Writing linear image equations in linear transformation

Let T: R2 → R3 be the linear transformation defined by T (x, y) = (y – x, −3x, −3y)

Write a linear equation that defines the subspace Im (T)

….. = 0

(Write your answer in the form ax + by + cz. For example "2x + 3y – 4z")

So, somehow I learned how to find the image or the nuclei, but I have no idea how to write a linear equation. Can anyone help me with that? At least please give me an answer to this question. I can figure out how to do it.

differential equations: how to define the boundary condition in 1D heat transfer

I am trying to calculate the head transfer between a 1-D rod, with an insulated end while the right end is submerged on a surface of constant temperature T = 0. Suppose the initial temperature of the rod is T = 1. The bar length is 5. I configured the equation like this:

$$
frac { partial ^ 2 u (x, t)} { partial x ^ 2} – frac { partial u (x, t)} { partial t} = 0 \
u (x, 0) = 1 \
frac { partial u (5, t)} { partial x} = 0 \
u (0, t) = 0
$$

sol = NDSolve({
       eqn = D(u(t, x), t) - D(u(t, x), {x, 2}) == 0,
       u(0, x) == 1,
       u(t, 0) == 0,
       (D(u(t, x), x) /. x -> 5) == 0
       }, u, {t, 0, 50}, {x, 0, 5})

    Plot3D(Evaluate(u(t, x) /. %), {t, 0, 50}, {x, 0, 5}, 
     PlotRange -> All)

Unfortunately, I got something like this:

NDSolve :: ibcinc: Warning: Initial limits and conditions are inconsistent.

Can anyone help me with the limit value problem?

differential equations: using WhenEvent to bind variables

I would like to use the WhenEvent (link here) to set a limit on a dynamic variable in an EDO set. That is, when the variables k21(t) Y k12(t) reach above the specific value aI would like to put them back at that value. Seems like a simplified version of the problem

w1 = 6/24.5;
w2 = 6/23.5;
a = 0.1;

Eqs = {
   x1'(t) == w1 + (k21(t)/2)*Sin(x2(t) - x1(t)),
   x2'(t) == w2 + (k12(t)/2)*Sin(x1(t) - x2(t)),
   k21'(t) == a*(Cos(x2(t) - x1(t) + (Pi)) + 1),
   k12'(t) == a*(Cos(x1(t) - x2(t) + (Pi)) + 1)};

ICs = {x1(0) == 3/2, x2(0) == 3/4, k21(0) == 0.0001, k12(0) == 0.0001};

events = {WhenEvent(Abs(k21(t)) > a, k21(t) -> a), WhenEvent(Abs(k12(t)) > a, k12(t) -> a)};

EqsICs = Join(Eqs, ICs, events);

SolutionValue(t_) = NDSolveValue(EqsICs, {x1(t), x2(t), k21(t), k12(t)}, {t, 0, 10^6});

And I trace the solutions as:

Show(
Plot(SolutionValue(t)((3)), {t, 0, tmax}, PlotRange -> {{0, tmax}, {-0.1, 0.1}}, AxesOrigin -> {0, 0}), 
Plot(SolutionValue(t)((4)), {t, 0, tmax}, PlotRange -> {{0, tmax}, {-0.1, 0.1}}, AxesOrigin -> {0, 0})
)

However, my WhenEventdoes not work The graph shows that the values ​​of kij continues to grow beyond a. Is my syntax incorrect? Thank you 🙂

Obtain the representation of {x, theta} with the given equations

I am working with a transfer function problem. I want to get the representation of {x, $ theta $} with the given equations (below).

to be solved eqns

In the equation x10 and x20 it means the initial value of x1 and x2. all other letter is undetermined.

I want to get the represented form of {x, theta} based on these two equations.

Here is my code:

ClearAll(x, (Theta), c);
{a, b} = {{1, -L/2}, {1, L/2}}.{x, (Theta)};
{Subscript(F, 1), Subscript(F, 
   2)} = {{k + c s, 0}, {0, k + c s}}.{a, 
     b} - {{k + c s, 0}, {0, k + c s}}.{Subscript(x, 1), Subscript(x, 
     2)} + c {x10, x20};
eqn = {x, (Theta)} - {{1/(M s^2), 1/(M s^2)}, {-L/(2 J), L/(
       2 J)}}.{Subscript(F, 1), Subscript(F, 2)} + {g/s^2, 0} == {0, 
    0};
Assuming({s != 0, M != 0, J != 0},
 Reduce(eqn // Thread, {x, (Theta)}, Reals))

This is the output:

Reduce :: nsmet: 无法 利用 Reduce 现有 的 方法 求解 该 系统.

Chinese characters mean "you cannot use the original methods in Reduce` to solve this system"

Thank you all.

How to solve these equations without errors?

I am a new user of Mathica 12. I have to plot real and imaginary solutions of two equations, eq1 and eq2, which contain the Dawson function, here is my program:

f(a_) := - 2 DawsonF(a)+I Exp(-a^2) Sqrt((Pi)) 

eq1(x1_, y_, z_) := 1 - (z y^2/x1^2) - (1/(Sqrt(2) x1 y)) f(x1/(Sqrt(2) y))
eq2(x2_, y_) := 1 - (1/(2 y^2)) f'(x2/(Sqrt(2) y))

slo1(y_, z_) := Re(x1 /. FindRoot(eq1(x1, y, z), {x1, Sqrt(1 + z y^2)}));
slo2(y_) := Re(x2 /. FindRoot(eq2(x2, y), {x2, Sqrt(1 + y^2)}));

Plot({slo1(y, 5000), slo2(y)}, {y, 0.0001, 0.5},PlotRange -> {{0.0001, 0.5}, {0, 4}}, PlotRangePadding -> 0)

When I plot the charts I get these errors

General::munfl: Exp(-4.11658*10^7) is too small to represent as a normalized machine number; precision may be lost.

General::munfl: Exp(-4.11658*10^7) is too small to represent as a normalized machine number; precision may be lost.

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

General::munfl: Exp(-4.11633*10^7) is too small to represent as a normalized machine number; precision may be lost.

General::stop: Further output of General::munfl will be suppressed during this calculation.

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

This means that the solutions plotted are incorrect. How to eliminate these errors to get the exact solutions?

How to draw imaginary solutions also without errors?

Thank you.

solving equations – Problem simulating FindRoot

I am writing a procedure that simulates the findRoot built-in function using the same code that appears in Chapter 6 of $ textit {Math programming} $ by Paul Wellin. the code is the following.

findRoot[expr_, {var_, init_}, [Epsilon]_] :=
 Module[{xi = init, fun = Function[fvar, expr]},
  While[Abs[fun[xi]] > [Epsilon],
   xi = N[xi - fun[xi]/fun'[xi]]];
  {var -> xi}]

In the text, the code is successfully tested, but when I test it it doesn't work. For example,

In[5]:= findRoot[x^2 - 2, {x, 2.0}, 0.0001]

Out[5]= {x -> 2.}

I tried to change the variable fvar to var, and the program works, but this is a reuse of names that makes Mathica color the variable var red.

findRoot[expr_ == val_, {var_, init_}, [Epsilon]_] :=
 Module[{xi = init, fun = Function[var, expr - val]},
  While[Abs[fun[xi]] > [Epsilon],
   xi = N[xi - fun[xi]/fun'[xi]]];
  {var -> xi}]

I need some suggestion to clarify this problem.

differential equations: analytical resolution of nonlinear ODE

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eigenvalues: different results when solving a system of equations analytically and numerically

I have the following system of differential equations:

begin {pmatrix} dfrac {d eta} {dt} \ dfrac {d eta ^ { dagger}} {dt} end {pmatrix} = begin {pmatrix} (1 + I kappa) (w-IJ) & R_1 + i R_1 kappa \ – R_1 + IR_1 kappa & (IJ + w) (- 1 + i kappa) \ end {pmatrix} begin {pmatrix} eta \ eta ^ { dagger} end {pmatrix}

To solve this system, I diagonalize the intermediate matrix and write the solution in terms of the normal modes. In Mathematica:

s = {{(1 + I (Kappa)) (w - I J), R1 + I R1 (Kappa)}, {-R1 +  I R1 (Kappa), -(1 - I (Kappa)) (w + I J)}};
{vals, vecs} = FullSimplify(Eigensystem(s ));
NormVector = Transpose(Normalize /@ vecs);
NormalModes = {{C1*Exp(I*vals((1))*t)}, {C1*Exp(I*vals((2))*t)}};
Solution = NormVector.NormalModes /. {w -> 0.27, (Kappa) -> 0.01, R1 -> 0.0025,
J -> 0.005}

In this case
$ Solution = left ( begin {array} {c}
(-0.00462882-0.0000462882 i) text {C1} e ^ {(0.0023 , -0.270038 i) t} – (0.999939 , +0.00999939 i)
text {CC1} e ^ {(0.0023 , +0.270038 i) t} \
0.999989 text {C1} e ^ {(0.0023 , -0.270038 i) t} +0.00462905 text {CC1} e ^ {(0.0023 , +0.270038 i) t}
\
end {array}
right) $

On the other hand, I define the constants before doing the calculation, in this way,

(Kappa) = 0.01
R1 = 0.025;
J = 0.005;
w = 0.27;
  s = {{(1 + I (Kappa)) (w - I J), R1 + I R1 (Kappa)}, {-R1 +  I R1 (Kappa), -(1 - I (Kappa)) (w + I J)}};
{vals, vecs} = FullSimplify(Eigensystem(s ));
NormVector = Transpose(Normalize /@ vecs);
NormalModes = {{C1*Exp(I*vals((1))*t)}, {C1*Exp(I*vals((2))*t)}};
Solution2 = NormVector.NormalModes /. {w -> 0.27, (Kappa) -> 0.01, R1 -> 0.0025,
J -> 0.005}

In this case $ Solution2 = left ( begin {array} {c}
(0.998926 , +0. I) text {CC1} e ^ {(0.0023 , +0.26889 i) t} – (0.0463375 , +0.000463375 i) text {C1}
e ^ {(0.0023 , -0.26889 i) t} \
(0.998926 , +0. I) text {C1} e ^ {(0.0023 , -0.26889 i) t} + (- 0.0463375 + 0.000463375 i) text {CC1}
e ^ {(0.0023 , +0.26889 i) t} \
end {array}
right) $

That is the correct result, I don't understand why I get different results. In the first solution, the imaginary part of the second row of the matrix is ​​missing.

Thank you

inequality – Equations and inequalities with a parameter!

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  • Please make sure answer the question. Please provide details and share your research!

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  • Make statements based on opinion; back them up with references or personal experience.

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differential equations: how to solve the natural frequency of the cantilever beam by mathematical method

I want to calculate the natural frequency of the cantilever beam according to the theoretical method.

enter the image description here

I find a post on how to solve the free vibration frequency of a cantilever beam with the finite element method.

(*https://mathematica.stackexchange.com/questions/99724/finite-
element-boundary-breaking*)
ps = {Inactive(
      Div)({{0, -((Y*ν)/(1 - ν^2))}, {-(Y*(1 - ν))/(2*(1 
- ν^2)), 0}}.Inactive(Grad)(v(x, y), {x, y}), {x, y}) + 
    Inactive(
      Div)({{-(Y/(1 - ν^2)), 
        0}, {0, -(Y*(1 - ν))/(2*(1 - ν^2))}}.Inactive(Grad)(
       u(x, y), {x, y}), {x, y}), 
   Inactive(
      Div)({{0, -(Y*(1 - ν))/(2*(1 - ν^2))}, {-((Y*ν)/(1 
- ν^2)), 0}}.Inactive(Grad)(u(x, y), {x, y}), {x, y}) + 
    Inactive(
      Div)({{-(Y*(1 - ν))/(2*(1 - ν^2)), 
        0}, {0, -(Y/(1 - ν^2))}}.Inactive(Grad)(
       v(x, y), {x, y}), {x, y})} /. {Y -> 10^3, ν -> 33/100}

{vals, funs} = 
 NDEigensystem({ps}, {u, v}, {x, y} ∈ 
   Rectangle({0, 0}, {5, 0.25}), 8)
theory = {0, 0, 0, 22/L^2 Sqrt((Y d^2)/(12 1)), 
   61.7/L^2 Sqrt((Y d^2)/(12 1)), 121/L^2 Sqrt((Y d^2)/(12 1)), 
   200/L^2 Sqrt((Y d^2)/(12 1)), π/L Sqrt(Y/1.)} /. {Y -> 10^3, 
   d -> 0.25, L -> 5}
TableForm(Transpose({Sqrt(Abs(vals)), theory}), 
 TableHeadings -> {Automatic, {"Calculated", "Theory"}})
bcs = DirichletCondition({u(x, y) == 0, v(x, y) == 0}, x == 0);

{vals, funs} = 
  NDEigensystem({ps, bcs}, {u, v}, {x, y} ∈ 
    Rectangle({0, 0}, {5, 0.25}), 5);
theory = {3.52 Sqrt((Y d^2)/(12 L^4)), 22 Sqrt((Y d^2)/(12 L^4)), 
    61.7 Sqrt((Y d^2)/(12 L^4)), π/2 Sqrt(Y/L^2), 
    121 Sqrt((Y d^2)/(12 L^4))} /. {Y -> 10^3, d -> 0.25, L -> 5.};
TableForm(Transpose({Sqrt(Abs(vals)), theory}), 
 TableHeadings -> {Automatic, {"Calculated", "Theory"}})
Needs("NDSolve`FEM`")
mesh = funs((1, 1))("ElementMesh");
Column(Table(uif = funs((n, 1));
  vif = funs((n, 2));
  dmesh = 
   ElementMeshDeformation(mesh, {uif, vif}, "ScalingFactor" -> 0.1);
  Show({mesh("Wireframe"), 
    dmesh("Wireframe"(
      "ElementMeshDirective" -> 
       Directive(EdgeForm(Red), FaceForm())))}), {n, 5}))

But I would like to know how to solve the natural frequency of the model in the previous post according to the partial differential equation solving method.

enter the image description here

Other related links:

How do you find the eigenvalues ​​of a PDE (Euler-Bernoulli dynamic beam)?

Euler-Bernoulli non-homogeneous dynamic beam equation with discontinuous parameters

Analytical solution of the Euler-Bernoulli dynamic beam equation with compatibility condition