Tag: ODEs

Differential equations – Solution system of ODEs with additional parameters

I would like to solve one $ 2 times $ 2 System of the form
$$ frac {d} {d theta} T = TA, quad T (0) = Id $$
from where $ theta $ is real and $ A $ is of the form
$$ A = begin {pmatrix} 0 & frac {e ^ {- i theta}} { lambda} \ frac {1} {36} e ^ {- i theta} left (9 Lambda + 2 ( lambda-1) ^ 2 (6 cos { theta} + cos {2 theta} + 6) right) & 0 end {pmatrix}, $$
With $ lambda $ a free parameter in the unit circle,

I am particularly interested in numerical solutions $ theta = 2 pi $ depending on the additional parameters $ lambda $, I'm pretty new with Mathematicaand I've tried it so far:

T[θ_] = {{T11[θ]T12[θ]}, {T21[θ]T22[θ]}};
ON[θ_] = {{0, E ^ (- Iθ) / λ}, {1/36 E ^ (- Iθ) (9λ + 2 (-1 + λ) ^ 2 (6 + 6 Cos[θ] + Cos[2 θ])), 0}};
sys = {T & # 39;[θ] == T[θ].ON[θ]};

The previous code sets the system I want to solve, and now I'm trying to solve numerically. I tried it first

NSol = NDSolve[{sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1}, {T11[θ]T12[θ]T21[θ]T22[θ]}, {θ}, {θ, 0, 2 Pi}];

that gives me the output

NDSolve :: dupv: "Double variable θ found in NDSolve[<[<[<[<<1>>], "

I tried too

Nsol2 = ParametricNDSolve[{sys, T11[0] == 1, T12[0] == 0, T21[0] == 0, T22[0] == 1}, {T11, T12, T21, T22}, {θ, 0, 2 Pi}, {λ}];

what gives me as an issue $ T_ {11}, dots, T_ {22} $ as Parametric Functions dependent on each other and on $ lambda $,

I do not know if this is the right approach and if so, how to extract a numeric expression depending on it $ lambda $ from the last issue – all I've seen in the documentation are examples that are recorded for specific parameter values. Any help is greatly appreciated.

Differential Equations – Problem solving the first ODEs with DSolve

The problem I'm facing is the following. I am trying to solve a system of two coupled first order ODEs with DSolve:

eqns = {A * f & # 39;[x] == -x * f[x]/ 2 + B * g[x], A * g & # 39;[x] == x * g[x]/ 2 + B * f[x]};
DSolve[EQNS{f[EQNS{f[eqns{f[eqns{f[x]G[x]}, x]

As output this only gives the same command that means from my collection that DSolve can not solve it.

But if I explicitly use the equations, it can suddenly be solved:

Fun = f[x] /. To solve[A*g&#39;[A*g'[A*g'[A*g'[x] == x * g[x]/ 2 + B * f[x]f[x]];
dfun = D[fun, x];
DSolve[A*f#39;[A*f'[A*f'[A*f'[x] == -x * f[x]/ 2 + B * g[x] /. {f[x] -> fun, f & # 39;[x] -> dfun},
G[x], x]

Which gives

{{G[x] ->
C[2] ParabolicCylinderD[B^2/A, (I x)/Sqrt[A]]+
C[1] ParabolicCylinderD[(-A - B^2)/A, x/Sqrt[A]]}}

That makes me think that I'm just doing something wrong on the first line, which is probably a stupid notation error. If someone could help me, I would be very grateful.

Differential equations – How do you solve these ODEs with NDSolve?

The calculation changes t = 0 there sin[ψ

eq1 = ω1
eq2 = ω2

and construct the linear combinations,

eq1n = Simplify[eq1Sin[eq1Sin[eq1Sin[eq1Sin[ψ
(* Cos[ψ

(If this were not possible, the equations themselves could not be solved in principle.)

Now replace eq1, eq2 by eq1n, eq2n,

I1 = 2; I2 = 3; I3 = 4;
s = NDSolveValue[{I1*ω1&#39;[{I1*ω1'[{I1*ω1'[{I1*ω1'
I2 * ω2 & # 39;
I3 * ω3 & # 39;
eq1n == 0, eq2n == 0,
ω3
ω1[0] == 2, ω2[0] == 3, ω3[0] == 4, ψ[0] == 0, φ[0] == 0, θ[0] == Pi / 6},
{ω1
plot[Evaluate@s[[1 ;; 3]], {t, 0, 120}, ImageSize -> Large]plot[Evaluate@s[[4 ;; 6]], {t, 0, 120}, ImageSize -> Large]

Enter the image description here

Enter the image description here

Incidentally, the original equations can be solved by slightly changing the initial state ψ[0] == 0 to ψ[0] == 10 ^ -6,

Another approach is to use the option.

Method -> {"EquationSimplification" -> "Residual"}

Everyone gives the same answer.

How do I solve these Odes with NDSolve?

I have six odes and can not use DSolve. So I tried NDSolve. But it does mean that there may be mistakes. The code looks like this:

I1 = 2; I2 = 3; I3 = 4;
NDSolve[{I1*ω1'
I2 * ω2 & # 39;
I3 * ω3 & # 39;
0, ω1[
t] == φ & # 39;
sin[ψ[ψ[ψ[ψ
t] == φ & # 39;
cos[ψ[ψ[ψ[ψ
t] == φ & # 39;
0] == 2, ω2[0] == 3, ω3[0] == 4, ψ[0] ==
0, φ[0] == 0, θ[0] ==
Pi / 6}, {ω1, ω2, ω3, ψ, φ,
θ}, {t, 0, 120}]

I want to know how to avoid this mistake.

Classical analysis and Odes – Algebraic Riccati and WKB

It is a one-liner to show that the algebraic Riccati equation (ARE) and the lowest-order form of WKB are the same for a linear ode. But I've looked everywhere on the Internet and it seems that this connection, despite its triviality is not recognized. It seems that there is a total separation between physicists who make many WKB and control people who do a lot of Riccati.

Another trivial point is that a linear ode in psi has the Lie symmetry psi -> lambda psi, so the Riccati variable is the Lie invariant, and this leads to a reduction in order (or more precisely, a reduced one Order) followed by a quadrature.) Is this so obvious that it can never be mentioned? This finding means that not only linear equations are used, but every equation that is homogeneous in degree 1 and whose order is reduced by the Riccati transformation.

Classical analysis and odes – Failure of a Falconer distance problem in one dimension

I am told that the Falconer distance guess in one dimension is trivial, but I really can not find any clue. I ask exactly the following question:

For a compact set $ E subseteq mathbb R ^ n $we define the set distance $ Delta (E) subseteq[0infty)$[0infty)$[0infty)$[0infty)$ his:
$$
Delta (E) = {| x-y |: x, y in E }.
$$
Then, when $ n = 1 $do we ask the following questions?

  1. Can we find a compact set? $ E subseteq [0,1]$ so that $ mathrm {dim} _H (E)> 1/2 $ but $ mathcal L ^ 1 ( Delta (E)) = 0 $?

  2. Can we find a compact set? $ E subseteq [0,1]$ so that $ mathrm {dim} _H (E) = 1 $ but $ mathcal L ^ 1 ( Delta (E)) = 0 $?

Differential Equations – Complex Solutions for ODEs

How do I solve the following IVP problem in Mathematica, so I get real solutions?

$ Q ((t) = b – dfrac {Q (t)} {100-t}; quad Q (0) = 250 $

I tried the following:

$ text {$ $$ assumptions} = b> 0; text {$ $$ assumptions} = t> 0; $

$ f = text {DSolve} left[left{Q'(t)=b-frac{Q(t)}{100-t},Q(0)=250right},Q,tright][[1,1,2]]$

$ f (t) $

which leads to the following:

$ frac {1} {2} (-2 i pi b t-2 bt log (100) -200 b log (t-100) +2 bt log (t-100) +200 i pi b + 200 b log (100) -5 t + 500) $

Any help would be appreciated. Many Thanks!!!

approx. classical analysis and odes – Sturm-Liouville equation with finite number of eigenvalues?

Consider the following Storm Liouville (SL) intrinsic value problem $ x in (- infty, 0) $ $$ (py & # 39;) & # 39; – qy = – lambda ^ 2wy, $$ in which $ p (x) = x ^ 2 $, $ w (x) = 1 $, and $ q (x) = (x / 2 + a) ^ 2 + a $ with parameters $ a> 0 $, It has a regular singularity $ x = 0 $, We basically hope for something like a homogeneous Dirichlet b.c.

It is solved by the substitution $ y (x) = e ^ {x / 2} x ^ {- frac {1} {2} + sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2}} u (x) $This leads directly to a standard confluent hypergeometric equation $$ xu & # 39; & # 39; (x) + ( gamma-x) u & # 39; (x) – alphau (x) = 0, $$ in which $ alpha = sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2} -a + frac {1} {2} $ and $ gamma = 1 + 2 sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2} $, There are two independent solutions (1st type & 2nd type).
Let's take a few ubiquitous arguments when solving the eigensystem related to confluent hypergeometric equations. Non-divergence required $ x = 0 $the 2nd kind is dropped. Non-divergence required $ infty $the 1st kind is reduced to a polynomial if $ – alpha $ is a non-negative integer and eigenvalue $ lambda ^ 2 $ is reached.

However, seen from this condition for $ alpha $obviously We have only a finite and finite series of eigenvalues ​​that is different from the infinite eigen spectrum that the SL theory claims.

What is wrong here? Do I miss a few solutions?

Differential Equations – Solving a Nonlinear System of 2nd Order ODEs with NDSolve

I'm trying to solve a second-order ODE system with five dependent variables.

The equations are:

eqn1 = D[z]* C & # 39; & # 39;[z] - u * C & # 39;[z] == a * r1[z]
eqn2 = k[z]* T & # 39; & # 39;[z] - q * T & # 39;[z] == a * (r1[z]* b + r2[z]* c + r3[z]* d)

from where A B C D are constants; k[z], D[z]ri[z] (from where i = 1,2,3) are dependent variables of T[z] and C[z], and C[z] and T[z] depend on each other and where

r1[z] = (kc1[z]* keb[z]* (y1[z]* p - (y3[z]* p + y4[z]* p) / keq[z]* P)) / (1 + keb[z]* y1[Z*p+kh2[Z*p+kh2[z*p+kh2[z*p+kh2[z]* y4[z]* p + kst[z]* y3[z]* p) ^ 2

r2 and r3 are like r1[z],

I have the following constraints:

initConds = {T[0] == 893.15, T & # 39;[L] == 0, C[0] 0.992, C & # 39;[L] == 0};

from where L = 9,

I use it to use NDSolve as shown below, but the code did not trigger the system, and the error message was also displayed.

s = NDSolve[{}{TeqnsinitConds[{}{TeqnsinitConds[{eqnsinitConds}{T[{eqnsinitConds}{T[z]C[z]}, {z, 0, 9, 0.01}]

NDSolve :: ntdv: Unable to resolve to find an explicit formula for the derivatives. Consider using the Method -> {"EquationSimplification" -> "Residual"} option.

Suppose Mathematica can solve my system, how could I write code? NDSolve or another method?