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This presentation is on an undergrad intermediate Quantum Mechanics topic. Tackling the problem within a computer algebra worksheet in the way shown below is actually the novelty, using the Physics package to formulate the problem with quantum operators and related algebra rules in tensor notation.

 

Quantization of the Lorentz Force

 

Pascal Szriftgiser1 and Edgardo S. Cheb-Terrab2 

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

 

We consider the case of a quantum, non-relativistic, particle with mass m and charge q evolving under the action of an arbitrary time-independent magnetic field "B=Curl(A(x,y,z)), "where `#mover(mi("A",mathcolor = "olive"),mo("→"))` is the vector potential. The Hamiltonian for this system is

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

where `#mover(mi("p",mathcolor = "olive"),mo("→"))` is the momentum of the particle, and the force acting in this particle, also called the Lorentz force, is given by

 

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t))

 

where `#mover(mi("v",mathcolor = "olive"),mo("→"))` is the quantized velocity of the particle, and all of  H, `#mover(mi("p",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("A",mathcolor = "olive"),mo("→"))` and `#mover(mi("F",mathcolor = "olive"),mo("→"))` are Hermitian quantum operators representing observable quantities.

 

In the classic (non-quantum) case, `#mover(mi("F"),mo("→"))` for such a particle in the absence of electrical field is given by

 

`#mover(mi("F"),mo("→"))` = `&x`(q*`#mover(mi("v"),mo("→"))`, `#mover(mi("B"),mo("→"))`) ,

 

Problem: Departing from the Hamiltonian, show that in the quantum case the Lorentz force is given by [1]

 

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`))

 

[1] Photons et atomes, Introduction à l'électrodynamique quantique, p. 179, Claude Cohen-Tannoudji, Jacques Dupont-Roc et Gilbert Grynberg - EDP Sciences janvier 1987.

 

Solution

 

We choose to tackle the problem in Heisenberg's picture of quantum mechanices, where the state of a system is static and only the quantum operators evolve in time according to

``

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`ℏ`

 

Also, the algebraic manipulations are simpler using tensor abstract notation instead of the standard 3D vector notation. We then start setting the framework for the problem, a system of coordinates X, indicating the dimension of the tensor space to be 3 and the metric Euclidean, and that we will use lowercaselatin letters to represent tensor indices. In addition, not necessary but for convenience, we set the lowercase latin i to represent the imaginary unit and we request automaticsimplification so that the output of everything comes automatically simplified in size.

 

restart; with(Physics); interface(imaginaryunit = i)

Setup(mathematicalnotation = true, automaticsimplification = true, coordinates = X, dimension = 3, metric = Euclidean, spacetimeindices = lowercaselatin, quiet)

[automaticsimplification = true, coordinatesystems = {X}, dimension = 3, mathematicalnotation = true, metric = {(1, 1) = 1, (2, 2) = 1, (3, 3) = 1}, spacetimeindices = lowercaselatin]

(1)

 

Next we indicate the letters we will use to represent the quantum operators with which we will work, and also the standard commutation rules between position and momentum, always the starting point when dealing with quantum mechanics problems

 

Setup(quantumoperators = {F}, hermitianoperators = {A, B, H, p, r, v, x}, realobjects = {`ℏ`, m, q}, algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0}, hermitianoperators = {A, B, H, p, r, v, x}, quantumoperators = {A, B, F, H, p, r, v, x}, realobjects = {`ℏ`, m, q, x1, x2, x3, %dAlembertian, Physics:-dAlembertian}]

(2)

 

Note that we start not indicating F as Hermitian, in order to arrive at that result. The quantum operators A, B, and F are explicit functions of X, so to avoid redundant display of this functionality on the screen we use

 

CompactDisplay((A, B, F)(X))

A(x1, x2, x3)*`will now be displayed as`*A

 

B(x1, x2, x3)*`will now be displayed as`*B

 

F(x1, x2, x3)*`will now be displayed as`*F

(3)

Define now as tensors the quantum operators that we will use with tensorial notation (recalling: for these, Einstein's sum rule for repeated indices will be automatically applied when simplifying)

 

Define(x, p, v, A, B, F, quiet)

{A, B, F, p, v, x, Physics:-Dgamma[a], Physics:-Psigma[a], Physics:-d_[a], Physics:-g_[a, b], Physics:-KroneckerDelta[a, b], Physics:-LeviCivita[a, b, c], Physics:-SpaceTimeVector[a](X)}

(4)

The Hamiltonian,

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

in tensorial notation, is given by

H = (p[n]-q*A[n](X))^2/(2*m)

H = (1/2)*Physics:-`^`(p[n]-q*A[n](X), 2)/m

(5)

Generally speaking to arrive at  ```#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`)) what we now need to do is

1) Express this Hamiltonian (5) in terms of the velocity

 

And, recalling that, in Heisenberg's picture, quantum operators evolve in time according to

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`ℏ`

 

2) Take the commutator of H with the velocity itself to obtain its time derivative and, from `#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t)) , that commutator is already the force up to some constant factors.

 

To get in contact with the basic commutation rules between position and momentum behind quantum phenomena, the quantized velocity itself can be computed as the time derivative of the position operator, i.e as the commutator of x[k] with H

I*Commutator(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, x[k])/`ℏ`

I*Physics:-Commutator(H, x[k])/`ℏ` = (1/2)*(I*q^2*Physics:-AntiCommutator(A[n](X), Physics:-Commutator(A[n](X), x[k]))-I*q*Physics:-AntiCommutator(p[n], Physics:-Commutator(A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics:-KroneckerDelta[k, n]*`ℏ`)/(`ℏ`*m)

(6)

This expression for the velocity, that involves commutators between the potential A[n](X), the position x[k] and the momentum p[n], can be simplified taking into account the basic quantum algebra rules between position and momentum. We assume that A[n](X)(X) can be decomposed into a formal power series (possibly infinite) of the x[k], hence all the A[n](X) commute between themselves as well as with all the x[k]

 

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

(7)

(Note: in some cases, this is not true, but those cases are beyond the scope of this worksheet.)

 

Add these rules to the algebra rules already set so that they are all taken into account when simplifying things

 

Setup(algebrarules = {%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(8)

Simplify(I*Physics[Commutator](H, x[k])/`ℏ` = (1/2)*(I*q^2*Physics[AntiCommutator](A[n](X), Physics[Commutator](A[n](X), x[k]))-I*q*Physics[AntiCommutator](p[n], Physics[Commutator](A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics[KroneckerDelta][k, n]*`ℏ`)/(`ℏ`*m))

I*Physics:-Commutator(H, x[k])/`ℏ` = (-A[k](X)*q+p[k])/m

(9)

The right-hand side of (9) is then the kth component of the velocity tensor quantum operator, the relationship is the same as in the classical case

v[k] = rhs(I*Physics[Commutator](H, x[k])/`ℏ` = (-A[k](X)*q+p[k])/m)

v[k] = (-A[k](X)*q+p[k])/m

(10)

and with this the Hamiltonian (5) can now be rewritten in term of the velocity completing step 1)

simplify(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, {SubstituteTensorIndices(k = n, (rhs = lhs)(v[k] = (-A[k](X)*q+p[k])/m))})

H = (1/2)*m*Physics:-`^`(v[n], 2)

(11)

For step 2), to compute

 `#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t)) and m*(diff(v(t), t)) = I*m*Physics:-Commutator(H, v(t)[k])/`ℏ` 

 

we need the commutator between the different components of the quantized velocity which, contrary to what happens in the classical case, do not commute. For this purpose, take the commutator between (10) with itself after replacing the free index

Commutator(v[k] = (-A[k](X)*q+p[k])/m, SubstituteTensorIndices(k = n, v[k] = (-A[k](X)*q+p[k])/m))

Physics:-Commutator(v[k], v[n]) = -q*(Physics:-Commutator(A[k](X), p[n])+Physics:-Commutator(p[k], A[n](X)))/m^2

(12)

To simplify (12), we use the fact that if f  is a commutative mapping that can be decomposed into a formal power series in all the complex plan (which is assumed to be the case for all A[n](X)(X)), then

Physics:-Commutator(p[k], f(x, y, z)) = -I*`ℏ`*`∂`[k](f(x, y, z))

where p[k]"=-i `ℏ` `∂`[k] " is the momentum operator along the x[k] axis. This relation reads in tensor notation:

Commutator(p[k], A[n](X)) = -I*`ℏ`*d_[k](A[n](X))

Physics:-Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X])

(13)

Add this rule to the rules previously set in order to automatically take it into account in (12)

Setup(Physics[Commutator](p[k], A[n](X)) = -I*`ℏ`*Physics[d_][k](A[n](X), [X]))

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X]), %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(14)

Physics[Commutator](v[k], v[n]) = -q*(Physics[Commutator](A[k](X), p[n])+Physics[Commutator](p[k], A[n](X)))/m^2

Physics:-Commutator(v[k], v[n]) = -I*q*`ℏ`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2

(15)

Also add this other rule so that it is taken into account automatically

Setup(Physics[Commutator](v[k], v[n]) = -I*q*`ℏ`*(Physics[d_][n](A[k](X), [X])-Physics[d_][k](A[n](X), [X]))/m^2)

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X]), %Commutator(v[k], v[n]) = -I*q*`ℏ`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(16)

Recalling now the expression of the Hamiltonian (11) as a function of the velocity, one can compute the components of the force operator  "()Component(v*B,k)=m (v[k])=(i m [H,v[k]][-])/`ℏ`"

F[k](X) = I*m*%Commutator(rhs(H = (1/2)*m*Physics[`^`](v[n], 2)), v[k])/`ℏ`

F[k](X) = I*m*%Commutator((1/2)*m*Physics:-`^`(v[n], 2), v[k])/`ℏ`

(17)

Simplify this expression for the quantized force taking the quantum algebra rules (16) into account

Simplify(F[k](X) = I*m*%Commutator((1/2)*m*Physics[`^`](v[n], 2), v[k])/`ℏ`)

F[k](X) = (1/2)*q*(-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])+Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))+Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X])))

(18)

It is not difficult to verify that this is the antisymmetrized vector product `&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`). Departing from `#mover(mi("B",mathcolor = "olive"),mo("→"))` = `&x`(VectorCalculus[Nabla], `#mover(mi("A",mathcolor = "olive"),mo("→"))`) expressed using tensor notation,

B[c](X) = LeviCivita[c, n, m]*d_[n](A[m](X))

B[c](X) = -Physics:-LeviCivita[c, m, n]*Physics:-d_[n](A[m](X), [X])

(19)

and taking into acount that

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`), k) = `ε`[b, c, k]*v[b]*B[c](X) 

multiply both sides of (19) by `ε`[b, c, k]*v[b], getting

LeviCivita[k, b, c]*v[b]*(B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = -Physics:-LeviCivita[b, c, k]*Physics:-LeviCivita[c, m, n]*Physics:-`*`(v[b], Physics:-d_[n](A[m](X), [X]))

(20)

Simplify(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = -Physics[LeviCivita][b, c, k]*Physics[LeviCivita][c, m, n]*Physics[`*`](v[b], Physics[d_][n](A[m](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[m], Physics:-d_[k](A[m](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(21)

Finally, replacing the repeated index m by n 

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[m], Physics[d_][k](A[m](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(22)

Likewise, for

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`), k) = `ε`[b, c, k]*B[b]*B[c](X) 

multiplying (19), this time from the right instead of from the left, we get

Simplify(((B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))*LeviCivita[k, b, c])*v[b])

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[m](X), [X]), v[m])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(23)

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[m](X), [X]), v[m])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(24)

Simplifying now the expression (18) for the quantized force taking into account (22) and (24) we get

simplify(F[k](X) = (1/2)*q*(-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n])+Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))+Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))), {(rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))), (rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))})

F[k](X) = (1/2)*q*Physics:-LeviCivita[b, c, k]*(Physics:-`*`(v[b], B[c](X))+Physics:-`*`(B[c](X), v[b]))

(25)

i.e.

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`))

in tensor notation. Finally, we note that this operator is Hermitian as expected

(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))-Dagger(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))

F[k](X)-Physics:-Dagger(F[k](X)) = 0

(26)



Download:  Quantization_of_the_Lorentz_force.mw,   Quantization_of_the_Lorentz_force.pdf


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

The presentation below is on undergrad Quantum Mechanics. Tackling this topic within a computer algebra worksheet in the way it's done below, however, is an exciting novelty and illustrates well the level of abstraction that is now possible using the Physics package.

 

Quantum Mechanics: Schrödinger vs Heisenberg picture

Pascal Szriftgiser1 and Edgardo S. Cheb-Terrab2 

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

 

Within the Schrödinger picture of Quantum Mechanics, the time evolution of the state of a system, represented by a Ket "| psi(t) >", is determined by Schrödinger's equation:

I*`ℏ`*(diff(Ket(psi, t), t)) = H*Ket(psi, t)

where H, the Hamiltonian, as well as the quantum operators O__S representing observable quantities, are all time-independent.

 

Within the Heisenberg picture, a Ket Ket(psi, 0) representing the state of the system does not evolve with time, but the operators O__H(t)representing observable quantities, and through them the Hamiltonian H, do.

 

Problem: Departing from Schrödinger's equation,

  

a) Show that the expected value of a physical observable in Schrödinger's and Heisenberg's representations is the same, i.e. that

Bra(psi, t)*O__S*Ket(psi, t) = Bra(psi, 0)*O__H(t)*Ket(psi, 0)

  

b) Show that the evolution equation of an observable O__H in Heisenberg's picture, equivalent to Schrödinger's equation,  is given by:

diff(O__H(t), t) = (-I*Physics:-Commutator(O__H(t), H))*(1/`ℏ`)

where in the right-hand-side we see the commutator of O__H with the Hamiltonian of the system.

Solution: Let O__S and O__H respectively be operators representing one and the same observable quantity in Schrödinger's and Heisenberg's pictures, and H be the operator representing the Hamiltonian of a physical system. All of these operators are Hermitian. So we start by setting up the framework for this problem accordingly, including that the time t and Planck's constant are real. To automatically combine powers of the same base (happening frequently in what follows) we also set combinepowersofsamebase = true. The following input/output was obtained using the latest Physics update (Aug/31/2016) distributed on the Maplesoft R&D Physics webpage.

with(Physics):

Physics:-Setup(hermitianoperators = {H, O__H, O__S}, realobjects = {`ℏ`, t}, combinepowersofsamebase = true, mathematicalnotation = true)

[combinepowersofsamebase = true, hermitianoperators = {H, O__H, O__S}, mathematicalnotation = true, realobjects = {`ℏ`, t}]

(1)

Let's consider Schrödinger's equation

I*`ℏ`*(diff(Ket(psi, t), t)) = H*Ket(psi, t)

I*`ℏ`*(diff(Physics:-Ket(psi, t), t)) = Physics:-`*`(H, Physics:-Ket(psi, t))

(2)

Now, H is time-independent, so (2) can be formally solved: psi(t) is obtained from the solution psi(0) at time t = 0, as follows:

T := exp(-I*H*t/`ℏ`)

exp(-I*t*H/`ℏ`)

(3)

Ket(psi, t) = T*Ket(psi, 0)

Physics:-Ket(psi, t) = Physics:-`*`(exp(-I*t*H/`ℏ`), Physics:-Ket(psi, 0))

(4)

To check that (4) is a solution of (2), substitute it in (2):

eval(I*`ℏ`*(diff(Physics[Ket](psi, t), t)) = Physics[`*`](H, Physics[Ket](psi, t)), Physics[Ket](psi, t) = Physics[`*`](exp(-I*H*t/`ℏ`), Physics[Ket](psi, 0)))

Physics:-`*`(H, exp(-I*t*H/`ℏ`), Physics:-Ket(psi, 0)) = Physics:-`*`(H, exp(-I*t*H/`ℏ`), Physics:-Ket(psi, 0))

(5)

Next, to relate the Schrödinger and Heisenberg representations of an Hermitian operator O representing an observable physical quantity, recall that the value expected for this quantity at time t during a measurement is given by the mean value of the corresponding operator (i.e., bracketing it with the state of the system Ket(psi, t)).

So let O__S be an observable in the Schrödinger picture: its mean value is obtained by bracketing the operator with equation (4):

Dagger(Ket(psi, t) = Physics[`*`](exp(-I*H*t/`ℏ`), Ket(psi, 0)))*O__S*(Ket(psi, t) = Physics[`*`](exp(-I*H*t/`ℏ`), Ket(psi, 0)))

Physics:-`*`(Physics:-Bra(psi, t), O__S, Physics:-Ket(psi, t)) = Physics:-`*`(Physics:-Bra(psi, 0), exp(I*t*H/`ℏ`), O__S, exp(-I*t*H/`ℏ`), Physics:-Ket(psi, 0))

(6)

The composed operator within the bracket on the right-hand-side is the operator O in Heisenberg's picture, O__H(t)

Dagger(T)*O__S*T = O__H(t)

Physics:-`*`(exp(I*t*H/`ℏ`), O__S, exp(-I*t*H/`ℏ`)) = O__H(t)

(7)

Analogously, inverting this equation,

(T*(Physics[`*`](exp(I*H*t/`ℏ`), O__S, exp(-I*H*t/`ℏ`)) = O__H(t)))*Dagger(T)

O__S = Physics:-`*`(exp(-I*t*H/`ℏ`), O__H(t), exp(I*t*H/`ℏ`))

(8)

As an aside to the problem, we note from these two equations, and since the operator T = exp((-I*H*t)*(1/`ℏ`)) is unitary (because H is Hermitian), that the switch between Schrödinger's and Heisenberg's pictures is accomplished through a unitary transformation.

 

Inserting now this value of O__S from (8) in the right-hand-side of (6), we get the answer to item a)

lhs(Physics[`*`](Bra(psi, t), O__S, Ket(psi, t)) = Physics[`*`](Bra(psi, 0), exp(I*H*t/`ℏ`), O__S, exp(-I*H*t/`ℏ`), Ket(psi, 0))) = eval(rhs(Physics[`*`](Bra(psi, t), O__S, Ket(psi, t)) = Physics[`*`](Bra(psi, 0), exp(I*H*t/`ℏ`), O__S, exp(-I*H*t/`ℏ`), Ket(psi, 0))), O__S = Physics[`*`](exp(-I*H*t/`ℏ`), O__H(t), exp(I*H*t/`ℏ`)))

Physics:-`*`(Physics:-Bra(psi, t), O__S, Physics:-Ket(psi, t)) = Physics:-`*`(Physics:-Bra(psi, 0), O__H(t), Physics:-Ket(psi, 0))

(9)

where, on the left-hand-side, the Ket representing the state of the system is evolving with time (Schrödinger's picture), while on the the right-hand-side the Ket `ψ__0`is constant and it is O__H(t), the operator representing an observable physical quantity, that evolves with time (Heisenberg picture). As expected, both pictures result in the same expected value for the physical quantity represented by O.

 

To complete item b), the derivation of the evolution equation for O__H(t), we take the time derivative of the equation (7):

diff((rhs = lhs)(Physics[`*`](exp(I*H*t/`ℏ`), O__S, exp(-I*H*t/`ℏ`)) = O__H(t)), t)

diff(O__H(t), t) = I*Physics:-`*`(H, exp(I*t*H/`ℏ`), O__S, exp(-I*t*H/`ℏ`))/`ℏ`-I*Physics:-`*`(exp(I*t*H/`ℏ`), O__S, H, exp(-I*t*H/`ℏ`))/`ℏ`

(10)

To rewrite this equation in terms of the commutator  Physics:-Commutator(O__S, H), it suffices to re-order the product  H  exp(I*H*t/`ℏ`) placing the exponential first:

Library:-SortProducts(diff(O__H(t), t) = I*Physics[`*`](H, exp(I*H*t/`ℏ`), O__S, exp(-I*H*t/`ℏ`))/`ℏ`-I*Physics[`*`](exp(I*H*t/`ℏ`), O__S, H, exp(-I*H*t/`ℏ`))/`ℏ`, [exp(I*H*t/`ℏ`), H], usecommutator)

diff(O__H(t), t) = I*Physics:-`*`(exp(I*t*H/`ℏ`), H, O__S, exp(-I*t*H/`ℏ`))/`ℏ`-I*Physics:-`*`(exp(I*t*H/`ℏ`), Physics:-`*`(H, O__S)+Physics:-Commutator(O__S, H), exp(-I*t*H/`ℏ`))/`ℏ`

(11)

Normal(diff(O__H(t), t) = I*Physics[`*`](exp(I*H*t/`ℏ`), H, O__S, exp(-I*H*t/`ℏ`))/`ℏ`-I*Physics[`*`](exp(I*H*t/`ℏ`), Physics[`*`](H, O__S)+Physics[Commutator](O__S, H), exp(-I*H*t/`ℏ`))/`ℏ`)

diff(O__H(t), t) = -I*Physics:-`*`(exp(I*t*H/`ℏ`), Physics:-Commutator(O__S, H), exp(-I*t*H/`ℏ`))/`ℏ`

(12)

Finally, to express the right-hand-side in terms of  Physics:-Commutator(O__H(t), H) instead of Physics:-Commutator(O__S, H), we take the commutator of the equation (8) with the Hamiltonian

Commutator(O__S = Physics[`*`](exp(-I*H*t/`ℏ`), O__H(t), exp(I*H*t/`ℏ`)), H)

Physics:-Commutator(O__S, H) = Physics:-`*`(exp(-I*t*H/`ℏ`), Physics:-Commutator(O__H(t), H), exp(I*t*H/`ℏ`))

(13)

Combining these two expressions, we arrive at the expected result for b), the evolution equation of a given observable O__H in Heisenberg's picture

eval(diff(O__H(t), t) = -I*Physics[`*`](exp(I*H*t/`ℏ`), Physics[Commutator](O__S, H), exp(-I*H*t/`ℏ`))/`ℏ`, Physics[Commutator](O__S, H) = Physics[`*`](exp(-I*H*t/`ℏ`), Physics[Commutator](O__H(t), H), exp(I*H*t/`ℏ`)))

diff(O__H(t), t) = -I*Physics:-Commutator(O__H(t), H)/`ℏ`

(14)


Download:    Schrodinger_vs_Heisenberg_picture.mw     Schrodinger_vs_Heisenberg_picture.pdf

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

I am solving "Fx=0" for geting "roots:x" using "solve(Fx,x)". Solution is in the form of "a+sqrt(b)", "a-sqrt(b)". One solution "f1" is given below.

f1:=1/2*(-8*R*d1^2*r^2*C+10*d1*r^2*C*R+5*d1*r^3*C+2*r*L*d1^2-2*C*r^3+2*R*L*d1^2-R*L*d1^3-r*L*d1^3-4*C*r^2*R+2*R*d1^3*r^2*C-4*r^3*d1^2*C+r^3*d1^3*C+sqrt(26*r^6*d1^4*C^2+41*r^6*d1^2*C^2-44*r^6*d1^3*C^2-20*C^2*r^6*d1+16*C^2*r^5*R-16*C*r^4*L-176*r^5*d1^3*C^2*R+164*r^5*d1^2*C^2*R-74*r^4*d1^4*C*L+136*r^4*d1^3*C*L-136*r^4*d1^2*C*L-80*C^2*r^5*d1*R+72*C*r^4*L*d1-64*C*r^3*R*L+104*R^2*d1^4*r^4*C^2-176*R^2*d1^3*r^4*C^2+164*R^2*d1^2*r^4*C^2-8*r^6*d1^5*C^2+r^2*L^2*d1^6-4*R^2*L^2*d1^5+104*r^5*d1^4*C^2*R+40*r*L*R^3*d1^5*C-72*r*L*R^3*d1^4*C+56*r*L*R^3*d1^3*C-16*r*L*R^3*d1^2*C+R^2*L^2*d1^6+20*r^4*L*C*d1^5-32*r^4*d1^5*C^2*R^2+2*R*L^2*d1^6*r-2*r^4*L*d1^6*C+4*R^2*d1^6*r^4*C^2+4*R*d1^6*r^5*C^2-306*r^3*d1^4*C*R*L+548*r^3*d1^3*C*L*R-544*r^3*d1^2*C*R*L+288*C*r^3*L*d1*R+16*C^2*r^4*R^2+4*R^2*L^2*d1^4-16*R^2*L*d1^6*r^2*C-10*R*L*d1^6*r^3*C+r^6*d1^6*C^2-32*r^5*d1^5*C^2*R-4*r^2*d1^5*L^2-352*R^2*d1^4*r^2*C*L+580*R^2*d1^3*r^2*C*L-552*R^2*d1^2*r^2*C*L-80*d1*r^4*C^2*R^2-8*R^3*d1^6*L*C*r+88*r^3*L*C*d1^5*R+116*r^2*L*R^2*d1^5*C+4*C^2*r^6-8*r*R*L^2*d1^5+288*d1*r^2*C*R^2*L-64*C*r^2*R^2*L+8*r*L^2*d1^4*R+4*r^2*L^2*d1^4)^(1/2))/(-3*r^2*d1*L*C-6*R*d1*L*C*r+2*L*C*r^2+r^2*d1^2*L*C+4*L*C*r*R+2*R*d1^2*L*C*r);

I used the following Maple syntax

patmatch(f1,XT::algebraic+sqrt(YT::algebraic),'q1');

the answer is "false"

Is there any modification in the syntax "patmatch" is required.

Here, my question is how to separate "a" and "b" in "a+sqrt(b)" (a, b are big expressions involving many variables).

Thanking you advance for your help.

 

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  kinematic analysis 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

``

``

``

``

``

``

``

``

``

``

``

``

``

``

``

``

 

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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