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dharr

Dr. David Harrington

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20 years, 339 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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Contact Dr. David Harrington
I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are replies submitted by dharr

diff_table...

dharr 8215
February 06 2025
0 0

@Carl Love True, but the standard way to abbreviate input for anything nontrivial would be with diff_table. For example

declare(u(x));
U:=diff_table(u(x));

then U[x,x] is used to abbreviate diff(u(x),x,x) and U[] for u(x).

Comment...

dharr 8215
February 06 2025
0 0

@FZ Sorry, I don't understand what they are doing.

first integral...

dharr 8215
February 06 2025
0 0

@FZ Please expand the image. As far as I can see, they use the first integral, which didn't work for your case unless you know the form for the integrating factor.

What about diff(phi(xi),xi) on rhs?...

dharr 8215
February 06 2025
0 0

@FZ Your equation has a derivative on the rhs, unlike your textbook example, which can be done like this:

> 

restart
> 

with(PDEtools); declare(U(xi), Phi(xi))

U(xi)*`will now be displayed as`*U

Phi(xi)*`will now be displayed as`*Phi

> 

de1 := diff(U(xi), xi, xi) = R__1*U(xi)+R__2*U(xi)^3

diff(diff(U(xi), xi), xi) = R__1*U(xi)+R__2*U(xi)^3

Ultimately the method consists of integrating the negative of the rhs, treating U as a simple variable, and then adding (1/2)*Phi^2

> 

subs(U(xi) = u, -rhs(de1)); int(%, u = 0 .. u); H := (1/2)*Phi(xi)^2+subs(u = U(xi), %)

-R__2*u^3-R__1*u

-(1/4)*R__2*u^4-(1/2)*R__1*u^2

(1/2)*Phi(xi)^2-(1/4)*R__2*U(xi)^4-(1/2)*R__1*U(xi)^2

> 

``

Download Hamiltonian3.mw

Not the same form?...

dharr 8215
February 05 2025
0 0

@FZ Sorry, I don't understand; r1 or r2 are not in the form of your first equation.

Hamiltonian2.mw

correct...

dharr 8215
February 04 2025
0 0

@Tom GIV Yes, that's correct.

another form...

dharr 8215
February 03 2025
0 0

@janhardo I managed to get various other solutions using various combinations of dchange and HINT=, but never sol11. Here's one attempt:

> 

restart;
> 

with(PDEtools): declare(u(x,t)); U := diff_table(u(x,t));

u(x, t)*`will now be displayed as`*u

table( [( ) = u(x, t) ] )

> 

PDE1 := a^2*U[x,x] + b*exp(beta*U[]) - U[t,t];
Sol11 := u(x,t) = 1/beta*ln(2*(B^2-a^2*A^2)/(b*beta*(A*x+B*t+C)^2));
Test11 := pdetest(Sol11,PDE1);

a^2*(diff(diff(u(x, t), x), x))+b*exp(beta*u(x, t))-(diff(diff(u(x, t), t), t))

u(x, t) = ln(2*(-A^2*a^2+B^2)/(b*beta*(A*x+B*t+C)^2))/beta

0

An obvious change of variables that would lead to this form is

> 

tr := u(x,t) = ln(w(x,t)/beta)/beta:
PDE2:=numer(dchange(tr,PDE1,[w],normal));

(diff(diff(w(x, t), x), x))*w(x, t)*a^2+b*w(x, t)^3-(diff(w(x, t), x))^2*a^2-(diff(diff(w(x, t), t), t))*w(x, t)+(diff(w(x, t), t))^2

> 

solw := pdsolve(PDE2);

w(x, t) = -2*(_C2^2*a^2-_C3^2)*tanh(_C2*x+_C3*t+_C1)^2/b+2*(_C2^2*a^2-_C3^2)/b

> 

sol2:=eval(tr, simplify(solw));

u(x, t) = ln((2*_C2^2*a^2-2*_C3^2)*sech(_C2*x+_C3*t+_C1)^2/(b*beta))/beta

> 

pdetest(sol2, PDE1);

0

> 

NULL

Download pdsolve.mw

comments...

dharr 8215
February 01 2025
0 0

@Andiguys You can manipulate the pieces with op, see: Q_12.mw

The empty otherwise just means there are no other possibilities.

bilinear form...

dharr 8215
January 31 2025
0 0

@salim-barzani I simply followed the logic of the paper. I don't know about bilinear forms in this context; perhaps the nonlinear parts hare are bilinear forms? Here the linear part was set to zero, which implies the nonlinear part is also zero.

underline...

dharr 8215
January 29 2025
0 0

@FZ This is _a, not a; just a name Maple chose for the dummy integration variable.

Good for subsets...

dharr 8215
January 29 2025
0 0

@vv Thanks. (I'm still waiting for this to finish.) But I am also playing around with subsets of the eigenvalues (in disguised form), for which this will work, and be faster.

Excellent...

dharr 8215
January 29 2025
0 0

@acer Thanks. Much faster than @vv's.

Solution for y=0...

dharr 8215
January 29 2025
0 0

@dharr 

> 

restart;
> 

with(PDEtools):
> 

interface(showassumed=0):
> 

#assume(x::real);assume(t::real);assume(lambda1::complex);assume(lambda2::complex);assume(A1::real);assume(A2::real);assume(B1::real);assume(B2::real);assume(delta1::real); assume(delta2::real);assume(y::real);assume(a::real);
> 

z := -2*delta2*exp((-2*I*a*delta1^3*y + (x + y + t)^2*(B2 + B1)*delta1^2 + (-2*I*a*y*delta2^2 + t*I)*delta1 + delta2^2*(x + y + t)^2*(B2 + B1))/(delta1^2 + delta2^2))/((exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B2 + 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)) + exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B1 - 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)))*(delta1^2 + delta2^2)):
> 

pde:=diff(u(x,t),t)-diff(u(x,t),x)=eval(z,y=0):
> 

infolevel[pdsolve]:=3;

3

> 

pdsolve(pde,u(x,t));

First set of solution methods (general or quasi general solution)

   -> trying characteristic strip method for first order PDEs

   <- characteristic strip method for first order PDEs successful

<- First set of solution methods successful

<- Returning a *general* solution

u(x, t) = 2*Intat(delta2*exp(B1*(x+t)^2*delta1^2/(delta1^2+delta2^2)+B1*(x+t)^2*delta2^2/(delta1^2+delta2^2)+B2*(x+t)^2*delta1^2/(delta1^2+delta2^2)+B2*(x+t)^2*delta2^2/(delta1^2+delta2^2)+I*delta1*(x+t)/(delta1^2+delta2^2)-I*delta1*_a/(delta1^2+delta2^2))/((delta1^2+delta2^2)*(exp(2*B2*(x+t)^2*delta1^2/(delta1^2+delta2^2)+2*B2*(x+t)^2*delta2^2/(delta1^2+delta2^2)+delta2*(x+t)/(delta1^2+delta2^2)-delta2*_a/(delta1^2+delta2^2))+exp(2*B1*(x+t)^2*delta1^2/(delta1^2+delta2^2)+2*B1*(x+t)^2*delta2^2/(delta1^2+delta2^2)-delta2*(x+t)/(delta1^2+delta2^2)+delta2*_a/(delta1^2+delta2^2)))), _a = x)+f__1(x+t)

> 

 

NULL

Download pdsolve1.mw

suggestions...

dharr 8215
January 29 2025
0 0

@FZ The assumptions about real are probably not effective. Since y is not involved, I would leave it out. If you know the general form of the solution you can give it as a hint. HINT=`*` is equivalent to HINT=f(x)*g(t) but you can put in other expressions here. 

> 

restart;
> 

with(PDEtools):
> 

interface(showassumed=0):
> 

#assume(x::real);assume(t::real);assume(lambda1::complex);assume(lambda2::complex);assume(A1::real);assume(A2::real);assume(B1::real);assume(B2::real);assume(delta1::real); assume(delta2::real);assume(y::real);assume(a::real);
> 

z := -2*delta2*exp((-2*I*a*delta1^3*y + (x + y + t)^2*(B2 + B1)*delta1^2 + (-2*I*a*y*delta2^2 + t*I)*delta1 + delta2^2*(x + y + t)^2*(B2 + B1))/(delta1^2 + delta2^2))/((exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B2 + 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)) + exp((2*(delta1^2 + delta2^2)*(x + y + t)^2*B1 - 2*(a*y*delta1^2 + a*y*delta2^2 + 1/2*t)*delta2)/(delta1^2 + delta2^2)))*(delta1^2 + delta2^2)):
> 

pde:=diff(u(x,t),t)-diff(u(x,t),x)=z:
> 

infolevel[pdsolve]:=3;

3

This one I canceled out of after a while.

> 

pdsolve(pde,u(x,t));

First set of solution methods (general or quasi general solution)

   -> trying characteristic strip method for first order PDEs

This one I canceled out of after a while.

> 

pdsolve(pde,u(x,t),HINT=strip);

HINT = strip

Next try jumps out after a short time, so no solution of this form

> 

pdsolve(pde,u(x,t),HINT=`*`);

HINT = *

Preparing a solution HINT ...

-> trying HINT = _F1(x)*_F2(t)

Trying travelling wave solutions as power series in tanh ...

   Trying travelling wave solutions as power series in ln ...

   Trying a general ODE solution to the travelling wave related ODE system using ln and tanh

> 

 

NULL

Download pdsolve1.mw

Challenge...

dharr 8215
January 29 2025
0 0

@Alfred_F Good; I'm always up for a challenge, especially if I learn something new in the process.

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