Maple 2024 Questions and Posts

These are Posts and Questions associated with the product, Maple 2024

I recently prepared a worksheet to teach vector fundamentals in one of my classes, and I wanted to share it with you all. It's nothing special, but I found Maple really helpful in demonstrating the concepts visually. Below is a breakdown of what the worksheet covers, with some Maple code examples included.

Feel free to take a look and use it if you find it useful! Any feedback or suggestions on how to improve it would be appreciated.

restart

NULL

v := `<|>`(`<,>`(2, 3)); w := `<|>`(`<,>`(4, 1))

Matrix(%id = 36893488152076804092)

(1)

Basic Vector Operations

• 

Addition and Subtraction

 

We  can add and subtract vectors easily if they are of the same dimension.

NULL

u_add := v+w; u_sub := v-w

Matrix(%id = 36893488152076803596)

(2)

NULL

NULL

Typesetting[delayDotProduct]((((Triangle(L)*a*w*o*f*A*d*d)*i*t*i)*o*n*o)*f, Vector(s), true)

"The famous triangle law can be used for the addition of vectors and this method is also called the head-to-tail method,As per this law,two vectors can be added together by placing them together in such away that the first vector's head joins the tail of the second vector. Thus,by joining the first vector's tail to the head of the second vector,we can obtain the resultant vector sum."

NULL

with(plots); display(arrow([0, 0], [2, 3], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([2, 3], [4, 1], color = green, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [6, 4], difference = true, color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), scaling = constrained, labels = [x, y], title = "
Triangle Law of Addition of Vectors", titlefont = [times, 20, bold])

 

NULL

NULL

Parallelogram Law of Addition of Vectors

"An other law that can be used for the addition of vectors is the parallelogram law of the addition of vectors*Let's take two vectors v and u,as shown below*They form the two adjacent sides of aparallelogram in their magnitude and direction*The sum v+u is represented in magnitude and direction by the diagonal of the parallelogram through thei rcommon point."

NULL

display(arrow([0, 0], [2, 3], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [6, 4], color = green, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [4, 1], difference = true, color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), scaling = constrained, labels = [x, y], title = "

Parallelogram Law of Addition of Vectors", titlefont = [times, 20, bold])

 

NULL

NULL

NULL

NULL

NULL

NULL

NULL

• 

Scalar Multiplication

We can multiply a vector by a scalar. To multiply a vector by a scalar (a constant), multiply each of its components by the constant.

Suppose we let the letter  λ represent a real number and  u = (x,y) be the vector

v_scaled := 3*v

Matrix(%id = 36893488152152005076)

(3)

NULL

NULL

• 

-Define*the*opposite*vector*v

Error, (in LinearAlgebra:-Multiply) invalid arguments

 

Two vectors are opposite if they have the same magnitude but opposite direction.

v_opposite := -v; vec1 := arrow([0, 0], [2, 3], color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1); vec2 := arrow([0, 0], [-2, -3], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1); display([vec1, vec2], scaling = constrained, title = "Original Vector and its Opposite (-v)")

 
• 

Dot Product

Sometimes the dot product is called the scalar product. The dot product is also an example of an inner product and so on occasion you may hear it called an inner product.

Given the two vectors  `#mover(mi("a"),mo("&rarr;"))` = (x__1, y__1) and `#mover(mi("b"),mo("&rarr;"))` = (x__2, y__2)
 the dot product is, `#mover(mi("a"),mo("&rarr;"))`.`#mover(mi("b"),mo("&rarr;"))` = x__1*x__2+y__1*y__2

`#mover(mi("a"),mo("&rarr;"))`.`#mover(mi("b"),mo("&rarr;"))` = x__1*x__2+y__1*y__2

(4)

`#mover(mi("a"),mo("&rarr;"))` := `<|>`(`<,>`(5, -8))

Matrix(%id = 36893488152255219820)

(5)

`#mover(mi("b"),mo("&rarr;"))` := `<|>`(`<,>`(1, 2))

Matrix(%id = 36893488152255215244)

(6)


display(arrow([0, 0], [5, -8], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [1, 2], difference = true, color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), scaling = constrained, labels = [x, y])

 

 

 

 

 

 

 

 

 

 

 

  with(LinearAlgebra)

dot_product := DotProduct(`#mover(mi("a"),mo("&rarr;"))`, `#mover(mi("b"),mo("&rarr;"))`)

-11

(7)
• 

Vector Norm (Magnitude)

To find the magnitude (or length) of a vector, use Norm.

norm_a := Norm(`#mover(mi("a"),mo("&rarr;"))`, 2); norm_b := Norm(`#mover(mi("b"),mo("&rarr;"))`, 2)

5^(1/2)

(8)
• 

Calculate the Cosine Between Two Vectors

cos_theta := dot_product/(norm_a*norm_b)

-(11/445)*89^(1/2)*5^(1/2)

(9)

angle_radians := arccos(cos_theta)

Pi-arccos((11/445)*89^(1/2)*5^(1/2))

(10)

angle_degrees := evalf(convert(angle_radians, degrees))

121.4295656*degrees

(11)

NULL

We can determine whether two vectors are parallel by using the scalar multiple method or the determinant (area of the parallelogram formed by the vectors) method.

``

• 

Scalar Multiple Method

a := Vector([5, -8]); b := Vector([10, -16]); k := a[1]/b[1]; is_parallel := a[2]/b[2] = k

1/2 = 1/2

(12)
• 

Determinant Method

determinant := a[1]*b[2]-a[2]*b[1]; result := determinant = 0

0 = 0

(13)

 

 

display(arrow([0, 0], [5, -8], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [10, -16], difference = true, color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), scaling = constrained, labels = [x, y], title = "Parallel Vectors")

 

 

 

 

Two vectors a and b are perpendicular (vertical)

NULL

• 

if and only if their dot product is zero

a1 := Vector([1, 2]); b1 := Vector([-2, 1]); dot_product := DotProduct(a1, b1); is_perpendicular := dot_product = 0

0 = 0

(14)

display(arrow([0, 0], [-2, 1], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), arrow([0, 0], [1, 2], difference = true, color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1), scaling = constrained, labels = [x, y], title = "Perpendicular Vectors")

 
• 

Slope Method

slope_a := a1[2]/a1[1]; slope_b := b1[2]/b1[1]; is_perpendicular := slope_a*slope_b = -1

-1 = -1

(15)

NULL

• 

Special case: If one vector is vertical (undefined slope) and the other is horizontal (zero slope), they are perpendicular.

a2 := Vector([0, 3]); b2 := Vector([4, 0]); is_perpendicular := a2[1] = 0 and b2[2] = 0

true

(16)

vec1 := arrow([0, 0], [4, 0], color = blue, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1); vec2 := arrow([0, 0], [0, 3], color = red, shape = double_arrow, width = 0.1e-1, border = false, head_width = .1, head_length = .1); display([vec1, vec2], scaling = constrained, title = "Perpendicular Vectors")

 

NULL

Download vectors.mw

The tab key, and the mouse can be used to trigger completion selection in Maple 2024.1 like previous versions. 

For interface resposiveness, a new option (disabled by default) has been added to the interface tab of Maple 2024.1 Options (available from the Tools menu). Users that prefer to use enter to trigger completion selection can enable the option. 

Change the option here and apply to the session or globally if using enter to trigger completion selection is preferred.

I have angled some test to be parallel to the slope of a line. It tuns out that the text is only parallel if scaling = constrained is used. That is not necasserily pratical for many plots. In prcttice I have a nice proedure for this. Is there any way to pickup the plot x-y scaling factor that maple decides on so I could built that into the procedure. I have manually rescaled her to demonstrate.

mmm...I see the website viewer does not rotate the text.

restart

with(plots):

P1:=[1,2]:P2:=[5/2,3]:

vec:=P2-P1

[3/2, 1]

(1)

ang:=arctan(vec[2],vec[1])

arctan(2/3)

(2)

plt1:=textplot([((P1+P2+[.2,.2])/2)[],"parallel piece of text",rotation=ang]):

plt2:=plottools:-line(P1,P2):

display(plt1,plt2)

 

display(plt1,plt2,scaling=constrained)

 

#scaling factor looks to be 1.5 i.e.delta(x)/delta(y) (2.5-1)/{3/2)

ang1:=arctan(1.5*vec[2],vec[1])

.7853981634

(3)

plt3:=textplot([((P1+P2+[.2,.2])/2)[],"parallel piece of text",rotation=ang1]):

display(plt3,plt2)

 
 

 

Download 2024-10-15_text_not_always_parallel.mw

restart;

{(-x-y+1)*t+n*y^2 = 0, -t*(x+y-1)*k+x = 0, (x+y-1)*(p-s)*k-p*(x-1) = 0, m*y^2+(n*x-m-t)*y-t*(x-1) = 0, n*x^2+(m*y-n-s)*x-s*(y-1) = 0, y*(n*x-n-s+1)+(1-x)*s+p*x-1 = 0, (x+y-1)*(p-s)*k+(-x-y+1)*p-m*x+s*y = 0, ((-x-y+1)*k+y)*t-m*y+m+x-1 = 0}

{0 < k, 0 < m, 0 < n, 0 < p, 0 < x, 0 < y, 0 < (-m*x+p)*t+(s-p)*(m*y-m+1), 0 < (n*x-n-p+1)*t+n*y*(p-s), 0 < (m*x-n*x+n-1)*t+(m*y-n*y-m+1)*s+(n*x-n+1)*(m*y-m+1)-n*y*m*x, 1 < x+y, k < 1, m < 1, n < 1, p < 1}

"solve(equalities union inequalities)", {k = .487116703, m = .656557610, n = .3562602382, p = .1863581607, s = .372716320, t = .8665379799, x = .4642487806, y = 1.635592831}

"------------------------------------------------------------------------------------------------------------------------------"

"solve(equalities)", {k = k, m = m, n = 0., p = 0., s = 0., t = 1., x = 0., y = 1.}, {k = 1., m = 0., n = 1., p = 0., s = 0., t = 1., x = 1., y = 1.}, {k = 2.130395435, m = 1.469396425, n = -1.299895929, p = 2.299895929, s = .8304995051, t = .1695004949, x = -.3611030805, y = .3611030805}, {k = .893308015, m = -1.942064138, n = .8632249760, p = -1.307713373, s = -2.615426746, t = 1.427148086, x = .5879938248, y = .8732201900}

"solve(solve(equalities) union inequalities"

"solve(solve(equalities) union inequalities"

"solve(solve(equalities) union inequalities"

"solve(solve(equalities) union inequalities"

 

NULL

Download problems_with_solve_15.10.24.mw

Hey guys, 

I'm working with Maple to solve sets of 8 equations and 14 inequalities. I use the command solve to get values for my 8 variables. Often there is no solution, sometimes we have one single solution or like a parametric solution. However sometimes this procedure fails to finish the calculation, that means I stop the calculation after a certain time (multiple houres). So now I am looking for some other ways to get the solution. For example I try a two-stage solve-attemp. In the first round I use the solve coammand only for the set of 8 equations. In the second round I take each of the solutions I found in the step before and combine them with the set of inequalities. Then I use solve again for this set. 

However I found out that the two ways I described above lead to diffrent solutions. In my opinion the solutions of solve(equations union inequalities) should be a subset of solve(equations), since all the solutions we find with equations union inequalities has to fullfill the 8 equations we want put into solve(equations). As you can see in the attached file that is not the case. The sets of solutions of solve(equalities union inequalites) and the set of solutions of solve(equalities) is disjoint, so no subset. 
Since I have disjoint sets in the step between, the solutions of solve(equations union inequalities) and the solutions of solve((solve equations) union inequalities) are disjoint as well, so I think the problem is already in the step before. 

I would be really glad if anyone can help me. Either explain me, where my argumentation above went wrong or why I find those solutions which dont fit together. 

Regards
Felix

I have currently a Maple session running with only one open worksheet.

The worksheet only contains an input line with the name "a" and an output line with the name "a".

The fan of my laptop is running full throttle and the task manager displays 10% CPU usage and a very high power usage (3 mserver.exe running, the process with the cpu load is javaw.exe. Suspending this process shuts down the fan).

After about an hour I decided to ask. While typing this post (in Firefox, probably unrelated) the CPU usage went down and the fan went quite. The total system CPU usage is now down to 2%.

Has anybody seen the same? What could have cause it? It's not the first time I observe this. Anything that I could check before restarting Maple? All on Windows 10 after system restart.

Update:

The thread that consumes cpu is called ucrtbase.dll!configthreadlocate

Starting Maplesim in parallel almost immediately turns off the fan while the displayed cpu load is still high but now variing.

I was not expecting to see step-by-step for this ode, but why does it give internal error instead of saying not supported? 

Am I doing something wrong in the call? Unfortunately, this Maple internal error can't be trapped at user level. Which means the whole program terminates also.

interface(version);

`Standard Worksheet Interface, Maple 2024.1, Windows 10, June 25 2024 Build ID 1835466`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1820 and is the same as the version installed in this computer, created 2024, September 28, 18:14 hours Pacific Time.`

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

ode:=3*diff(y(x),x$2)+x*diff(y(x),x)-4*y(x)=0;

3*(diff(diff(y(x), x), x))+x*(diff(y(x), x))-4*y(x) = 0

Student:-ODEs:-ODESteps(ode,y(x))

Error, (in anonymous procedure called from Student:-ODEs:-ODESteps) too many levels of recursion

try
   Student:-ODEs:-ODESteps(ode,y(x));
catch:
  print("cought the error ");
end try;

Error, (in anonymous procedure called from Student:-ODEs:-ODESteps) too many levels of recursion

 


 

Download too_many_Levels_odesteps_oct_14_2024.mw

 

I have a procedure that I am trying to run that would be an improvement/more sophisticated way of solving a problem that I have previously solved. When I try and run my procedure I am getting an error, and from what I gather with the error is that there are some values when inserted into my procedure that cannot be evaluated. Just for context it is a procedure that contains numerical solutions to a system of DEs and and contains inequalities. 

I would like to know is there an easy method to figure out what values are giving me this error? 

Or a follow up, is there something wrong with my procedure that is giving me this error? I have included some commentary in my workshet as well to hopefully make everything clear. 

Thanks. 

Proc_Error.mw

I never really understood Intat. Help says 

"The intat command expresses an integral evaluated at a point; it is analogous to using the D command to express a derivative evaluated at a point."

But slope at a point is clear what it is and one can visualize it.

I do not understand what integral at single point means.

If one thinks of integration as area under the curve of the function, so what does area at single point mean? Should not integration (definite) always have lower and upper limits?

But my main question is not the above. I am sure there is a valid reason for Intat, otherwise it will not be in Maple.

But my quesiton is, in the context of solution to ode, can one replace result given using Intat by Int such that the lower limit starts from zero, and using same upper limit?

ie change Intat(...., a_ = something)  by Int( ... , a_ = 0 ... something) without changing the semantics or the correctness of the solution?

Because in  Intat, the lower limit is empty, and this always bothered me. At school the teacher says definite integration should have both lower and upper limits.

I tried it few places, and odetest verifies the solution of ode when using Intat or when using Int with lower limit zero:

ode:=diff(y(x), x) = B + C*f(a*x + b*y(x));

diff(y(x), x) = B+C*f(a*x+b*y(x))

sol_1:=Intat(1/(C*f(_a*b)*b+b*B+a),_a = (a*x+b*y(x))/b)*b-x+_C1 = 0;
sol_2:=Int(1/(C*f(_a*b)*b+b*B+a),_a = 0..(a*x+b*y(x))/b)*b-x+_C1 = 0;

Intat(1/(C*f(_a*b)*b+b*B+a), _a = (a*x+b*y(x))/b)*b-x+_C1 = 0

(Int(1/(C*f(_a*b)*b+b*B+a), _a = 0 .. (a*x+b*y(x))/b))*b-x+_C1 = 0

odetest(sol_1,ode);

0

odetest(sol_2,ode);

0

 

 

Download intat_vs_int.mw

Is there a case you know, where solution of ode which has Intat, when replaced by Int with lower limit 0, will no longer verifies the ode?  I am not able to find one so far. But may be there is.
 

Update

fyi, I found case where it makes difference. Not for odetest, but when using value. When using Int(...,tau=0..upper) vs   Intat(....,tau=upper)

value was able to find the value when using Intat, but not when using Int (for this example). 

So I think I will just stick to Intat even though both verified the ode as valid solution as it is better to be able to find value for integral if possible. I knew there must be good reason why Intat was invented.

 

ode:=diff(y(x),x)= sin(x-y(x));

diff(y(x), x) = sin(x-y(x))

sol_1:=Int(1/(1 - sin(tau)), tau = 0..x - y(x)) = x + _C1;
sol_2:=Intat(1/(1 - sin(tau)), tau = x - y(x)) = x + _C1;

Int(1/(1-sin(tau)), tau = 0 .. x-y(x)) = x+_C1

Intat(1/(1-sin(tau)), tau = x-y(x)) = x+_C1

odetest(sol_1,ode);

0

odetest(sol_2,ode);

0

value(sol_1);

int(1/(1-sin(tau)), tau = 0 .. x-y(x)) = x+c__1

value(sol_2);

-2/(tan((1/2)*x-(1/2)*y(x))-1) = x+c__1

 


 

Download int_vs_intat_v2.mw

 

I have expression which can have many different functions  in it. Assume the expression is `+` type.

I do not know before what these function names are. 

Is there a way to make Maple collect on these functions automatically? Simplify does not do it. So if there is something like b*g(x)+g(x) in the expression, to simplify this to (1+b)*g(x) automatically?

I noticed if expression is   just b*g(x)+g(x) then simplify does give (1+b)*g(x) but once I add a new term, like 3+b*g(x)+g(x) then it no longer does it! which for me is very strange. I was expecting to get 3+(1+b)*g(x) 

Here is an example

restart;

T:=g(x)+b*g(x);
simplify(T);

g(x)+b*g(x)

g(x)*(b+1)

T:=3+g(x)+b*g(x);
simplify(T);

3+g(x)+b*g(x)

3+g(x)+b*g(x)

T:=3+g(x)+b*g(x)+f(x);
simplify(T);

3+g(x)+b*g(x)+f(x)

3+g(x)+b*g(x)+f(x)

 

 

Download simplification_oct_9_2024.mw

What do I need to do to get this output using another software

Ofcourse, I could go and add code to find the names of each function and then use collect. But it seems to me simplify should have done this automatically as the above shows. So I am wondering if there is an option I've overlooked which will do this more easily.

Maple 2024.1

In the rectangular Cartesian coordinate system, three straight lines gA, gB, gC are given, which are not all parallel to each other. Another straight line g and the points Oa, Ob, Oc on it are given. A triangle ABC is to be constructed, one of whose vertices lies on gA, gB or gC and the triangle sides a, b and c (or their extensions) each run through Oa, Ob or Oc.
We are looking for the coordinates of the vertices A, B, C.
In a purely constructive solution, the calculation can be omitted.

Hello,

I noticed that the Linearly Implicit Euler method (also known as the Semi-Implicit Euler method) is not available in Maple's built-in ODE solvers. This method is useful for stiff ODEs, where part of the function is treated implicitly (for the linear term) and part is treated explicitly (for the non-linear term).

I know that the Linearly Implicit Euler method is a specialized method that probably does not find enough widespread use to justify its inclusion as a standard feature in Maple, especially given Maple's focus on numerical methods such as Runge-Kutta methods and fully implicit methods for rigid equations.

I’m wondering:

  1. Why isn’t this method included in Maple’s standard set of numerical solvers?
  2. How can I implement this method in my own code in Maple to solve stiff ODEs?

Any guidance or examples of implementation would be greatly appreciated!

Thank you!

Linearly_Implicit_Method.pdf

Hello,

I need to order a set of polynomials based on a given ranking. For example:

F:=[F1,F2,F3,F4,F5,F6]:
F1 := x1*x4^2+x4^2-x1*x2*x4-x2*x4+x1*x2+3*x2;
F2 := x1*x4-x3-x1*x2;
F3 := x3*x4-2*x2^2-x1*x2-1;
F4 := x1*x3^2+x3^2-x1^2*x2*x3-x1*x2*x3+x1^3*x2+3*x1^2*x2;
F5 := -x3^2 + x1*x2*x3 -2*x1*x2^2-x1^2*x2-x1;
F6 := 2*x1*x2^2+2*x1^2*x2^2-2*x1^2*x2+x1^2+x1;

The order used to rank the polynomials is based on the variables in the following order: x1 < x2 < x3 < x4.

The ranking criteria are:

  1. The first polynomial should be the one involving x4 with the lowest degree of nonlinearity in x4 and the fewest number of terms (in this case, F2).
  2. The next polynomial should also involve x4, but with the lowest degree of nonlinearity and the next fewest number of terms and so on.
  3. Once the polynomials in x4 are exhausted, the ranking continues similarly for the next variable x3, followed by x2, and finally x1.

How can I do this efficiently?

Obs.:If necessary, I can share my solution to the problem, though it's far from efficient.

Hello,

I am attempting to solve what appears to be a simple system of two ordinary differential equations (ODEs) (please find the attached file). However, I encountered an error message from Maple stating: "division by zero."

Could anyone suggest an approach to obtain a closed-form solution for this system?

Please note that the system includes two variables: S(t) and K(t). All other symbols represent positive parameters.

Thank you in advance for your assistance!

Best regards,

Dmitry

Download ODE.mw

A classic task from surveying that is unfortunately no longer taught in our GPS age and is worth remembering:

A hiker has lost his way and wants to know where he is. He has a map, a compass, paper, pen and calculator in his bag. From his position he sees three distant objects from left to right: a radio mast F, a chimney S and a church tower K. He also finds these objects on his map. Using his compass he aims at the three objects and measures the angles at which the distances FS and SK appear: angle for FS=alpha, angle for SK=beta. The hiker also manages to get the approximate coordinates of the three objects from the map to scale: F=(xf;yf), S=(xs;ys) and K=(xk;yk).
Question:
What are the hiker's coordinates?

Hello, I am having problem in solving the pde in the image using maple. Due to its nonlinear nature it has been solved for small value of v using first order perturbation technique and seperation of variable method into radial and angular part in many papers. I am having trouble in proceeding as Maple complains about Boundary/Initial condition.Please tell me if Maple can provide any help in improving existing solution or providing new solution? I can post the full solution procedure by the method i mentioned if needed.

restart

ode0 := (diff(xi^2*(diff(theta[E](xi), xi)), xi))/xi^2 = -theta[E](xi)^n

(2*xi*(diff(theta[E](xi), xi))+xi^2*(diff(diff(theta[E](xi), xi), xi)))/xi^2 = -theta[E](xi)^n

(1)

bc0 := theta[E](0) = 1, (D(theta[E]))(0) = 0

theta[E](0) = 1, (D(theta[E]))(0) = 0

(2)

base := dsolve({bc0, ode0}, theta[E](xi), series)

theta[E](xi) = series(1-(1/6)*xi^2+((1/120)*n)*xi^4+O(xi^6),xi,6)

(3)

pde1 := (diff(xi^2*(diff(psi(xi, mu), xi)), xi))/xi^2+(diff((-mu^2+1)*(diff(psi(xi, mu), mu)), mu))/xi^2 = -psi(xi, mu)^n+v

(2*xi*(diff(psi(xi, mu), xi))+xi^2*(diff(diff(psi(xi, mu), xi), xi)))/xi^2+(-2*mu*(diff(psi(xi, mu), mu))+(-mu^2+1)*(diff(diff(psi(xi, mu), mu), mu)))/xi^2 = -psi(xi, mu)^n+v

(4)

bc1 := psi(0, mu) = 1, (D[1](psi))(0, mu) = 0, (D[2](psi))(0, mu) = 0, limit(psi(xi, mu), v = 0) = rhs(base)

psi(0, mu) = 1, (D[1](psi))(0, mu) = 0, (D[2](psi))(0, mu) = 0, psi(xi, mu) = series(1-(1/6)*xi^2+((1/120)*n)*xi^4+O(xi^6),xi,6)

(5)

pdsolve(pde1, [bc1], psi(xi, mu))

Error, (in pdsolve/sys) too many arguments; some or all of the following are wrong: [[psi(xi,mu)], [psi(0,mu) = 1, D[1](psi)(0,mu) = 0, D[2](psi)(0,mu) = 0, psi(xi,mu) = series(1-1/6*xi^2+1/120*n*xi^4+O(xi^6),xi,6)]]

 

``

Download Nonlinear_Elliptic_PDE_in_Spherical_Coordinate.mw

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