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RSS Feed for Wolfram Community showing any discussions in tag Equation Solving sorted by activeNo output in solving system of trigonometric equations
https://community.wolfram.com/groups/-/m/t/2243285
Hello everybody, I hope you are doing all well today.
I'm new to Mathematica, I have a system of equations that I want to solve using Mathematica: <br><br>
$\begin{equation}
\begin{cases}
sin(x)cos( \epsilon y)cosh(\epsilon z) + sin(\epsilon x)cos(y)cosh(z) - sin(x(1 +
\epsilon)) = 0 \\
sin(x)sin(\epsilon y)sinh(\epsilon z) + sin(\epsilon x)sin(y)sinh(z) = 0
\end{cases}
\end{equation}$ <br><br>
I'm only interested in {x, y, z} values in the domain [0, 2*Pi] <br>
$\epsilon$ is a parameter that can take any value between 1.8 to 3.
I tried solving the problem first by setting $\epsilon = 2$, I used methods like `Solve` and `Nsolve`, but it didn't work. it took too much time to execute without any result. after a long day I found out about `Reduce` function, i used it and it works it gave me the full answer. but that's only for the case of $\epsilon = 2$. When I change $\epsilon$ to 2.5 it didn't gave me any answer. I left it to run to about 10 hours or more without any result. <br>
This is my fifth day with this problem, and I'm still stuck.
can anyone provide me with any advices. I really appreciate your help. Thank you so much <br><br>
This is my code for the case of $\epsilon = 2$
Reduce[{
Sin[x]*Cos[2.5*y]*Cosh[2.5*z] + Sin[2.5*x]*Cos[y]*Cosh[z] -
Sin[3.5*x] == 0,
Sin[x]*Sin[2.5*y]*Sinh[2.5*z] + Sin[2.5*x]*Sin[y]*Sinh[z] == 0,
0 <= x <= 2*Pi, 0 <= y <= Pi, 0<= z <= 2*Pi}, {x, w, y}, Reals]Mohamed El Ghafiani2021-04-13T17:04:13ZSolving an equation with complex parameters in Wolfram|Alpha
https://community.wolfram.com/groups/-/m/t/2244163
If one has an equation of the form: (y + i /2) g(u, w) = 0, where i denotes √(-1), where y is real while u & v are complex, and where g(v, w) is an algebraic function of v and w; then: is the correct solution for y of this equation: 'no solution for y' {as is given by Wolfram Alpha Pro for a special case of the algebraic function denoted by g(u, w)} ... or else is the correct solution: 'y = any real number'? This strange problem arose during my recent correspondence with a Prof. of Mathematics at my alma mater of UC Berkeley, who specializes in algebra & number theory & who argued that one could divide both sides of this equation by (y + i / 2) since y is defined to be real and so (y + i / 2) cannot vanish; and then he argued further that therefore one obtains (after this division) the algebraic equation: g(v, w) = 0 which is, of course, trivially satisfied for all real y since y does not appear in it, and so this equation places no constraints on (assumed real) y at all ... whatsoever. However, this seems to me to be very poor methodology, does it to you as well? (And please give me a good supporting reference or two to your answer to this question.)Dr. Dennis P Allen Jr.2021-04-14T19:02:50ZPreserve arguments' patterns while taking a Derivative?
https://community.wolfram.com/groups/-/m/t/2242146
Hello!
I've encountered a problem with loss of pattern information after using "function-producing" `Derivative` function.
Assume I have a scalar-valued function defined on vectors:
H[r_?VectorQ] := r.r
(evidently, `r` is assumed to be a three-dimentional vector, but id does not matter)
And then I would like to solve a differential equation like *dr / dt == grad(H)*. Note that I would like to keep the vectorial notation i.e. the solution must be a vector-valued function. I've tried the following:
NDSolveValue[{D[r[t], t] == Table[Derivative[xspec][H][r[t]], {xspec, IdentityMatrix[3]}],
r[0] == {1, 1, 1}}, r, {t, 0, 100}]
This returns an interpolated function, but it does not handle the derivatives of `H` properly. For example, if in the output of `Derivative` appeared term like `#1[[1]]` then it accepts `r[t]` as argument and evaluates simply to `t` and if there's `#1[[2]]` it throws a message that this part does not exist. It seems like `Derivative` loses argument check and the returned function does not formally require vectorial input anymore.
So, the question is: can I somehow tell WL that this derivative is still a function of a vector?Nikolay Shilov2021-04-12T12:32:31ZKernel quits while using DSolve[ ]?
https://community.wolfram.com/groups/-/m/t/2241599
Why does my DSolve command does not solve the Non-linear ODEs?
Clear["Global`*"]
(*Parameters*)
length = 1.5;
width = 0.25;
height = 0.25;
area = width*height;
cforce = -400;
dforce = (500*(X1^2));
stress = 10*displacement'[X1] + 100000*displacement'[X1]^3;
DE1 = D[stress*area, X1] + dforce;
DE2 = 10*displacement'[X1] + 100000*displacement'[X1]^3 + 6400;
(* Solving for the Exact Displacement Function *)
solution =
DSolve[{DE1 == 0, displacement[0] == 0, DE2 == 0 /. X1 -> length},
displacement[X1], X1];
displacement = displacement[X1] /. solution[[1]];
Print["The Exact Displacement Function of the Bar is"]
Print[displacement];
Print[Plot[{displacement}, {X1, 0, 1.5}, AspectRatio -> 0.4,
AxesLabel -> {"Bar Length (m)", "Axial Displacement (m)"},
PlotLegends -> {"Exact Displacement of the Bar"}]];
Print[" "]Aslam Kittur2021-04-10T18:10:30ZNo output from Solve[ ]?
https://community.wolfram.com/groups/-/m/t/2241678
Could someone please look at this and tell me what's wrong? It's probably a simple mistake. I think this system of equations should return {{pd -> -3}, {pd -> 10}}. At least, that's what I get when I use a pencil and paper. But my input below doesn't return anything. Not even an error message.
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/9a795edc-6893-4a6d-9925-c543fd8bbafcJay Gourley2021-04-12T05:45:56ZPlot 2 parameter solution from ParametricNDSolve with Table
https://community.wolfram.com/groups/-/m/t/2242051
I have a 2-D differential equation which I solve with ```ParametricNDSolve``` with initial time given as a parameters. I solve the equation using
twoeqnpara = ParametricNDSolve[{x'[t]==-3x[t] - y[t],y'[t] == x[t],x[0]==a,
y[0]==b},{x,y},{t,0,100},{a,b}].
When I try to plot with
pp = ParametricPlot[Evaluate@Table[{x[a,b][t],y[a,b][t]}/.twoeqnpara,
{a,-2,2,0.5},{b,-2,2,0.5}],{t,0,40},PlotRange->All]
I don't get any plot. But when I do for single parameter
pp = ParametricPlot[Evaluate@Table[{x[a,a][t],y[a,a][t]}/.twoeqnpara,{a,-2,2,0.4}],
{t,0,40},PlotRange->All,PlotLegends->Range[-2,2,0.4]]
It works and give a plot.
What I think the issue is when I use the command
Table[ x+y, {x,-2,2,0.5},{y,-2,2,0.5}]
, what the command is taking a single value of x and all the values of y. What I need is all the pairs possible from the 2 arrays so that can be used as a initial condition for plotting.Sahil Goyal2021-04-11T17:01:59ZDistance between a point and others is an integer
https://community.wolfram.com/groups/-/m/t/2241420
I have 4 points in a 2D plane
{{0, 0}, {20, 0}, {138/5, (24 Sqrt[6])/5}, {25, 10 Sqrt[6]}}
and wish to find a 5th point such that the distances to the 4 points is an integer.
I have tried solving for a point to the point {138/5, (24 Sqrt[6])/5}, specifying Integer elements and reals and cannot get any specific point as an answer, it's all conditionally based. Satisfying 4 points simultaneously seems unobtainable. My question is, is this solvable using Mathematica's built in functions? Or do I resort to a programmable approach?
Btw the pairwise distances between those 4 points are Integer.Paul Cleary2021-04-10T15:50:20ZEmpty plot using NDSolve[ ] ?
https://community.wolfram.com/groups/-/m/t/2239857
Hey guys,
I'm quite new here and hope i'm in the right section. In trying to solve a PDE system with NDSolve the plots I obtain in the end are empty. I read through quite a few posts regarding this issue but haven't found a solution that I could apply to my code.
The PDE system:
deqs = {D[m[x, t], t] == D[vm[x]/am[x] D[vm[x] m[x, t], x], x] + f[{m[x, t], c[x, t]}],
D[c[x, t], t] == D[vc[x]/ac[x]*D[vc[x]*c[x, t], x], x] - f[{m[x, t], c[x, t]}],
m[x, 0] == M, c[x, 0] == C0, m[0, t] == m[L, t] , c[0, t] == c[L, t]};
Most important code:
sol1 = NDSolve[deqs, {m, c}, {x, 0, L}, {t, 0, 10}]
Plot3D[m[x, t] /. sol1, {x, 0, L}, {t, 0, 10}]
I assume my mistake already is somewhere in these lines, f is a polynome, vc and vm are constants and ac and am are arctan(x) like functions.
To be precise:
f[{a_, b_}] = g*a^2*b - d*a
am[x_]:= (A(ArcTan[B(x-L/2)]+ Pi/2))^(-1);
ac[x_]:=(A (ArcTan[B(-x+L/2)]+Pi/2))^(-1);
vm[x_] := vM;
vc[x_] := vC;
Does anyone have an Idea why this doesn't work?
Thanks a lot in advance!Janis Köhler2021-04-08T13:39:37Z