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What is the funcional difference between individual, academic, and professional editions of Maple?

Hi There,

Im currently on using a trial version of maple for engineering students. I am looking to see how you can view a step by step method of solving equations. I can't seem to find a way of doing so and I keep just getting the final answer. Can you assist?

 

Also I can't find the quadratic equation option.

 

thanks,

 

stuart

Dear All,

I have a problem solving the attached nonlinear system of equations using shooting method.
I will be grateful if you could help me finding the solutions out.

 

restart; Shootlib := "C:/Shoot9"; libname := Shootlib, libname; with(Shoot);
with(plots);
N1 := 1.0; N2 := 2.0; N3 := .5; Bt := 6; Re_m := N1*Bt; gamma1 := 1;
FNS := {f(eta), fp(eta), fpp(eta), g(eta), gp(eta), m(eta), mp(eta), n(eta), np(eta), fppp(eta)};
ODE := {diff(f(eta), eta) = fp(eta), diff(fp(eta), eta) = fpp(eta), diff(fpp(eta), eta) = fppp(eta), diff(g(eta), eta) = gp(eta), diff(gp(eta), eta) = N1*(2.*g(eta)+(eta-2.*f(eta)).gp(eta)+2.*g(eta)*fp(eta)+2.*N2.N3.(m(eta).np(eta)-n(eta).mp(eta))), diff(m(eta), eta) = mp(eta), diff(mp(eta), eta) = Re_m.(m(eta)+(eta-2.*f(eta)).mp(eta)+2.*m(eta)*fp(eta)), diff(n(eta), eta) = np(eta), diff(np(eta), eta) = Re_m.(2.*n(eta)+(eta-2.*f(eta)).np(eta)+2.*N2/N3.m(eta).gp(eta)), diff(fppp(eta), eta) = N1*(3.*fpp(eta)+(eta-2.*f(eta)).fppp(eta)-2.*N2.N2.m(eta).(diff(mp(eta), eta)))};
blt := 1.0; IC := {f(0) = 0, fp(0) = 0, fpp(0) = alpha1, g(0) = 1, gp(0) = beta1, m(0) = 0, mp(0) = beta2, n(0) = 0, np(0) = beta3, fppp(0) = alpha2};
BC := {f(blt) = .5, fp(blt) = 0, g(blt) = 0, m(blt) = 1, n(blt) = 1};
infolevel[shoot] := 1;
S := shoot(ODE, IC, BC, FNS, [alpha1 = 1.425, alpha2 = .425, beta1 = -1.31, beta2 = 1.00, beta3 = 1.29]);
Error, (in isolate) cannot isolate for a function when it appears with different arguments
p := odeplot(S, [eta, fp(eta)], 0 .. 15);
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution
display(p);
Error, (in plots:-display) expecting plot structure but received: p
p2 := odeplot(S, [eta, theta(eta)], 0 .. 10);
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution
display(p2);
Error, (in plots:-display) expecting plot structure but received: p2

 

 

      General description of the method of solving underdetermined systems of equations. As a particular application of the idea proposed a universal method of calculation for all kinds of linkage (lever) mechanisms. With the description and examples.
      The method can be used for powerful CAD linkages.

Description: Calculation_method_of_linkages.pdf

Attachment:
figure_1.mw
figure_2.mw

Or all in one
Calculation_method_of_linkages_(with_attach.).pdf


        Some examples of a much larger number calculated by the proposed method. Examples gathered here not to look for them on the forum and opportunity to demonstrate the method.  Among the examples, I think, there are very complicated.

https://vk.com/doc242471809_408704758
https://vk.com/doc242471809_408704572
https://vk.com/doc242471809_376439263
https://vk.com/doc242471809_402619761
https://vk.com/doc242471809_402610228
https://vk.com/doc242471809_401188803
https://vk.com/doc242471809_400465891
https://vk.com/doc242471809_400711315
https://vk.com/doc242471809_387358164
https://vk.com/doc242471809_380837279
https://vk.com/doc242471809_379935473
https://vk.com/doc242471809_380217387
https://vk.com/doc242471809_363266817
https://vk.com/doc242471809_353980472
https://vk.com/doc242471809_375452868
https://vk.com/doc242471809_353988163 
https://vk.com/doc242471809_353986884 
https://vk.com/doc242471809_353987119
https://vk.com/doc242471809_324249241
https://vk.com/doc242471809_324102889
https://vk.com/doc242471809_322219275
https://vk.com/doc242471809_437298137
https://vk.com/doc242471809_437308238
https://vk.com/doc242471809_437308241
https://vk.com/doc242471809_437308243
https://vk.com/doc242471809_437308245
https://vk.com/doc242471809_437308246
https://vk.com/doc242471809_437401651
https://vk.com/doc242471809_437664558

 

 


 Hello,every one,i want to solve system of equations but i recieve an error ,how can i find the coeffecients c1,c2,c3,c4?thank.

``

restart

``

``

A := 45*x*c4+72*c3 = 0:

 

B := 56*c2*c4+28*c3^2 = 0:

C := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c4+42*beta*c1*c4+42*beta*c2*c3 = 0:

E := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c3+30*beta*c1*c3+15*beta*c2 = 0:

F := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c2+20*beta*c1*c2-20*beta*c1*c4-20*beta*c2*c4-20*beta*c3*c4-20*beta*c4^2-20*beta*c4-20*c4 = 0:

G := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c1+6*beta*c1^2-12*beta*c1*c3-12*beta*c2*c3-12*beta*c3^2-12*beta*c3*c4-12*beta*c3-12*c3 = 0:

``

beta*c1+beta*c2^2+beta*c2*c3+beta*c2*c4+beta*c2+c2 = 0:

M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n = 0:

with(SolveTools):

``

PolynomialSystem({{45*c4*x+72*c3 = 0}, {30*beta*c1*c3+15*beta*c2 = 0}, {42*beta*c1*c4+42*beta*c2*c3 = 0}, {20*beta*c1*c2-20*beta*c1*c4-20*beta*c2*c4-20*beta*c3*c4-20*beta*c4^2-20*beta*c4-20*c4 = 0}, {6*beta*c1^2-12*beta*c1*c3-12*beta*c2*c3-12*beta*c3^2-12*beta*c3*c4-12*beta*c3-12*c3 = 0}}, {c1, c2, c3, c4}, {beta = 2, x = 1/5})

Error, invalid input: too many and/or wrong type of arguments passed to SolveTools:-PolynomialSystem; first unused argument is {beta = 2, x = 1/5}

 

NULL

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Download Numerical.mw

Hello,how can i find the lambda in this equation? and x=0..2 , t=0..2

Hello,how can i find the roots of this equation? "5lambda.tan(lambda)-1" and lambda=0..10

Hi,
The latest update to the differential equations Maple libraries (this week, can be downloaded from the Maplesoft R&D webpage for Differential Equations and Mathematical functions) includes new functionality in pdsolve, regarding whether the solution for a PDE or PDE system is or not a general solution.

In brief, a general solution of a PDE in 1 unknown, that has differential order N, and where the unknown depends on M independent variables, involves N arbitrary functions of M-1 arguments. It is not entirely evident how to extend this definition in the case of a coupled, possibly nonlinear PDE system. However, using differential algebra techniques (automatically used by pdsolve when tackling a PDE system), that extension to define a general solution for a DE system is possible, and also when the system involves ODEs and PDEs, and/or algebraic (that is, non-differential) equations, and/or inequations of the form algebraic*expression <> 0 involving the unknowns, and all of this in the presence of mathematical functions (based on the use of Maple's PDEtools:-dpolyform). This is a very nice case were many different advanced developments come together to naturally solve a problem that otherwise would be rather difficult.

The issues at the center of this Maple development/post are then:

        a) How do you know whether a PDE or PDE system solution returned is a general solution?

        b) How could you indicate to pdsolve that you are only interested in a general PDE or PDE system solution?

The answer to a) is now always present in the last line of the userinfo. So input infolevel[pdsolve] := 3 before calling pdsolve, and check what the last line of the userinfo displayed tells.


The answer to b) is a new option, generalsolution, implemented in pdsolve so that it either returns a general solution or otherwise it returns NULL. If you do not use this new option, then pdsolve works as always: first it tries to compute a general solution and if it fails in doing that it tries to compute a particular solution by separating the variables in different ways, or computing a traveling wave solution or etc. (a number of other well known methods).

 

The examples that follow are from the help page pdsolve,system, and show both the new userinfo telling whether the solution returned is a general one and the option generalsolution at work.The examples are all of differential equation systems but the same userinfos and generalsolution option work as well in the case of a single PDE.

 

 

Example 1.

Solve the determining PDE system for the infinitesimals of the symmetry generator of example 11 from Kamke's book . Tell whether the solution computed is or not a general solution.

infolevel[pdsolve] := 3

3

(1.1)

The PDE system satisfied by the symmetries of Kamke's ODE example number 11 is

sys__1 := [diff(xi(x, y), y, y) = 0, diff(eta(x, y), y, y)-2*(diff(xi(x, y), y, x)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(eta(x, y), y, x))-(diff(xi(x, y), x, x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(eta(x, y), x, x) = 0]

This is a second order linear PDE system, with two unknowns {eta(x, y), xi(x, y)} and four equations. Its general solution is given by the following, where we now can tell that the solution is a general one by reading the last line of the userinfo. Note that because the system is overdetermined, a general solution in this case does not involve any arbitrary function

sol__1 := pdsolve(sys__1)

-> Solving ordering for the dependent variables of the PDE system: [xi(x,y), eta(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
<- Returning a *general* solution

 

{eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}

(1.2)

Next we indicate to pdsolve that n and r are parameters of the problem, and that we want a solution for n <> 1, making more difficult to identify by eye whether the solution returned is or not a general one. Again the last line of the userinfo tells that pdsolve's solution is indeed a general one

`sys__1.1` := [op(sys__1), n <> 1]

[diff(diff(xi(x, y), y), y) = 0, diff(diff(eta(x, y), y), y)-2*(diff(diff(xi(x, y), x), y)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(diff(eta(x, y), x), y))-(diff(diff(xi(x, y), x), x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(diff(eta(x, y), x), x) = 0, n <> 1]

(1.3)

`sol__1.1` := pdsolve(`sys__1.1`, parameters = {n, r})

-> Solving ordering for the dependent variables of the PDE system: [r, n, xi(x,y), eta(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
<- Returning a *general* solution

 

{n = 2, r = -5, eta(x, y) = y*(_C1*x+3*_C2), xi(x, y) = x*(_C1*x+_C2)}, {n = 2, r = -20/7, eta(x, y) = -(2/343)*(-6*_C1*x^2-98*x^(8/7)*_C1*a*y-147*_C2*a*x*y)/(x*a), xi(x, y) = _C1*x^(8/7)+_C2*x}, {n = 2, r = -15/7, eta(x, y) = -(1/343)*(-49*_C2*a*x*y-147*x^(6/7)*_C1*a*y+12*_C1*x)/(x*a), xi(x, y) = _C1*x^(6/7)+_C2*x}, {n = 2, r = r, eta(x, y) = -_C1*y*(r+2), xi(x, y) = _C1*x}, {n = -r-3, r = r, eta(x, y) = ((_C1*x+_C2)*r+4*_C1*x+2*_C2)*y/(r+4), xi(x, y) = x*(_C1*x+_C2)}, {n = n, r = r, eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}

(1.4)

map(pdetest, [`sol__1.1`], `sys__1.1`)

[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

(1.5)

 

Example 2.

Compute the solution of the following (linear) overdetermined system involving two PDEs, three unknown functions, one of which depends on 2 variables and the other two depend on only 1 variable.

sys__2 := [-(diff(F(r, s), r, r))+diff(F(r, s), s, s)+diff(H(r), r)+diff(G(s), s)+s = 0, diff(F(r, s), r, r)+2*(diff(F(r, s), r, s))+diff(F(r, s), s, s)-(diff(H(r), r))+diff(G(s), s)-r = 0]

The solution for the unknowns G, H, is given by the following expression, were again determining whether this solution, that depends on 3 arbitrary functions, _F1(s), _F2(r), _F3(s-r), is or not a general solution, is non-obvious.

sol__2 := pdsolve(sys__2)

-> Solving ordering for the dependent variables of the PDE system: [F(r,s), H(r), G(s)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [r, s]
tackling triangularized subsystem with respect to F(r,s)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
differential factorization successful.
First set of solution methods successful
tackling triangularized subsystem with respect to H(r)
tackling triangularized subsystem with respect to G(s)
<- Returning a *general* solution

 

{F(r, s) = _F1(s)+_F2(r)+_F3(s-r)-(1/12)*r^2*(r-3*s), G(s) = -(diff(_F1(s), s))-(1/4)*s^2+_C2, H(r) = diff(_F2(r), r)-(1/4)*r^2+_C1}

(1.6)

pdetest(sol__2, sys__2)

[0, 0]

(1.7)

Example 3.

Compute the solution of the following nonlinear system, consisting of Burger's equation and a possible potential.

sys__3 := [diff(u(x, t), t)+2*u(x, t)*(diff(u(x, t), x))-(diff(u(x, t), x, x)) = 0, diff(v(x, t), t) = -v(x, t)*(diff(u(x, t), x))+v(x, t)*u(x, t)^2, diff(v(x, t), x) = -u(x, t)*v(x, t)]

We see that in this case the solution returned is not a general solution but two particular ones; again the information is in the last line of the userinfo displayed

sol__3 := pdsolve(sys__3, [u, v])

-> Solving ordering for the dependent variables of the PDE system: [v(x,t), u(x,t)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
tackling triangularized subsystem with respect to v(x,t)
tackling triangularized subsystem with respect to u(x,t)
First set of solution methods (general or quasi general solution)
Second set of solution methods (complete solutions)
Trying methods for second order PDEs
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Trying methods for second order linear PDEs
Preparing a solution HINT ...
Trying HINT = _F1(x)*_F2(t)
Fourth set of solution methods
Preparing a solution HINT ...
Trying HINT = _F1(x)+_F2(t)
Trying travelling wave solutions as power series in tanh ...
* Using tau = tanh(t*C[2]+x*C[1]+C[0])
* Equivalent ODE system: {C[1]^2*(tau^2-1)^2*diff(diff(u(tau),tau),tau)+(2*C[1]^2*(tau^2-1)*tau+2*u(tau)*C[1]*(tau^2-1)+C[2]*(tau^2-1))*diff(u(tau),tau)}
* Ordering for functions: [u(tau)]
* Cases for the upper bounds: [[n[1] = 1]]
* Power series solution [1]: {u(tau) = tau*A[1,1]+A[1,0]}
* Solution [1] for {A[i, j], C[k]}: [[A[1,1] = 0], [A[1,0] = -1/2*C[2]/C[1], A[1,1] = -C[1]]]
travelling wave solutions successful.
tackling triangularized subsystem with respect to v(x,t)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
Trying methods for PDEs "missing the dependent variable" ...
Second set of solution methods (complete solutions)
Trying methods for second order PDEs
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Trying methods for second order linear PDEs
Preparing a solution HINT ...
Trying HINT = _F1(x)*_F2(t)
Third set of solution methods successful
tackling triangularized subsystem with respect to u(x,t)
<- Returning a solution that *is not the most general one*

 

{u(x, t) = -_C2*tanh(_C2*x+_C3*t+_C1)-(1/2)*_C3/_C2, v(x, t) = 0}, {u(x, t) = -_c[1]^(1/2)*((exp(_c[1]^(1/2)*x))^2*_C1-_C2)/((exp(_c[1]^(1/2)*x))^2*_C1+_C2), v(x, t) = _C3*exp(_c[1]*t)*_C1*exp(_c[1]^(1/2)*x)+_C3*exp(_c[1]*t)*_C2/exp(_c[1]^(1/2)*x)}

(1.8)

pdetest(sol__3, sys__3)

[0, 0, 0]

(1.9)

This example is also good for illustrating the other related new feature: one can now request to pdsolve to only compute a general solution (it will return NULL if it cannot achieve that). Turn OFF userinfos and try with this example

infolevel[pdsolve] := 1

This returns NULL:

pdsolve(sys__3, [u, v], generalsolution)

Example 4.

Another where the solution returned is particular, this time for a linear system, conformed by 38 PDEs, also from differential equation symmetry analysis

sys__4 := [diff(xi[1](x, y, z, t, u), u) = 0, diff(xi[1](x, y, z, t, u), x)-(diff(xi[2](x, y, z, t, u), y)) = 0, diff(xi[2](x, y, z, t, u), u) = 0, -(diff(xi[1](x, y, z, t, u), y))-(diff(xi[2](x, y, z, t, u), x)) = 0, diff(xi[3](x, y, z, t, u), u) = 0, diff(xi[1](x, y, z, t, u), x)-(diff(xi[3](x, y, z, t, u), z)) = 0, -(diff(xi[3](x, y, z, t, u), y))-(diff(xi[2](x, y, z, t, u), z)) = 0, -(diff(xi[1](x, y, z, t, u), z))-(diff(xi[3](x, y, z, t, u), x)) = 0, diff(xi[4](x, y, z, t, u), u) = 0, diff(xi[3](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), z)) = 0, diff(xi[2](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), y)) = 0, diff(xi[1](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), x)) = 0, -(diff(xi[1](x, y, z, t, u), x))+diff(xi[4](x, y, z, t, u), t) = 0, diff(eta[1](x, y, z, t, u), y, y)+diff(eta[1](x, y, z, t, u), z, z)-(diff(eta[1](x, y, z, t, u), t, t))+diff(eta[1](x, y, z, t, u), x, x) = 0, diff(eta[1](x, y, z, t, u), u, u) = 0, diff(eta[1](x, y, z, t, u), u, x)+diff(xi[1](x, y, z, t, u), x, x) = 0, diff(xi[1](x, y, z, t, u), x, y)+diff(eta[1](x, y, z, t, u), u, y) = 0, -(diff(xi[1](x, y, z, t, u), y, y))+diff(eta[1](x, y, z, t, u), u, x) = 0, diff(xi[1](x, y, z, t, u), x, z)+diff(eta[1](x, y, z, t, u), u, z) = 0, diff(xi[1](x, y, z, t, u), y, z) = 0, -(diff(xi[1](x, y, z, t, u), z, z))+diff(eta[1](x, y, z, t, u), u, x) = 0, -(diff(eta[1](x, y, z, t, u), t, u))-(diff(xi[1](x, y, z, t, u), t, x)) = 0, diff(xi[1](x, y, z, t, u), t, y) = 0, diff(xi[1](x, y, z, t, u), t, z) = 0, diff(xi[1](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, x) = 0, -(diff(xi[2](x, y, z, t, u), z, z))+diff(eta[1](x, y, z, t, u), u, y) = 0, diff(xi[2](x, y, z, t, u), t, z) = 0, diff(xi[2](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, y) = 0, diff(xi[3](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, z) = 0, diff(eta[1](x, y, z, t, u), u, x, x) = 0, diff(eta[1](x, y, z, t, u), u, x, y) = 0, diff(eta[1](x, y, z, t, u), u, y, y) = 0, diff(eta[1](x, y, z, t, u), u, x, z) = 0, diff(eta[1](x, y, z, t, u), u, y, z) = 0, diff(eta[1](x, y, z, t, u), u, z, z) = 0, diff(eta[1](x, y, z, t, u), t, u, x) = 0, diff(eta[1](x, y, z, t, u), t, u, y) = 0, diff(eta[1](x, y, z, t, u), t, u, z) = 0]

There are 38 coupled equations

nops(sys__4)

38

(1.10)

When requesting a general solution pdsolve returns NULL:

pdsolve(sys__4, generalsolution)

A solution that is not a general one, is however computed by default if calling pdsolve without the generalsolution option. In this case again the last line of the userinfo tells that the solution returned is not a general solution

infolevel[pdsolve] := 3

3

(1.11)

sol__4 := pdsolve(sys__4)

-> Solving ordering for the dependent variables of the PDE system: [eta[1](x,y,z,t,u), xi[1](x,y,z,t,u), xi[2](x,y,z,t,u), xi[3](x,y,z,t,u), xi[4](x,y,z,t,u)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
tackling triangularized subsystem with respect to eta[1](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(x,y,z,t), _F2(x,y,z,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
tackling triangularized subsystem with respect to _F1(x,y,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F3(x,y,z), _F4(x,y,z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y, z, t]
tackling triangularized subsystem with respect to _F3(x,y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F4(x,y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F5(y,z), _F6(y,z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [y, z, x]
tackling triangularized subsystem with respect to _F5(y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F6(y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F7(z), _F8(z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [z, y]
tackling triangularized subsystem with respect to _F7(z)
tackling triangularized subsystem with respect to _F8(z)
tackling triangularized subsystem with respect to _F2(x,y,z,t)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
Trying methods for PDEs "missing the dependent variable" ...
Second set of solution methods (complete solutions)
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Preparing a solution HINT ...
Trying HINT = _F3(x)*_F4(y)*_F5(z)*_F6(t)
Third set of solution methods successful

tackling triangularized subsystem with respect to xi[1](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(x,z,t), _F2(x,z,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z, y]
tackling triangularized subsystem with respect to _F1(x,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F2(x,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful

-> Solving ordering for the dependent variables of the PDE system: [_F3(x,t), _F4(x,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z]
tackling triangularized subsystem with respect to _F3(x,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F4(x,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F5(x), _F6(x)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
tackling triangularized subsystem with respect to _F5(x)
tackling triangularized subsystem with respect to _F6(x)
tackling triangularized subsystem with respect to xi[2](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(t), _F2(t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, z]
tackling triangularized subsystem with respect to _F1(t)
tackling triangularized subsystem with respect to _F2(t)
tackling triangularized subsystem with respect to xi[3](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to xi[4](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
<- Returning a solution that *is not the most general one*

 

{eta[1](x, y, z, t, u) = (_C13*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*cos((-_c[1]-_c[2]-_c[3])^(1/2)*t)+_C12*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*sin((-_c[1]-_c[2]-_c[3])^(1/2)*t)+u*exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)*(_C1*t+_C2*x+_C3*y+_C4*z+_C5))/(exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)), xi[1](x, y, z, t, u) = -(1/2)*_C2*x^2+(1/2)*(-2*_C1*t-2*_C3*y-2*_C4*z+2*_C17)*x+(1/2)*(-t^2+y^2+z^2)*_C2+_C16*t+_C15*z+_C14*y+_C18, xi[2](x, y, z, t, u) = -(1/2)*_C3*y^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C4*z+2*_C17)*y+(1/2)*(-t^2+x^2+z^2)*_C3+_C20*t+_C19*z-_C14*x+_C21, xi[3](x, y, z, t, u) = -(1/2)*_C4*z^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C3*y+2*_C17)*z+(1/2)*(-t^2+x^2+y^2)*_C4+_C22*t-_C19*y-_C15*x+_C23, xi[4](x, y, z, t, u) = -(1/2)*_C1*t^2+(1/2)*(-2*_C2*x-2*_C3*y-2*_C4*z+2*_C17)*t+(1/2)*(-x^2-y^2-z^2)*_C1+_C20*y+_C22*z+_C16*x+_C24}

(1.12)

pdetest(sol__4, sys__4)

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

(1.13)

Example 5.

Finally, the new userinfos also tell whether a solution is or not a general solution when working with PDEs that involve anticommutative variables  set using the Physics  package

with(Physics, Setup)

[Setup]

(1.14)

Set first theta and Q as suffixes for variables of type/anticommutative  (see Setup )

Setup(anticommutativepre = {Q, theta})

`* Partial match of  'anticommutativepre' against keyword 'anticommutativeprefix'`

 

[anticommutativeprefix = {Q, _lambda, theta}]

(1.15)

A PDE system example with two unknown anticommutative functions of four variables, two commutative and two anticommutative; to avoid redundant typing in the input that follows and redundant display of information on the screen let's use PDEtools:-diff_table   PDEtools:-declare

PDEtools:-declare(Q(x, y, theta[1], theta[2]))

Q(x, y, theta[1], theta[2])*`will now be displayed as`*Q

(1.16)

q := PDEtools:-diff_table(Q(x, y, theta[1], theta[2]))

table( [(  ) = Q(x, y, theta[1], theta[2]) ] )

(1.17)

Consider the system formed by these two PDEs (because of the q diff_table just defined, we can enter derivatives directly using the function's name indexed by the differentiation variables)

pde[1] := q[x, y, theta[1]]+q[x, y, theta[2]]-q[y, theta[1], theta[2]] = 0

Physics:-diff(diff(diff(Q(x, y, theta[1], theta[2]), x), y), theta[1])+Physics:-diff(diff(diff(Q(x, y, theta[1], theta[2]), x), y), theta[2])-Physics:-diff(Physics:-diff(diff(Q(x, y, theta[1], theta[2]), y), theta[1]), theta[2]) = 0

(1.18)

pde[2] := q[theta[1]] = 0

Physics:-diff(Q(x, y, theta[1], theta[2]), theta[1]) = 0

(1.19)

The solution returned for this system is indeed a general solution

pdsolve([pde[1], pde[2]])

-> Solving ordering for the dependent variables of the PDE system: [_F4(x,y), _F2(x,y), _F3(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to _F4(x,y)
tackling triangularized subsystem with respect to _F2(x,y)
tackling triangularized subsystem with respect to _F3(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
HINT = _F6(x)+_F5(y)
Trying HINT = _F6(x)+_F5(y)
HINT is successful
First set of solution methods successful
<- Returning a *general* solution

 

Q(x, y, theta[1], theta[2]) = _F1(x, y)*_lambda1+(_F6(x)+_F5(y))*theta[2]

(1.20)

NULL

This solution involves an anticommutative constant `_&lambda;2`, analogous to the commutative constants _Cn where n is an integer.

 

Download PDE_general_solutions.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft


Hi

New developments (after the release of Maple 2016) happened in the project on exact solutions for "Partial Differential Equations & Boundary Conditions". This is work in collaboration with Katherina von Bulow and the improvements are of wide range, representing a noticeable step forward in the capabilities of the Maple system for this kind of problem. As usual, these improvements can be installed in current Maple 2016 by downloading the updated library from the Maplesoft R&D webpage for Differential Equations and Mathematical functions (the update is distributed merged with the updates of the Physics package)

 

The improvements cover:

 

• 

PDE&BC in semi-infinite domains for which a bounded solution is sought

• 

PDE & BC problems in bounded spatial domains via eigenfunction (Fourier) expansions

• 

Implementation of another algebraic method for tackling linear PDE & BC

• 

Improvements in solving PDE & BC solutions by first finding the PDE's general solution.

• 

Improvements in solving PDE & BC problems by using a Fourier transform.

• 

PDE & BC problems that used to require the option HINT = `+` are now solved automatically

 

What follows is a set of examples solved now with these new developments, organized in sections according to the kind of problem. Where relevant, the sections include a subsection on "How it works step by step".

PDE&BC in semi-infinite domains for which a bounded solution is sought can now also be solved via Laplace transforms

 

Maple is now able to solve more PDE&BC problems via Laplace transforms.

 

How it works: Laplace transforms act to change derivatives with respect to one of the independent variables of the domain into multiplication operations in the transformed domain. After applying a Laplace transform to the original problem, we can simplify the problem using the transformed BC, then solve the problem in the transformed domain, and finally apply the inverse Laplace transform to arrive at the final solution. It is important to remember to give pdsolve any necessary restrictions on the variables and constants of the problem, by means of the "assuming" command.

 

A new feature is that we can now tell pdsolve that the dependent variable is bounded, by means of the optional argument HINT = boundedseries.

 

restart

 

Consider the problem of a falling cable lying on a table that is suddenly removed (cf. David J. Logan's Applied Partial Differential Equations p.115).

pde[1] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))-g
iv[1] := u(x, 0) = 0, u(0, t) = 0, (D[2](u))(x, 0) = 0

 

If we ask pdsolve to solve this problem without the condition of boundedness of the solution, we obtain:

`assuming`([pdsolve([pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

u(x, t) = -invlaplace(exp(s*x/c)*_F1(s), s, t)+invlaplace(exp(-s*x/c)*_F1(s), s, t)-(1/2)*g*t^2+invlaplace(exp(s*x/c)/s^3, s, t)*g

(1.1)

New: If we now ask for a bounded solution, by means of the option HINT = boundedseries, pdsolve simplifies the problem accordingly.

ans[1] := `assuming`([pdsolve([pde[1], iv[1]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

u(x, t) = (1/2)*g*(Heaviside(t-x/c)*(c*t-x)^2-c^2*t^2)/c^2

(1.2)

 

And we can check this answer against the original problem, if desired:

`assuming`([pdetest(ans[1], [pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

[0, 0, 0, 0]

(1.3)

How it works, step by step

 

 Let us see the process this problem undergoes to be solved by pdsolve, step by step.

 

First, the Laplace transform is applied to the PDE:

with(inttrans)

transformed_PDE := laplace((lhs-rhs)(pde[1]), t, s)

s^2*laplace(u(x, t), t, s)-(D[2](u))(x, 0)-s*u(x, 0)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

(1.1.1)

and the result is simplified using the initial conditions:

simplified_transformed_PDE := eval(transformed_PDE, {iv[1]})

s^2*laplace(u(x, t), t, s)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

(1.1.2)

Next, we call the function "laplace(u(x,t),t,s)" by the new name U:

eq_U := subs(laplace(u(x, t), t, s) = U(x, s), simplified_transformed_PDE)

s^2*U(x, s)-c^2*(diff(diff(U(x, s), x), x))+g/s

(1.1.3)

And this equation, which is really an ODE, is solved:

solution_U := dsolve(eq_U, U(x, s))

U(x, s) = exp(-s*x/c)*_F2(s)+exp(s*x/c)*_F1(s)-g/s^3

(1.1.4)

Now, since we want a BOUNDED solution, the term with the positive exponential must be zero, and we are left with:

bounded_solution_U := subs(coeff(rhs(solution_U), exp(s*x/c)) = 0, solution_U)

U(x, s) = exp(-s*x/c)*_F2(s)-g/s^3

(1.1.5)

Now, the initial solution must also be satisfied. Here it is, in the transformed domain:

Laplace_BC := laplace(u(0, t), t, s) = 0

laplace(u(0, t), t, s) = 0

(1.1.6)

Or, in the new variable U,

Laplace_BC_U := U(0, s) = 0

U(0, s) = 0

(1.1.7)

And by applying it to bounded_solution_U, we find the relationship

simplify(subs(x = 0, rhs(bounded_solution_U))) = 0

(_F2(s)*s^3-g)/s^3 = 0

(1.1.8)

isolate((_F2(s)*s^3-g)/s^3 = 0, indets((_F2(s)*s^3-g)/s^3 = 0, unknown)[1])

_F2(s) = g/s^3

(1.1.9)

so that our solution now becomes

bounded_solution_U := subs(_F2(s) = g/s^3, bounded_solution_U)

U(x, s) = exp(-s*x/c)*g/s^3-g/s^3

(1.1.10)

to which we now apply the inverse Laplace transform to obtain the solution to the problem:

`assuming`([u(x, t) = invlaplace(rhs(bounded_solution_U), s, t)], [0 < x, 0 < t, 0 < c])

u(x, t) = (1/2)*g*(-t^2+Heaviside(t-x/c)*(c*t-x)^2/c^2)

(1.1.11)

Four other related examples

 

A few other examples:

pde[2] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
iv[2] := u(x, 0) = 0, u(0, t) = g(t), (D[2](u))(x, 0) = 0

ans[2] := `assuming`([pdsolve([pde[2], iv[2]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

u(x, t) = Heaviside(t-x/c)*g((c*t-x)/c)

(1.2.1)

`assuming`([pdetest(ans[2], [pde[2], iv[2]])], [0 < t, 0 < x, 0 < c])

[0, 0, 0, 0]

(1.2.2)

pde[3] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[3] := u(x, 0) = 0, u(0, t) = 1

ans[3] := `assuming`([pdsolve([pde[3], iv[3]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

u(x, t) = 1-erf((1/2)*x/(t^(1/2)*k^(1/2)))

(1.2.3)

pdetest(ans[3], [pde[3], iv[3][2]])

[0, 0]

(1.2.4)

pde[4] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[4] := u(x, 0) = mu, u(0, t) = lambda

ans[4] := `assuming`([pdsolve([pde[4], iv[4]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

u(x, t) = (-lambda+mu)*erf((1/2)*x/(t^(1/2)*k^(1/2)))+lambda

(1.2.5)

pdetest(ans[4], [pde[4], iv[4][2]])

[0, 0]

(1.2.6)

 

The following is an example from page 76 in Logan's book:

pde[5] := diff(u(x, t), t) = diff(u(x, t), x, x)
iv[5] := u(x, 0) = 0, u(0, t) = f(t)

ans[5] := `assuming`([pdsolve([pde[5], iv[5]], HINT = boundedseries)], [0 < t, 0 < x])

u(x, t) = (1/2)*x*(int(f(_U1)*exp(-x^2/(4*t-4*_U1))/(t-_U1)^(3/2), _U1 = 0 .. t))/Pi^(1/2)

(1.2.7)

More PDE&BC problems in bounded spatial domains can now be solved via eigenfunction (Fourier) expansions

 

The code for solving PDE&BC problems in bounded spatial domains has been expanded. The method works by separating the variables by product, so that the problem is transformed into an ODE system (with initial and/or boundary conditions) problem, one of which is a Sturm-Liouville problem (a type of eigenvalue problem) which has infinitely many solutions - hence the infinite series representation of the solutions.

restart

 

Here is a simple example for the heat equation:

pde__6 := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv__6 := u(0, t) = 0, u(l, t) = 0

ans__6 := `assuming`([pdsolve([pde__6, iv__6])], [0 < l])

u(x, t) = Sum(_C1*sin(_Z1*Pi*x/l)*exp(-k*Pi^2*_Z1^2*t/l^2), _Z1 = 1 .. infinity)

(2.1)

pdetest(ans__6, [pde__6, iv__6])

[0, 0, 0]

(2.2)

 

Now, consider the displacements of a string governed by the wave equation, where c is a constant (cf. Logan p.28).

pde__7 := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
iv__7 := u(0, t) = 0, u(l, t) = 0

ans__7 := `assuming`([pdsolve([pde__7, iv__7])], [0 < l])

u(x, t) = Sum(sin(_Z2*Pi*x/l)*(sin(c*_Z2*Pi*t/l)*_C1+cos(c*_Z2*Pi*t/l)*_C5), _Z2 = 1 .. infinity)

(2.3)

pdetest(ans__7, [pde__7, iv__7])

[0, 0, 0]

(2.4)

Another wave equation problem (cf. Logan p.130):

pde__8 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = 0; iv__8 := u(0, t) = 0, (D[2](u))(x, 0) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

ans__8 := `assuming`([pdsolve([pde__8, iv__8], u(x, t))], [0 <= x, x <= l])

u(x, t) = Sum(2*(Int(f(x)*sin((1/2)*Pi*(2*_Z3+1)*x/l), x = 0 .. l))*sin((1/2)*Pi*(2*_Z3+1)*x/l)*cos((1/2)*c*Pi*(2*_Z3+1)*t/l)/l, _Z3 = 1 .. infinity)

(2.5)

pdetest(ans__8, [pde__8, iv__8[1 .. 3]])

[0, 0, 0, 0]

(2.6)

 

Here is a problem with periodic boundary conditions (cf. Logan p.131). The function u(x, t) stands for the concentration of a chemical dissolved in water within a tubular ring of circumference 2*l. The initial concentration is given by f(x), and the variable x is the arc-length parameter that varies from 0 to 2*l.

pde__9 := diff(u(x, t), t) = M*(diff(u(x, t), x, x))
iv__9 := u(0, t) = u(2*l, t), (D[1](u))(0, t) = (D[1](u))(2*l, t), u(x, 0) = f(x)

ans__9 := `assuming`([pdsolve([pde__9, iv__9], u(x, t))], [0 <= x, x <= 2*l])

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*sin(_Z4*Pi*x/l), x = 0 .. 2*l))*sin(_Z4*Pi*x/l)+(Int(f(x)*cos(_Z4*Pi*x/l), x = 0 .. 2*l))*cos(_Z4*Pi*x/l))*exp(-M*Pi^2*_Z4^2*t/l^2)/l, _Z4 = 1 .. infinity)

(2.7)

pdetest(ans__9, [pde__9, iv__9[1 .. 2]])

[0, 0, 0]

(2.8)

 

The following problem is for heat flow with both boundaries insulated (cf. Logan p.166, 3rd edition)

pde__10 := diff(u(x, t), t) = k*(diff(u(x, t), x, x))
iv__10 := (D[1](u))(0, t) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

ans__10 := `assuming`([pdsolve([pde__10, iv__10], u(x, t))], [0 <= x, x <= l])

u(x, t) = Sum(2*(Int(f(x)*cos(_Z6*Pi*x/l), x = 0 .. l))*cos(_Z6*Pi*x/l)*exp(-k*Pi^2*_Z6^2*t/l^2)/l, _Z6 = 1 .. infinity)

(2.9)

pdetest(ans__10, [pde__10, iv__10[1 .. 2]])

[0, 0, 0]

(2.10)

 

This is a problem in a bounded domain with the presence of a source. A source term represents an outside influence in the system and leads to an inhomogeneous PDE (cf. Logan p.149):

pde__11 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = p(x, t)
iv__11 := u(0, t) = 0, u(Pi, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

ans__11 := pdsolve([pde__11, iv__11], u(x, t))

u(x, t) = Int(Sum(2*(Int(p(x, tau1)*sin(_Z7*x), x = 0 .. Pi))*sin(_Z7*x)*sin(c*_Z7*(t-tau1))/(Pi*_Z7*c), _Z7 = 1 .. infinity), tau1 = 0 .. t)

(2.11)

Current pdetest is unable to verify that this solution cancells the pde__11 mainly because it currently fails in identifying that there is a fourier expansion in it, but its subroutines for testing the boundary conditions work well with this problem

pdetest_BC := `pdetest/BC`

pdetest_BC({ans__11}, [iv__11], [u(x, t)])

[0, 0, 0, 0]

(2.12)

 

 

Consider a heat absorption-radiation problem in the bounded domain 0 <= x and x <= 2, t >= 0:

pde__12 := diff(u(x, t), t) = diff(u(x, t), x, x)
iv__12 := u(x, 0) = f(x), (D[1](u))(0, t)+u(0, t) = 0, (D[1](u))(2, t)+u(2, t) = 0

ans__12 := `assuming`([pdsolve([pde__12, iv__12], u(x, t))], [0 <= x and x <= 2, 0 <= t])

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*cos((1/2)*_Z8*Pi*x), x = 0 .. 2))*cos((1/2)*_Z8*Pi*x)+(Int(f(x)*sin((1/2)*_Z8*Pi*x), x = 0 .. 2))*sin((1/2)*_Z8*Pi*x))*exp(-(1/4)*Pi^2*_Z8^2*t), _Z8 = 1 .. infinity)

(2.13)

pdetest(ans__12, pde__12)

0

(2.14)

Consider the nonhomogeneous wave equation problem (cf. Logan p.213, 3rd edition):

pde__13 := diff(u(x, t), t, t) = A*x+diff(u(x, t), x, x)
iv__13 := u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

ans__13 := pdsolve([pde__13, iv__13])

u(x, t) = Int(Sum(2*A*(Int(x*sin(Pi*_Z9*x), x = 0 .. 1))*sin(Pi*_Z9*x)*sin(Pi*_Z9*(t-tau1))/(Pi*_Z9), _Z9 = 1 .. infinity), tau1 = 0 .. t)

(2.15)

pdetest_BC({ans__13}, [iv__13], [u(x, t)])

[0, 0, 0, 0]

(2.16)

 

Consider the following Schrödinger equation with zero potential energy (cf. Logan p.30):

pde__14 := I*h*(diff(f(x, t), t)) = -h^2*(diff(f(x, t), x, x))/(2*m)
iv__14 := f(0, t) = 0, f(d, t) = 0

ans__14 := `assuming`([pdsolve([pde__14, iv__14])], [0 < d])

f(x, t) = Sum(_C1*sin(_Z10*Pi*x/d)*exp(-((1/2)*I)*h*Pi^2*_Z10^2*t/(d^2*m)), _Z10 = 1 .. infinity)

(2.17)

pdetest(ans__14, [pde__14, iv__14])

[0, 0, 0]

(2.18)

Another method has been implemented for linear PDE&BC

 

This method is for problems of the form

 

 "(&PartialD;w)/(&PartialD;t)=M[w]"", w(`x__i`,0) = f(`x__i`)" or

 

"((&PartialD;)^2w)/((&PartialD;)^( )t^2)="M[w]", w(`x__i`,0) = f(`x__i`), (&PartialD;w)/(&PartialD;t)() ? ()|() ? (t=0) =g(`x__i`)"

 

where M is an arbitrary linear differential operator of any order which only depends on the spatial variables x__i.

 

Here are some examples:

pde__15 := diff(w(x1, x2, x3, t), t)-(diff(w(x1, x2, x3, t), x2, x1))-(diff(w(x1, x2, x3, t), x3, x1))-(diff(w(x1, x2, x3, t), x3, x3))+diff(w(x1, x2, x3, t), x3, x2) = 0
iv__15 := w(x1, x2, x3, 0) = x1^5*x2*x3NULL

pdsolve([pde__15, iv__15])

w(x1, x2, x3, t) = 20*(((1/20)*x2*x3-(1/20)*t)*x1^2+(1/4)*t*(x2+x3)*x1+t^2)*x1^3

(3.1)

pdetest(%, [pde__15, iv__15])

[0, 0]

(3.2)

 

Here are two examples for which the derivative with respect to t is of the second order, and two initial conditions are given:

pde__16 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
iv__16 := w(x1, x2, x3, 0) = x1^3*x2^2+x3, (D[4](w))(x1, x2, x3, 0) = -x2*x3+x1

pdsolve([pde__16, iv__16])

w(x1, x2, x3, t) = x1^3*x2^2+x3-t*x2*x3+t*x1+3*t^2*x2*x1^2+(1/6)*t^3+(1/2)*t^4*x1

(3.3)

pdetest(%, [pde__16, iv__16])

[0, 0, 0]

(3.4)

pde__17 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
iv__17 := w(x1, x2, x3, 0) = x1^3*x3^2+sin(x1), (D[4](w))(x1, x2, x3, 0) = cos(x1)-x2*x3

pdsolve([pde__17, iv__17])

w(x1, x2, x3, t) = (1/2)*t^4*x1+t^2*x1^3+3*t^2*x1^2*x3+x1^3*x3^2+(1/6)*t^3-t*x2*x3+cos(x1)*t+sin(x1)

(3.5)

pdetest(%, [pde__17, iv__17])

[0, 0, 0]

(3.6)

More PDE&BC problems are now solved via first finding the PDE's general solution.

 

The following are examples of PDE&BC problems for which pdsolve is successful in first calculating the PDE's general solution, and then fitting the initial or boundary condition to it.

pde__18 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0
iv__18 := u(0, y) = sin(y)/y

If we ask pdsolve to solve the problem, we get:

ans__18 := pdsolve([pde__18, iv__18])

u(x, y) = (sin(-y+I*x)+_F2(y-I*x)*(y-I*x)+(-y+I*x)*_F2(y+I*x))/(-y+I*x)

(4.1)

and we can check this answer by using pdetest:

pdetest(ans__18, [pde__18, iv__18])

[0, 0]

(4.2)

How it works, step by step:

 

The general solution for just the PDE is:

gensol := pdsolve(pde__18)

u(x, y) = _F1(y-I*x)+_F2(y+I*x)

(4.1.1)

Substituting in the condition iv__18, we get:

u(0, y) = sin(y)/y

(4.1.2)

gensol_with_condition := eval(rhs(gensol), x = 0) = rhs(iv__18)

_F1(y)+_F2(y) = sin(y)/y

(4.1.3)

We then isolate one of the functions above (we can choose either one, in this case), convert it into a function operator, and then apply it to gensol

_F1 = unapply(solve(_F1(y)+_F2(y) = sin(y)/y, _F1(y)), y)

_F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc)

(4.1.4)

eval(gensol, _F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc))

u(x, y) = (-_F2(y-I*x)*(y-I*x)-sin(-y+I*x))/(y-I*x)+_F2(y+I*x)

(4.1.5)

 

 

Three other related examples

 

pde__19 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
iv__19 := u(0, y) = sin(y)/y

pdsolve([pde__19, iv__19])

u(x, y) = (2*sin(-y+((1/2)*I)*2^(1/2)*x)+(-I*2^(1/2)*x+2*y)*_F2(y-((1/2)*I)*2^(1/2)*x)+(I*2^(1/2)*x-2*y)*_F2(y+((1/2)*I)*2^(1/2)*x))/(I*2^(1/2)*x-2*y)

(4.2.1)

pdetest(%, [pde__19, iv__19])

[0, 0]

(4.2.2)

pde__20 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
iv__20 := u(x, 0) = sin(x)/x

pdsolve([pde__20, iv__20])

u(x, y) = (sinh((1/2)*(I*2^(1/2)*x-2*y)*2^(1/2))*2^(1/2)-(I*2^(1/2)*x-2*y)*(_F2(-y+((1/2)*I)*2^(1/2)*x)-_F2(y+((1/2)*I)*2^(1/2)*x)))/(I*2^(1/2)*x-2*y)

(4.2.3)

pdetest(%, [pde__20, iv__20])

[0, 0]

(4.2.4)

pde__21 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
iv__21 := u(3, t) = sin(6*t)

ans__21 := pdsolve([pde__21, iv__21])

u(r, t) = -_F2(-(2*I)*ln(3)+I*ln(r)+t)+sin(-(6*I)*ln(3)+(6*I)*ln(r)+6*t)+_F2(-I*ln(r)+t)

(4.2.5)

pdetest(ans__21, [pde__21, iv__21])

[0, 0]

(4.2.6)

More PDE&BC problems are now solved by using a Fourier transform.

 

restart

Consider the following problem with an initial condition:

pde__22 := diff(u(x, t), t) = diff(u(x, t), x, x)+m
iv__22 := u(x, 0) = sin(x)

 

pdsolve can solve this problem directly:

ans__22 := pdsolve([pde__22, iv__22])

u(x, t) = sin(x)*exp(-t)+m*t

(5.1)

And we can check this answer against the original problem, if desired:

pdetest(ans__22, [pde__22, iv__22])

[0, 0]

(5.2)

How it works, step by step

 

Similarly to the Laplace transform method, we start the solution process by first applying the Fourier transform to the PDE:

with(inttrans)

transformed_PDE := fourier((lhs-rhs)(pde__22) = 0, x, s)

-2*m*Pi*Dirac(s)+s^2*fourier(u(x, t), x, s)+diff(fourier(u(x, t), x, s), t) = 0

(5.1.1)

Next, we call the function "fourier(u(x,t),x,s1)" by the new name U:

transformed_PDE_U := subs(fourier(u(x, t), x, s) = U(t, s), transformed_PDE)

-2*m*Pi*Dirac(s)+s^2*U(t, s)+diff(U(t, s), t) = 0

(5.1.2)

And this equation, which is really an ODE, is solved:

solution_U := dsolve(transformed_PDE_U, U(t, s))

U(t, s) = (2*m*Pi*Dirac(s)*t+_F1(s))*exp(-s^2*t)

(5.1.3)

Now, we apply the Fourier transform to the initial condition iv__22:

u(x, 0) = sin(x)

(5.1.4)

transformed_IC := fourier(iv__22, x, s)

fourier(u(x, 0), x, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))

(5.1.5)

Or, in the new variable U,

trasnformed_IC_U := U(0, s) = rhs(transformed_IC)

U(0, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))

(5.1.6)

Now, we evaluate solution_U at t = 0:

solution_U_at_IC := eval(solution_U, t = 0)

U(0, s) = _F1(s)

(5.1.7)

and substitute the transformed initial condition into it:

eval(solution_U_at_IC, {trasnformed_IC_U})

I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s)

(5.1.8)

Putting this into our solution_U, we get

eval(solution_U, {(rhs = lhs)(I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s))})

U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)

(5.1.9)

Finally, we apply the inverse Fourier transformation to this,

solution := u(x, t) = invfourier(rhs(U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)), s, x)

u(x, t) = sin(x)*exp(-t)+m*t

(5.1.10)

PDE&BC problems that used to require the option HINT = `+` to be solved are now solved automatically

 

The following two PDE&BC problems used to require the option HINT = `+` in order to be solved. This is now done automatically within pdsolve.

pde__23 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
iv__23 := u(1, t) = 0, u(2, t) = 5

ans__23 := pdsolve([pde__23, iv__23])

u(r, t) = 5*ln(r)/ln(2)

(6.1)

pdetest(ans__23, [pde__23, iv__23])

[0, 0, 0]

(6.2)

pde__24 := diff(u(x, y), y, y)+diff(u(x, y), x, x) = 6*x-6*y

iv__24 := u(x, 0) = x^3+11*x+1, u(x, 2) = x^3+11*x-7, u(0, y) = -y^3+1, u(4, y) = -y^3+109

ans__24 := pdsolve([pde__24, iv__24])

u(x, y) = x^3-y^3+11*x+1

(6.3)

pdetest(ans__24, [pde__24, iv__24])

[0, 0, 0, 0, 0]

(6.4)

``



Download PDE_and_BC_update.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

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Авторский опыт использования математической системы Maple и других компьютерных инструментов в школьном научном обществе

 Арина Козлова

E-mail: k_arina99@mail.ru; МБОУ «Школа  № 57» Кировского района г.Казани, 10 класс

 Научный руководитель –

Гибадуллина Алсу, учитель математики МБОУ «Школа  № 57» Кировского района г.Казани;

е-mail: gialid@mail.ru

 Аннотация. Рассмотрен авторский опыт использования математической системы Maple и других компьютерных инструментов для создания научно-популярных проектов физико-математического направления в рамках школьного научного общества.

 

На протяжении более 10 лет наша школа наряду с различными информационными технологиями работает с системой компьютерной математики Maple. Один из аспектов этой деятельности  –  научное общество учащихся «ГЕОДРОМчик», научным руководителем которого является учитель математики Гибадуллина А.И. Направления деятельности ученического научного общества – знакомство с пакетом Maple; освоение компьютерных инструментов, позволяющих работать с графикой, видео, создавать интерактивные меню; работа над индивидуальными научно-популярными проектами и создание авторских тематических электронных журналов, содержащих элементы научного исследования и математического моделирования. Компьютерная математика находит все более широкое применение – от научных исследований до продукции масскультур. Математическое моделирование проникло и в сферу создания рисунка, и в киноиндустрию. Изучение и использование учащимися нашего школьного общества символьных систем, в частности Maple, – это попытка приобщиться к современной мировой культуре компьютерного математического моделирования.

В данной статье описывается личный опыт автора, как одного из членов школьного НОУ.  

Знакомство с математической системой Maple началось с работы над проектом «Построение анимированной математической 3D-модели открывающейся книги» в 6-ом классе. Этот проект представляет собой создание пространственного анимированного изображения открывающейся книги средствами аналитической геометрии. В среде Maple была построена поэтапная программа получения этого изображения (таблицы 1 и 2). 

Таблица 1. Фрагмент программы получения анимированного изображения. 

> restart:

Подключение к дополнительным библиотекам

> with(plots):

> with(plottools):

Построение одной из страниц:

s1:= polygon([[0,0.01,0],[1,0.01,0],[1,1,0],[0,1,0]], thickness=1,color=orange):

Визуализация совокупных элементов книги:

display(k11,s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,k0,k27, title="KNIGA",scaling=constrained);

Поворот и анимация открывания обложки:

r_k11:=rotate(k11,10*Pi/9,[[0,0,0.29],[1,0,0.29]]):

plots[display](r_k11,kn_1,k0,scaling=constrained);

> anm:=seq(rotate(k11,t*Pi/9,[[0,0,0.29],[1,0,0.29]]),t=0..10):

> anim:=plots[display](anm,insequence=true):

> plots[display](kn_1,anim,scaling=constrained);

 

a)    

c)   

 

e)   

b)   

d)   

f)   

 Рис. 1. Кадры анимации книги

Следующий проект, выполненный в среде Maple совместно с Нигометзяновой Эльзой в 7-ом классе, – короткометражный мультфильм «Колобок в лесу».

a)      b)   

 Рис. 2. Кадры анимации мультфильма

В 8-ом классе велась работа по техническому переводу сайта компании Waterloo Maple Inc. [3]. Как известно, такой перевод имеет свои особенности, которые не предусмотрены в школьной программе по изучению английского языка, поэтому опыт такой работы способствует совершенствованию владения английским языком.

В 9-ом классе началась работа над электронным журналом по космологии «Вселенная: теория и факты». Черные дыры Вселенной – один из самых загадочных и любопытных для человека объектов. Их изучение привело к интересу к астрофизике вообще. Знакомство с понятием черной дыры неизбежно вынудило изучать строение Вселенной и ее геометрии [9, 10, 11, 12]. Пришлось осмысливать сложнейшие фундаментальные понятия, теории, а также элементы высшей математики [1, 5, 6, 7, 8]. Чтобы хотя бы попытаться понять огромный объем, казалось бы, беспорядочной информации, нужно было ее анализировать и систематизировать. И тогда возникла идея проекта – авторского электронного журнала. Тем более складывается парадоксальная ситуация: астрофизика бурно развивается, проникая практически во все сферы нашей жизни, а предмета астрономии в школе нет. Поэтому такой проект мог бы восполнить этот досадный пробел и помочь школьникам – и не только – в познании Вселенной. Журнал имеет следующие разделы: Вселенная, черные дыры, белые дыры, глоссарий, теории, неевклидовы геометрии, видео-опыты, интересные факты, ссылки, использованные ресурсы. Один из разделов журнала составляют Maple-разработки, в частности, визуализированная модель искривления пространства.

Далее приводится Maple-программа (табл. 2) построения визуализации деформации плоскости под шаром определенного размера. Используются библиотеки <plots> и <plottools> пакета.

 Таблица 2.  Maple–программа визуализации деформации плоскости. 

Комментарий

Команда и результат

Функция глубины "ямы"

( a - ширина "ямы", b - глубина )

f:=(x,a,b)->(-b*exp(-x^2/a^2));

 

Вводим параметры:

h - влияет на размеры тела-шарика и связывает их с шириной "ямы" ;

 k - влияет на диапазон площади вокруг "ямы"

h:=1:  k:=1:

 

Задание параметрическое прямой на поверхности (плоскости)

L0:=(m,n)->plot3d([0,r,f(r,m,n)], phi = -2*Pi ..2*Pi, r = -10k*h..10+k*h, scaling=CONSTRAINED,

numpoints=10000, color=blue,thickness=4):

Задание параметрическое поверхности (плоскости) путем кручения прямой

P0:=(m,n)->plot3d([r*cos(phi),r*sin(phi),f(r,m,n)], phi= 0..2*Pi,r=-8k*h..8+k*h, scaling=CONSTRAINED, numpoints=3000, style=POINT, color=blue):

Задание анимации искривления прямой

L:=plots[display](seq(L0(h,i),i=0..10+k*h), insequence=true):                    l:=plots[display](L,insequence=true):

Задание анимации искривления плоскости

p:=plots[display](seq(P0(h,i),i=0..10+k*h), insequence=true):

 p:=plots[display](P,insequence=true):

Задание анимация шарика ( тела, обладающего массой )

 

with(plottools):sp:=seq(sphere([0,0,-i-1.5*f(h,h,h)], f(h,h,h), style=HIDDEN,color=red),i=0..10+k*h):     

s:=plots[display](sp,insequence=true,

scaling=CONSTRAINED):

Совмещение всех компонентов модели визуализации

plots[display](p,s,l,scaling=CONSTRAINED);

 При h:=1:  k:=1:

 1)      2)      3)   

4)      5)      6)   

 

При h:=5:  k:=1:

7)      8)      9)   

Рис. 3. Кадры анимации при заданных параметрах.

Долго подбиралась функция глубины "ямы". Наконец, была найдена – это стало понятно после просмотра лекции А.Линде, где говорится об экспоненциальных процессах [13].

Меняя только параметры h и k (задающие размеры шара и ширины «ямы») и прокручивая программу снова, меняется и визуализация. Надо заметить, что построена всего лишь математическая модель визуализации, а не самого процесса.

Этот раздел предполагается пополнять новыми разработками, выполненными в среде Maple.

Журнал имеет удобную систему ссылок и организован так, что его можно оперативно обновлять. Астрофизика бурно развивается, поэтому журнал не потеряет своей актуальности.

 Заключение.

В течение 4-х лет занятий в научном обществе авторские проекты были представлены на различных сайтах, конкурсах, конференциях, форумах федерального и международного уровней:

  • сайт еxponenta.ru в разделе студенческих работ [4];
  • Конкурс исследовательских и творческих работ «Нобелевские надежды КНИТУ»
  • Республиканский конкурс «Арт-дебют»
  • V Международная ассамблея школьников (участие и публикация) [2]
  • Всероссийский Горчаковский форум в г.Санкт-Петербург
  • Поволжская научной конференция учащихся им. Н.И.Лобачевского
  • Всероссийский фестиваль «Нескучная наука» в г.Санкт-Петербург
  • Пост н.р. Гибадуллиной А.И. на сайте компании Maplesoft  http://www.mapleprimes.com/users/Alsu

  Использованная литература

 [1] Матросов А.В. Maple 6: Решение задач высшей математики и механики: Практическое руководство. – СПб.: БХВ – Петербург, 2001 г. – 528 с.

[2] V Международная Интеллектуальная Ассамблея школьников: сборник научно-исследовательских работ / Отв. ред. М. В. Волкова – Чебоксары: НИИ педагогики и психологии, 2012 – 136с. (с. 44–45)

[3] Сайт компании Maplesoft. – Режим доступа:  http://www.maplesoft.com

[4] Сайт <exponenta.ru> / Архив студенческих работ – Режим доступа:

http://www.exponenta.ru/educat/referat/XXIVkonkurs/5/index.asp

[5] Высшая математика: Учеб. Пособие для студентов пед. ин-тов по спец. 2120 «Общетехн. дисциплины и труд» / Г. Луканкин, Н. Мартынов, Г. Шадрин, Г. Яковлев; Под. ред. Г.Н. Яковлева. – М.: Просвещение, 1988. – 431 с.: ил.

[6] Справочник по высшей математике / М. Я. Выгодский. – М.: ООО «Издательство Астрель»: ООО «Издательство АСТ», 2002. – 992 с.: ил.

[7] Математический словарь высшей школы: Общ. часть/В. Т. Воднев, А. Ф. Наумович, Н.Ф. Наумович; Под ред. Ю.С. Богданова. – 2-е изд. – М.:Изд-во МПИ, 1988 – 527 с., ил.

[8] Толковый математический словарь. Основные термины: около 2500 терминов. – М.: Рус. яз., 1989. – 244 с., 186 ил.

[9] Открываем неевклидову геометрию. Кн. для внеклас. чтения учащихся 9-10 кл. сред. шк. – М.: Просвещение, 1988. – 126 с.: ил. – (Мир Знаний).

[10] Геометрия: Учебник для вузов. – СПб.: Издательство «Лань», 2003. – 416 с., ил. – (Учебники для вузов. Специальная литература)

[11] Основания геометрии: Учебн. пособие для вузов. – М.: Наука. Гл. ред. физ.-мат. лит., 1987. – 288 с.

[12] Обзорные лекции по геометрии к государственному экзамену по математике, Х семестр, курс лекций с примерами решений задач (в помощь выпускнику), проф. Ю.Г. Игнатьева. Программный продукт BIBLIO профессора Ю.Г. Игнатьева, Казань 2002 г.

[13] Видеозапись лекции Андрея Дмитриевича Линде, Стэнфордский университет (США), профессор «Многоликая Вселенная», прямая ссылка: http://elementy.ru/lib/430484

I'm an educator (physicist) who has migrated to Maple because of the lower "activation barrier" to get something of interest produced by the student. The students in my courses are exposed to several language (Python, C++, Java) and mathematical systems (Mathematica, Maple, MATLAB.) Many claim that unless forced to used a particular language or system, their first choice is Python and Maple for the reason I cite. 

As a consequence, it is my experience that students truly perfer the math-like appearance of the 2-D Math notation as opposed to the Maple notation. They see it as more natural - again with a lower activation barrier. Hence I see no reason to change. However, I would be interested in reasons why it might be beneficial.

My ultimate question is: do I start them with worksheet mode or documents mode? I'm use to worksheet mode and have found the call and response method easy for them to understand. But document mode has many valuable benefits. Is it worth the increase in learning (and frustration) for the benefits if the students use the software only a few times per semester? Or for some, every week?

I would be interested in hearing about the experiences of other educators.

 

I have encountered a behavior of Maple that I find hard to explain and I am hoping for help. The command

sum(floor((exp(Pi)-Pi)*n)/3^n,n=0..infinity);

was meant to be an example of "High-precision fraud" as in the 1992 paper of Borwein and Borwein, and indeed it gives 29/2 to within 531 digits. But I am unable to make Maple see this; indeed I get with evalf(%,1000)

14.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

I find it hard to guess why Maple gets this wrong, actually. The point of the example is that floor((exp(Pi)-Pi)*n)=20*n-1 for many n, but Maple has no problems finding the first failure at n=1112. It must hence be trying something more advanced than just adding up all summands until the tail sum is small enough to satisfy the precision? I guess an alternative approach would indeed be possible since what is being "floored" is relatively simple, but then that seems to be buggy?

Would be very grateful for any assistance!

Best,

Soren Eilers

 

 

 

Below is the worksheet with the whole material presented yesterday in the webinar, “Applying the power of computer algebra to theoretical physics”, broadcasted by the “Institute of Physics” (IOP, England). The material was very well received, rated 4.5 out of 5 (around 30 voters among the more than 300 attendants), and generated a lot of feedback. The webinar was recorded so that it is possible to watch it (for free, of course, click the link above, it will ask you for registration, though, that’s how IOP works).

Anyway, you can reproduce the presentation with the worksheet below (mw file linked at the end, or the corresponding pdf also linked with all the input lines executed). As usual, to reproduce the input/output you need to have installed the latest version of Physics, available in the Maplesoft R&D Physics webpage.

Why computer algebra?

 

 

 

... and why computer algebra?


We can concentrate more on the ideas instead of on the algebraic manipulations

 

We can extend results with ease

 

We can explore the mathematics surrounding a problem

 

We can share results in a reproducible way

 

Representation issues that were preventing the use of computer algebra in Physics

 

 


Notation and related mathematical methods that were missing:


coordinate free representations for vectors and vectorial differential operators,

covariant tensors distinguished from contravariant tensors,

functional differentiation, relativity differential operators and sum rule for tensor contracted (repeated) indices

Bras, Kets, projectors and all related to Dirac's notation in Quantum Mechanics

 

Inert representations of operations, mathematical functions, and related typesetting were missing:

 

inert versus active representations for mathematical operations

ability to move from inert to active representations of computations and viceversa as necessary

hand-like style for entering computations and textbook-like notation for displaying results

 

Key elements of the computational domain of theoretical physics were missing:

 

ability to handle products and derivatives involving commutative, anticommutative and noncommutative variables and functions

ability to perform computations taking into account custom-defined algebra rules of different kinds

(commutator, anticommutator and bracket rules, etc.)

 

 

Examples

 

The Maple computer algebra environment

   

Classical Mechanics

 

Inertia tensor for a triatomic molecule

   

Classical Field Theory

 

*The field equations for the lambda*Phi^4 model

   

*Maxwell equations departing from the 4-dimensional Action for Electrodynamics

   

*The Gross-Pitaevskii field equations for a quantum system of identical particles

   

Quantum mechanics

 

*The quantum operator components of  `#mover(mi("L",mathcolor = "olive"),mo("&rarr;",fontstyle = "italic"))` satisfy "[L[j],L[k]][-]=i `&epsilon;`[j,k,m] L[m]"

   

Quantization of the energy of a particle in a magnetic field

   

Unitary Operators in Quantum Mechanics

 

*Eigenvalues of an unitary operator and exponential of Hermitian operators

   

Properties of unitary operators

 

 

Consider two set of kets " | a[n] >" and "| b[n] >", each of them constituting a complete orthonormal basis of the same space.


One can always build an unitary operator U that maps one basis to the other, i.e.: "| b[n] >=U | a[n] >"

*Verify that "U=(&sum;) | b[k] >< a[k] |" implies on  "| b[n] >=U | a[n] >"

   

*Show that "U=(&sum;) | b[k] > < a[k] | "is unitary

   

*Show that the matrix elements of U in the "| a[n] >" and  "| b[n] >" basis are equal

   

Show that A and `&Ascr;` = U*A*`#msup(mi("U"),mo("&dagger;"))`have the same spectrum

   

````

Schrödinger equation and unitary transform

 

 

Consider a ket "| psi[t] > " that solves the time-dependant Schrödinger equation:

 

"i `&hbar;` (&PartialD;)/(&PartialD;t) | psi[t] >=H(t) | psi[t] >"

and consider

"| phi[t] > =U(t) | psi[t] >",

 

where U(t) is a unitary operator.

 

Does "| phi[t] >" evolves according a Schrödinger equation

 "i*`&hbar;` (&PartialD;)/(&PartialD;t) | phi[t] >=`&Hscr;`(t) | phi[t] >"

and if yes, which is the expression of `&Hscr;`(t)?

 

Solution

   

Translation operators using Dirac notation

 

In this section, we focus on the operator T[a] = exp((-I*a*P)*(1/`&hbar;`))

Settings

   

The Action (translation) of the operator T[a]"=(e)^(-i (a P)/(`&hbar;`))" on a ket

   

Action of T[a] on an operatorV(X)

   

General Relativity

 

*Exact Solutions to Einstein's Equations  Lambda*g[mu, nu]+G[mu, nu] = 8*Pi*T[mu, nu]

   

*"Physical Review D" 87, 044053 (2013)

 

Given the spacetime metric,

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -exp(lambda(r)), (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -r^2, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -r^2*sin(theta)^2, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = exp(nu(r))}))

a) Compute the Ricci and Weyl scalars

 

b) Compute the trace of

 

"Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

 

where `&equiv;`(Phi, Phi(r)) is some function of the radial coordinate, R[alpha, `~beta`] is the Ricci tensor, `&Dscr;`[alpha] is the covariant derivative operator and T[alpha, `~beta`] is the stress-energy tensor

 

T[alpha, beta] = (Matrix(4, 4, {(1, 1) = 8*exp(lambda(r))*Pi, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 8*r^2*Pi, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 8*r^2*sin(theta)^2*Pi, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 8*exp(nu(r))*Pi*epsilon}))

c) Compute the components of "W[alpha]^(beta)"" &equiv;"the traceless part of  "Z[alpha]^(beta)" of item b)

 

d) Compute an exact solution to the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)" obtained in c)

 

Background: paper from February/2013, "Withholding Potentials, Absence of Ghosts and Relationship between Minimal Dilatonic Gravity and f(R) Theories", by P. Fiziev.

 

a) The Ricci and Weyl scalars

   

b) The trace of "  Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

   

b) The components of "W[alpha]^(beta)"" &equiv;"the traceless part of " Z[alpha]^(beta)"

   

c) An exact solution for the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)"

   

*The Equivalence problem between two metrics

 

 

From the "What is new in Physics in Maple 2016" page:

  

In the Maple PDEtools package, you have the mathematical tools - including a complete symmetry approach - to work with the underlying [Einstein’s] partial differential equations. [By combining that functionality with the one in the Physics and Physics:-Tetrads package] you can also formulate and, depending on the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation.

Example from: A. Karlhede, "A Review of the Geometrical Equivalence of Metrics in General Relativity", General Relativity and Gravitation, Vol. 12, No. 9, 1980

   

*Equivalence for Schwarzschild metric (spherical and Krustal coordinates)

   

Tetrads and Weyl scalars in canonical form

 

 

Generally speaking a canonical form is obtained using transformations that leave invariant the tetrad metric in a tetrad system of references, so that theWeyl scalars are fixed as much as possible (conventionally, either equal to 0 or to 1).