using ParametricNDSolveValue. The boundary conditions are: x == 0 at r == R, x == 0 at r == 0, and x to R == 0 .c0 is a parameter.

R = 512

a = 100


my function:

X & # 39; & # 39;[r] + x & # 39;[r]/ r == -c0 Exp[-X[-x[-X[-x[r]]+357/25 for 0 <= r <= a,
X & # 39; & # 39;[r] + x & # 39;[r]/ r == -c0 Exp[-X[-x[-X[-x[r]]for a <r <= R


Help?

Posted on

## differential equations – coupled pde with second-order spatial derivative

I'm trying to solve a PDE system using NDsolve but the following errors come with it.

NDSolveValue :: femcmsd: The spatial derivative order of the EDP can not
exceed two.

equations:  Qm and Q are constant.

                pde1 = {
re[qr[r, x, t], t]+ tq * D[RE[D[RÉ[D[qr[r, x, t], t], t]==
alpha * D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), r]+
alpha * ta *
re[RE[D[RÉ[D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), r], t]+
alpha * wrc * D[Te[r, x, t], r]+ alpha * wrc * ta * D[RE[D[RÉ[D[Te[r, x, t], t], r]};

pde2 = {
re[qx[r, x, t], t]+ tq * D[RE[D[RÉ[D[qx[r, x, t], t], t]==
alpha * D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), X]+
alpha * ta *
re[RE[D[RÉ[D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), X], t]+
alpha * wrc * D[Te[r, x, t], X]+ alpha * wrc * ta * D[RE[D[RÉ[D[Te[r, x, t], t], X]};

pde3 = {
ro * c * d[Te[r, x, t],
t]== - (1 / r) * (D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]+
wrc * (Tb - Te[r, x, t]) + Qm
};

sol = NDSolveValue[{pde1, pde2, pde3, BC1, BC2, BC3, BC4, BC5, IC1,
IC2, IC3, IC4, IC5, IC6},
You[r, x, t], {t, 0, 5}, {r, x} [Element] [CapitalOmega]];


Is it possible to handle this error?

Posted on