MaplePrimes Questions

Is it possible to solve DDE with dsolve?

restart:

Eq1 := diff(x(t), t) = 1-.1*x(t)-0.5e-3*x(t)*v(t)/(1+0.1e-5*v(t));

Eq2 := diff(y(t), t) = 0.5e-3*x(t-10)*v(t-10)/(1+0.1e-5*v(t-10))-.3*y(t)-y(t)*z(t);

Eq3 := diff(v(t), t) = 200*y(t-10)-8*v(t);

Eq4 := diff(z(t), t) = 2*y(t)*z(t)-.15*z(t);

ics := x(0) = 1, y(0) = 1, v(0) = 5, z(0) = 1;

Thanks

I am trying to solve the Morrison equation for a normal force acting on a cylinder in a viscous fluid by using the potentional theory. With my limited experience in Maple I do not get why my dubble integral cannot be computed. Any help would be appreciated, tips are also very welkom as I am trying to expand my knowledge of Maple. 

Hi all,

I have lots of contstraint equations group and I want to fund a group of parameters which can fit them. 

For example, these are a simple constraint eqqations group:
eqs:{x1>0, x2>0 x1<1000, x2<1000, x1+x2>300,x1+x2<700}

Through SolveTools library, I can determine whether there is a group of parameters.


with(SolveTools[Inequality]);

LinearMultivariateSystem({x1 > 0, x1+x2 > 300, x2 < 1000, x1+x2 < 700, x2 > 0*x1 and 0*x1 < 1000}, [x1, x2]);

{[{x1 <= 300, 0 < x1}, {x2 < -x1 + 700, 300 - x1 < x2}],[{300 < x1, x1 < 700}, {0 < x2, x2 < -x1 + 700}]}

 

Then, if I want to find a group of parameter a group of parameters (ex, x1=300, x2=200 in this case), how should I do?

I've got the following four differential equations :

v_x:=diff(x(t),t);
v_y:=diff(y(t),t);
d2v_x:=-((C_d)*rho*Pi*(r^2)*(v_x)*sqrt((v_x)^2 +(v_y)^2))/(2*m);
d2v_y:=-((C_d)*rho*Pi*(r^2)*(v_y)*sqrt((v_x)^2 +(v_y)^2))/(2*m)-g;


and the following initial value conditions:

x(0)=0,y(0)=0,v_x(0)=v0/sqrt(2),v_y(0)=v0/sqrt(2) given v0=65 

I need to solve these using the numeric type and then draw overlaid plots

(i) setting C_d=0

(ii) leaving C_d as a variable

before plotting y(t) vs x(t). The hint for this last part is that the path can be seeing using [x(t),y(t)] instead of [t,y(t)]

I've tried to do it but seemed to have several syntax errors.

 

 

I've got the following diff.eq

y'(x)=sin(x*y(x)) given y(0)=1 

and need to solve it numerically which is why I've used:

dy4:=diff(y(x),x);
eqn4:=dy4=sin(x*y(x));
ic1:=y(0)=1;
ans3:=dsolve({eqn4,ic1},y(x),type=numeric);

This code doesn't return a value though and in fact, ans3 is being displayed as a procedure

"ans3:=proc(x_rkf45) ... end proc"

I don't quite understand why and what I need to do to get the required numerical solution

 

Hello

Could you plese help me to plot my function. i want to plot a function in which variables are have able formatting. I have attache my file.

Thank you.


restart

L[b] := 400:

L[c] := 400:

q := 5:

M[ab] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), theta[b]):``

M[ba] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), VectorCalculus:-`*`(2, theta[b])):

M[bc] := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[b]), 1/L[b]), VectorCalculus:-`+`(VectorCalculus:-`*`(2, theta[b]), theta[c])), VectorCalculus:-`*`(VectorCalculus:-`*`(q, L[b]^2), 1/12)):

M[cb] := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[b]), 1/L[b]), VectorCalculus:-`+`(theta[b], VectorCalculus:-`*`(2, theta[c]))), VectorCalculus:-`-`(VectorCalculus:-`*`(VectorCalculus:-`*`(q, L[b]^2), 1/12))):

M[cd] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), VectorCalculus:-`*`(2, theta[c])):

M[dc] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), theta[c]):

eq1 := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`+`(M[ab], M[ba]), 1/L[c]), VectorCalculus:-`*`(VectorCalculus:-`+`(M[cd], M[dc]), 1/L[c])) = 0:

eq2 := VectorCalculus:-`+`(M[ba], M[bc]) = 0:

solve([eq1, eq2], {theta[b], theta[c]})

{theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}

(1)

theta[b] := rhs({theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}[1]):

theta[c] := rhs({theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}[2]):

subs(theta[b] = theta[b], M[ba])

-(400000/3)*I[c]/(I[b]+2*I[c])

(2)

subs(theta[b] = theta[b], M[bc]):

``

EXplore(plot(M[ba], I[b] = 1 .. 10), parameters = [I[c] = 1 .. 20])

Error, (in plot) unexpected option: I[b] = 1 .. 10

 

``


Download shiboft2.mw

Matlab seems to be pretty strong at doing color plots with separate color bars, e.g.

Is this also possible in Maple and somehow in combination with `plots[surfdata](...,color=zhue,...)`?

In a 3d coordinate system I have a circular spacecurve with z-minimum -4 and z-maximum +4. In the same 3d coordinate system I have a 3d surface plot with z-minimum -0.5 and z-maximum +1.3 . When I choose the color option "Z(Hue)" in order to color-code the z-values on the 3d surface and make the topography more clear, I mostly get a totally green 3d surface. It seems that the color scaling is coupled with the spacecurve with z-values of +-4 . How can I uncouple the color scaling from the spacecurve and couple it with the z-range of the surface, while the color-limits shall be at +-1.3 ?

I have a nice family of functions of the form:

W:=(p,n,mu,w)->sum(w[k] * (n-k)* mu(n-k),k=1..n)

which can be evaluated for different p's using the operator mu*diff(...,mu)

The recursion begins with p=0 and proceeds using mu*diff(W(p,n,mu,w),mu) = W(p+1,n,mu,w).

Can anybody implement this procedure in Maple

Thank you 

How can I make transformations like:

sum(a*u[k]+b*v[k],k=1..n) into a*sum(u[k],k=1..n)+b*sum(v[k],k=1..n)

and in the opposite direction?

Dear,

I paste below a simple code illustrating what I want to do: interpolate a function from two lists . I wonder it is a very simple task but I the function and the graph I get don't match with the correct ones in the end (both are plotted below).

> y_data:=[0.5266426348e-1, 0.7090942099e-1, 0.9392192453e-1, .1228458380, .1598545751, .2107200296, .3127241088, .4144428927, .5679723063, .6817484378, .7973388763, .9326799232, 1.393259472, 1.493936979, 1.566845149, 1.624353545, 1.670898228, 1.708874880, 1.739919717, 1.765298377, 1.786042765, 1.803007037, 1.816897481, 1.828294327, 1.837670808, 1.845410531, 1.851823072, 1.857157571, 1.861614283, 1.865354162, 1.868506707, 1.871176289, 1.873447229, 1.875387839, 1.877053637, 1.878489894, 1.879733654, 1.880815343, 1.881760039, 1.882588494, 1.883317935, 1.883962726, 1.884534877, 1.885044474, 1.885500010, 1.885908668, 1.886276537, 1.886608800, 1.886909882, 1.887183573, 1.887433127, 1.887661351, 1.887870669, 1.888063180, 1.888240711, 1.888404854, 1.888557000, 1.888698365, 1.888830022, 1.888952911, 1.889067865];

> x_data:=[10, 53/5, 56/5, 59/5, 62/5, 13, 68/5, 68/5, 13, 62/5, 59/5, 56/5, 56/5, 59/5, 62/5, 13, 68/5, 71/5, 74/5, 77/5, 16, 83/5, 86/5, 89/5, 92/5, 19, 98/5, 101/5, 104/5, 107/5, 22, 113/5, 116/5, 119/5, 122/5, 25, 128/5, 131/5, 134/5, 137/5, 28, 143/5, 146/5, 149/5, 152/5, 31, 158/5, 161/5, 164/5, 167/5, 34, 173/5, 176/5, 179/5, 182/5, 37, 188/5, 191/5, 194/5, 197/5, 40];

inverted_pairs:=[seq([y_[i], x_[i]], i = 1 .. nops(y_))]; 

interp := LeastSquares(y_data, x_data, k_, curve = a*k_^3+b*k_^2+c*k_+d);

 

I've tried different kinds of interpolation methods. In this code I use LeastSquares() function. I'm sure it has a quick solution but I'm really stuck and would appreciate any help/advice.  

Thanks in advance!

 

 

PS: I didn't mention in above, but it is really important for me to get the associated function of the curve (e.g. ak_^3+bk_^2+ck_+d). I can get it using LeastSquares() function but with your code (where you use ArrayInterpolation()) the function is not generated in the end. Could you please indicate some way to do that? Thanks!

I need to find the local maxima and minima of f(x,y)=x(x+y)*e^(y-x). I have tried to look for an appropriate method that I could use to achieve this, but got stuck. I also don't quite understand the math behind tying to obtain the local maxima and  minima for a function of this type.

I'm trying to evaluate the multidimensional limit: 

(1+y)^(x-1)-1/(1-cos((x-1)^2+y^2)^(1/4)) as (x,y)->(1,0) using the limit command :

limit((1+y)^(x-1)/(1-cos((x-1)^2+y^2)^(1/4)),{x=1,y=0});

 

but don't seem to get any output. Also, I think the limit for this function doesn't exist or is indeterminate on R2. Where am I wrong?

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