baustamm1

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These are replies submitted by baustamm1

I am still trying to convert the expression to a sum. However, 

convert(Simp(rho(i)),Sum,include  = powers) ;

gives binomials. It seems possible to write the expression as a sum of sums. Is it possible to do this with Maple?

@rlopez Thank you!

Thank you two very much, quite helpful. Is it possible to find a form for f(i) without radicals?

Or to rewrite f(i) as a finite sum?

@Preben Alsholm Sorry you are right, I have to equations, this is the other one but they are quite similar. As already mentioned: 

simplify(eval(%,i=1),assume=positive);
normal(%,expanded);

this works and gives A (and B cancels out). I think so because s stands for the time derivative and it would be odd to have fractional derivates in my problem. Furthermore, as you can see in the post of  vv 2189 there are no roots for i = 1..5 (of course it does not proof it).

@Rouben Rostamian  Thank you. Unfortunately, my expression is a bit longer and there your advise does not work:

f(i) = -(2*(((A*sqrt(s)*a^(2*i+2)-a^(2*i)*B*L*dz*(sqrt(s)*c+s^(3/2)))*2^i*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)-(s+c)*(A*L*s*dz*2^i-B*2^(i+1))*a^(2*i+2)+(2^(i+1)*c*s+2^i*(c^2+s^2))*a^(2*i)*dz^2*L^2*B*s)*(1/(-dz*L*sqrt(s*(s+c)*(L^2*c*dz^2*s+L^2*dz^2*s^2+4*a^2))+L^2*s*(s+c)*dz^2+2*a^2))^i+(dz*L*sqrt(s*(s+c)*(L^2*c*dz^2*s+L^2*dz^2*s^2+4*a^2))+L^2*s*(s+c)*dz^2+2*a^2)^(-i)*((A*sqrt(s)*a^(2*i+2)-a^(2*i)*B*L*dz*(sqrt(s)*c+s^(3/2)))*2^i*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)+(s+c)*(A*L*s*dz*2^i-B*2^(i+1))*a^(2*i+2)-(2^(i+1)*c*s+2^i*(c^2+s^2))*a^(2*i)*dz^2*L^2*B*s)))*a^2/(sqrt(L^2*s*(s+c)*dz^2+4*a^2)*sqrt(s+c)*sqrt(s)*(dz*L*sqrt(s)*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)-L^2*s*(s+c)*dz^2-2*a^2)*(dz*L*sqrt(s)*sqrt(s+c)*sqrt(L^2*s*(s+c)*dz^2+4*a^2)+L^2*s*(s+c)*dz^2+2*a^2))

where i is an integer.

In the previous example, f was evaluated for i = 1. I am pretty sure that the square roots here can be simplified and are not necessary.

 

 

 

@Markiyan Hirnyk What does SCR means? And why do you state that criticism should be constructive? Do you feel critized by me? Hopefully not but if yes, why?

Thank you two so far. Is there a way to avoid this error in Maple? My calculation is a bit lenghtier than the example :)

@Axel Vogt Thank you very much

@Preben Alsholm 

 

Initial conditions:

p(x, 0) = 1e5; # could be scaled to 0

mdot(x, 0) = 0;

Boundary conditions:

p(0, t) = 1.1e5; # or better some arbitrary (smooth) function (could be scaled)

p(L, t) = 1e5; # could be scaled to 0

It would be nice, if other combinations of BCs are also possible.

 

@Thomas Dean 

You mean from here:

http://www.maplesoft.com/support/help/Maple/view.aspx?path=examples/pdsolve_boundaryconditions&term=pdsolve[boundaryconditions]

that is where I got the second example in my first post from. Thank you

@Preben Alsholm 

You mean building blocks as in the sense of the Fourier coefficients in the solution of the heat equation?

How do you get this piece of information from:

-> Solving ordering for the dependent variables of the PDE system: [mdot(x,t), p(x,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
tackling triangularized subsystem with respect to mdot(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" ...
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying characteristic strip method for first order PDEs
First set of solution methods successful
Laplace's method successful.
First set of solution methods successful
tackling triangularized subsystem with respect to p(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

?

How to continue with this solution:

sol1 := {mdot(x, t) = _C1*exp(_c[1]*A*(x*sqrt(1/(RL*T))+t)*d/(A*d+4*_c[1]))*(exp(_c[1]^2*(x*sqrt(1/(RL*T))+t)/(A*d+4*_c[1])))^4*(exp(A*d*_c[1]*((1/2)*x-(1/2)*t/sqrt(1/(RL*T)))/(sqrt(RL)*sqrt(T)*(A*d+4*_c[1]))))^2*_C2, p(x, t) = -(1/2)*_C1*_C2*exp(4*_c[1]*(sqrt(RL)*sqrt(T)*_c[1]*t+(1/2)*x*(A*d+2*_c[1]))/(sqrt(RL)*sqrt(T)*(A*d+4*_c[1])))*sqrt(T)*sqrt(RL)*(A*d+2*_c[1])/(_c[1]*A)+_C3}

?

First I tried to solve it numerically by finite differences in MATLAB and with the "numeric" option for pdesolve, in both cases I have different kind of nonphysical oscillations which are dependent on the mesh/solver but do not vanish even for very small elements. Therefore, I am trying to find an analytical solution.

@Preben Alsholm 

Thank you for your answer and advise. The "problem" for me is this step:

_F1:=y->piecewise(y<0,exp(v*y),sin(y));

I do not really need Maple to come up with this solution. Therefore, I thought, it could be done more by itself without giving this piece of explicit information.

Furthermore, the ultimate goal is to solve a system of PDEs (two coupled transport equations, one is damped):

sys1 := [diff(p(x,t),x) + 1/A*diff(mdot(x,t),t) + d*mdot(x,t) = 0, diff(mdot(x,t),x) + A/RL/T*diff(p(x,t),t) = 0];

sol1 := pdsolve(sys1);

The solution contains the constants(?) _C1, _C2, _C3 and _c[1]. If these are not constants but functions then the argument is missing. I do not know how to incorporate my ICs and BCs here. It works, if undamped equations are used:

sys2 := [diff(p(x,t),x) + 1/A*diff(mdot(x,t),t) = 0, diff(mdot(x,t),x) + A/RL/T*diff(p(x,t),t) = 0];

sol2 := pdsolve(sys2);

Thanks!

@Preben Alsholm 
But one more thing, why does this work?

P:=a*b;
op(P);
subs(a*b=y,P);

@Preben Alsholm Thank you very much, this helped me to understand and to solve the problem

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