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Transformation of some variables with ":-"...

Asked by: Bendesarts 450
trigonometric
June 30 2016
0  3

Hello,

In some trigonometric equations, I have variables depending of time like Psi(t). I don't why after some manipulation my variable is transformed in :-Psi(t).

Have some ideas why some variables can be change with :- before ?

Here an example

ResolTrig.mw

It may come because of the code "Local Psi". I added this because it seemed to me that Psi was protected. Do I right? If not can I remove Local Psi?

Thanks a lot for your help

running the code in maple...

Asked by: sadeq 20
June 29 2016
0  0

hello friend

please run this code and resend result to me.

thank you.

 

restart;
with(LinearAlgebra); with(plots);
c[1] := 0; c[2] := 1/2;
a[2, 1] := 1/8;
b[-1] := -1/3; b[1] := 2/3; b[2] := 5/6;
bb[2] := 5/12;
s := 2;
e[1] := Matrix(s+1, 1); e[2] := Matrix(s+1, 1); e[3] := Matrix(s+1, 1); e[4] := Matrix(s+1, 1);
c[0] := 0;
for ii to s+1 do e[1][ii, 1] := 1; e[2][ii, 1] := -1; e[3][ii, 1] := c[ii-1] end do;
e[1][1, 1] := 0; e[2][2, 1] := 0; e[4] := -e[3];
A := Matrix(s+1, s+1);
for i from 3 to s+1 do for j from 2 to i-1 do A[i, j] := a[i-1, j-1] end do; A[i, 1] := A[i, 2] end do;
Id := Matrix(s+1, s+1, shape = identity);
N := Matrix(s+1, s+1);
for i to s+1 do for j to s+1 do N[i, j] := -H*A[i, j]+Id[i, j] end do end do;
Bb := Matrix(s+1, 1);
for i from 3 to s+1 do Bb[i, 1] := bb[i-1] end do;
B := Matrix(s+1, 1);
for i from 2 to s+1 do B[i, 1] := b[i-1] end do; B[1, 1] := b[-1];
Z := Multiply(Transpose(Bb), 1/N);
for i to s+1 do Z[1, i] := H*Z[1, i] end do;
Q := Multiply(Transpose(B), 1/N);
for i to s+1 do Q[1, i] := H*Q[1, i] end do;
Dh := proc (H) options operator, arrow; Matrix([[1+Multiply(Z, e[1]), Multiply(Z, e[2]), 3/2+Multiply(Z, e[3]), -1/2+Multiply(Z, e[4])], [1, 0, 0, 0], [Multiply(Q, e[1]), Multiply(Q, e[2]), 1+Multiply(Q, e[3]), Multiply(Q, e[4])], [0, 0, 1, 0]]) end proc;
k := Matrix(4, 4);
for i to 4 do for j to 4 do if i = j then k[i, j] := ep-Dh(H)[i, j] else k[i, j] := -Dh(H)[i, j] end if end do end do;
sp := unapply(collect(Determinant(k), ep), H, ep);
sol := solve(sp(H, ep), ep, explicit);
sol := simplify(map(allvalues, {sol}));
l := numelems(sol);
inequal({evalc(abs(subs(H = x+I*y, sol[1]))) <= 1, evalc(abs(subs(H = x+I*y, sol[2]))) <= 1, evalc(abs(subs(H = x+I*y, sol[3]))) <= 1, evalc(abs(subs(H = x+I*y, sol[4]))) <= 1}, x = -3 .. 3, y = -3 .. 3, color = "Nautical 1");

 

maple code of heat transfer for CM and LSM method...

Asked by: Tom2586 10
June 29 2016
1  5

hi every one
please help me for this problem
f

New version of the Multivariate Calculus Study...

by: eithne 2072 Maple , Maple Toolboxes
June 29 2016
2  0

We have just released a new version of the Multivariate Calculus Study Guide.  It provides a new section on Vector Calculus, with over 100 additional worked problems, and makes extensive use of Maple’s Clickable Math tools as well as commands.

Existing study guide customers can get the new content via a free update, available through the Check for Updates system or from our website. See Multivariate Calculus Study Guide 2016 Update for details.

For more information about this guide, including a full table of contents, visit Multivariate Calculus Study Guide.

 

eithne

Maple can't solve it....wolframalpha does...

Asked by: Giulianot 25
June 29 2016
2  14

Hi, i got this diff ecuation to answer. It is an ecuation of a proyectile, it give the angle of speed with respect to time. I typed it on wolfram; got an aswer. When i tried on maple without any initial condition, got a good answer. With an initial condition, didn't get nothing. I even tried with the interactive solver, got a 'can't solve'. So this is my ecuation:

dx/dt = -[g cos(x)] / [vo - g t sin(x)]

I had to find the initial coindition for myself after that, i supect what the problem is. Maple gave an aswer in terms of arctan (y,x), and i think, based on my own try, it doesn't know how to find a term if it's inside the arctan function, typed in that way. So i rewrited it like arctan (y/x), anf found my constant C1.

 

Another problem i got was about this system of diff ecuations:

dx/dt = (-k/m) (x srqt(x^2 + y^2))

dy/dt = (-k/m) (y srqt(x^2 + y^2))

Again was wolfram who gave me a good answer to the problem. So you are going to understand im thinking wolfram is better on symbolic calculation. Thanks anyway for your help and effort.

 

 

 

 

Maple Worksheet frustration ...

Asked by: Wtolrud 75
June 29 2016
1  4

How do I turn off Worksheet mode in a Maple Document.  I copied a calculation from a Worksheet into my Document, now I have to do calculation in 1-D math. 

Any advice would be helpful. 

Why can't I use this while in a procedure?...

Asked by: AmirHosein Sadeghimanesh 225
procedure while
June 29 2016
0  4

I want to use while in proc but Maple shows an error saying

 

"Error, (in exa) illegal use of a formal parameter"

 

When I click on it, as usual it says "There is no help page available for this error". I tried a very simple example of while which you can see it below and it again shows same error. The while works well but when I put it inside of a procedure it makes an error. Note: the following example is just an example, this is not what I'm doing so please don't reply this while proc is not giving anything. My question is how can I use a while in procedure without encountering the above error.

 

i := 10;
while i > 1 do
if i mod 2 = 0 then i := i/2; else i := i+1; end if;
end do;

 

Now with proc

 

exa := proc (i::integer)::integer;
while 1 < i do
if i mod 2 = 0 then i :=i/2; else i := i+1; end if
end do; 
end proc;

exa(10);

"Error, (in exa) illegal use of a formal parameter"

How to find a "smart" form of the converse of a bo...

Asked by: sand15 1189
logic inequality
June 29 2016
1  3



 
No need to hurry, esthetics is not a vital issue ... but thanks in advance.

PS : sorry for the syntax errors "waves" generated in the original Word document

Symmetries of strange PDE...

Asked by: MrYouMath 40
pde liealgebra symmetry
June 29 2016
2  11

I thought about the following PDE:

 

EDIT: The u(0,t) is not a typo! It is really meant to be part of the PDE!

 

Latex/Matjax: $$\dfrac{\partial u(x,t)}{\partial t}=\alpha \dfrac{\partial^2 u(x,t)}{\partial x^2}+u(x=0,t).$$

Maple: diff(u(x,t),t)=alpha*diff(u(x,t),x$2)+u(0,t)

 

 

How can i determine the symmetries of this PDE with Maple?

Help me fix my mathematics maple code problem...

Asked by: candy898 60
procedure programming module
June 28 2016
0  13

I need some help fix my mathematics maple code problem , I define everything my Procedures,and procedures run is ok, but when I using call functions for another procedures  I can not operator my program . For sure procedures is work and correct. if somebody can help me , I really appreciate you help.

Please help me with this DTM code...

Asked by: kencom1 115
numapprox
June 28 2016
1  1

restart; Digits := 10; Ha := 2; R := 2; `&theta;r` := .5; Rt := 1; B := 1.5; Xi := 0; U[0] := 0; U[1] := alpha; U[2] := -6+(1/16)*Rt*Xi; U[3] := (1/6)*beta; Theta[0] := -(1/2)*Rt Theta[1]:=phi:   delta:=(k)->`if`(k=0, 1,0);   for k  from 0 to 20do    U[k+4]:=((Ha^(2)+R)*U[k+2]+Xi*Theta[k+2])/(4(k+3)*(k+4)) 

Theta[k+2] := (-4*R*`&theta;r`*(`&theta;r`*Theta[k]+1)^2*(sum((i+1)*Theta[i+1]*(k-i+1)*Theta[k-i+1], i = 0 .. k))-B*(sum((i+1)*U[i+1]*(k-i+1)*U[k-i+1], i = 0 .. k))+(1/4)*Ha^2*B*(sum(U[i]*(k-i)*U[k-i], i = 0 .. k)))/((1+(4/3)*R*(`&theta;r`*Theta[k]+1)^3)*(k+1)*(k+2))

 od:  u:=0:  theta:=0:  for k from 0 to 20 do  u:=u+U[k]*y^k:  theta:=theta+Theta[k]*y^k:  od:  print(expand(u)):

with(numapprox); pade(u, y, [3, 3])

pade(diff(u, `$`(y, 2)), y, [3, 3])

pade(theta, y, [3, 3])

solve({limit(pade(theta, y, [3, 3]), y = 1) = (1/2)*Rt, limit(pade(u, y, [3, 3]), y = 1) = 0, limit(pade(diff(u, `$`(y, 2)), y, [3, 3]), y = 1) = -12-(1/8)*Rt*Xi}, [alpha, beta, phi])

Symbolic resolution of trigonometric equations sys...

Asked by: Bendesarts 450
symbolic resolution trigonometric
June 28 2016
0  4

Hello,

I would like to determine a closed form solution (=analytical solution) of the following trigonometric equations system.

TrigonometricSystem2.mw

The unknowns are :

ListAllUnknowns := [Psi(t), Theta[1](t), Theta[2](t), x[1](t), x[2](t), z[1](t), z[2](t)]

Do you have ideas so as to conduct the symbolic resolution of this trigonometric equations system ?

I have been told that the use of Grobner basis could be useful but I have never try this.

Thanks a lot for yours feedbacks.

PDE solutions: when are they "general"?

by: ecterrab 14630 Maple
pde differential-equations
June 27 2016
4  0

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

New developments on exact solutions for PDEs with...

by: ecterrab 14630 Maple
pde differential-equations
June 27 2016
4  3

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

help about text mode and math mode...

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2dmath
June 27 2016
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