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These are questions asked by Jamie128

I want to write an if statement with a matrix R:
such that maple executes command 1 if R is Matrix(4, 4, [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]])

and command 2 if R is not zero.

What is the easiest way to go about this?

I was wondering if there was a command that could tell me if an expression consisting of undermined functions was always positive (with real domain).

For example f1:=f(x)^2+g(y)^4 is always positive.

but f(x)^2-g(y)^2 is only positive on a restricted domain.

I have a lists within lists.  I needed to search for a specific element in the sub-list and need to extract each such sublist that has an element.

I am only familiar with the search tool. Could someone suggest a simple way to do this.

For example, I have the following.

s := [{f0(t, r) = 0, n(t) = n(t)}, {f0(t, r) = 0, n(t) = 0}, {f0(t, r) = 0, n(t) = 0}]

I need to extract the sublist that has n(t)=n(t) so only the first element of the list s.

I want to plot this function in cylindrical coordinates in a way that it is clear where the function is greater than zero and where it is less that zero. I was wondering what would be the best way to do it. It has two arbritary constants. Usually when I have only one constant, I animate the plot for different values of that constant. But I am not sure what to do for this case.

-2*M^2*sin(theta)^2 - 2*k*r^2

I wanted to solve a pde of two variables in spherical coordinates with a boundary condition at a specific radius. I am doing something wrong because the pdsolve just evaluates for several minutes and does not actually give anything.

Could someone have a look and let me know what I am doing wrong.




Setup(mathematicalnotation = true)

[mathematicalnotation = true]


eq3 := Laplacian(Phi_out(r, theta)) = 0

((r^2*(diff(diff(Phi_out(r, theta), r), r))+2*r*(diff(Phi_out(r, theta), r))+diff(diff(Phi_out(r, theta), theta), theta))*sin(theta)+(diff(Phi_out(r, theta), theta))*cos(theta))/(r^2*sin(theta)) = 0


bc1 := eval(Phi_out(r, theta)-Omega*mu*(-1+cos(2*theta))/(2*r), r = R) = 0

Phi_out(R, theta)-(1/2)*Omega*mu*(-1+cos(2*theta))/R = 0



pdsolve([eq3, bc1])


Download laplacian.mw

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