Jjjones98

35 Reputation

4 Badges

2 years, 311 days

MaplePrimes Activity


These are replies submitted by Jjjones98

@dharr I have obtained a solution but at some points I get in the final result Dirac(2,t), can I take this to be a product of distributions Dirac(2) times Dirac(t)?  (I'll assume such a product is well-defined in my context).

@dharr  I think I have resolved my problem but I realise that there are points when I am using the two-sided Laplace transform instead of one-side when I use invlaplace, is it possible to specify that I need the two-sided transform?

@dharr When I try just invfourier(AA,k) to work with the frequency only it says that a third algebraic variable is required, also Maple Help says the calling sequence is the following where w is the variable with respect to which the expression is transformed and t is the parameter of the transform:

 

invfourier(expr, w, t)

@dharr I have tried to modify the file you sent me but doesn't work for some reason so I have just attached the relevant bit, it leads on from the file which you sent

Example.mw

 

 

@dharr I have tried inverse Fourier transforming the result but do not an exact expression, is this correct?

Linear_System.mw

@dharr Is there also a function invfourier which takes the inverse Fourier transform?  I essentially need to take the one-dimensional inverse Laplace transform of each variable as you say, but then I need to follow that with the one-dimensional inverse Fourier transform (ie. the result of what has been inverse Laplace transformed then needs to be inverse Fourier transformed).

@Christian Wolinski That doesn't really work, perhaps I should explain what I am attempting a bit more clearly?

@Rouben Rostamian  I think it might be impossible but would be willing to listen if anyone else has any ideas.

@Rouben Rostamian  

I will explain the context a bit more, essentially I am looking for some inverse Fourier transforms in 2D.  As a specific example I would like the 2D inverse Fourier transform of

 

((2*i*k_1)/(k*(1 + 2*k*lambda)))*exp(-z*k);

 

where

k=sqrt(k_1^2 + k_2^2)

A trick you can use with some complex analysis is the fact that the 2D Fourier transform of exp(-z*k) is (z/r^3).

Then differentiating wrt to k_1 corresponds to multiplication by x, so after multiplying by -i and dividing by z you get  

 

FT[x/r^3] = (i*k_1/k)*exp(-z*k);

 

 If I could get(1/((1 + 2*k*lambda)))*exp(-z*k) using the same process I suppose I can then get the full inverse Fourier transform by convolution, but I need a substitution to be able to transform this part, let me know if that's not clear what I mean.

@Carl Love I realised that I made a mistake and that my original system should in fact have been a system of 3 equations for 3 unknowns.  I have solved this now for A,B,C but the first unknown ends up being very bulky again (although I can see that some terms can be collected).  So for the system we have 

-sqrt(Pi/2)*H/(2*k)*exp(-k*h)*I + B*exp(-k*h) + B*sqrt(Pi/2)*K*k*exp(-k*h) = h/(4*Pi*K)*H*exp(-k*h)*I/k - sqrt(Pi/2)*h/(2*Pi)*H*exp(-k*h)*I,

-sqrt(Pi/2)*J/(2*k)*exp(-k*h)*I + C*exp(-k*h) + C*sqrt(Pi/2)*K*k*exp(-k*h) = h/(4*Pi*K)*J*exp(-k*h)*I/k - sqrt(Pi/2)*h/(2*Pi)*J*exp(-k*h)*I,

0 = -H*(B*exp(-k*z) + H*A*I/(2*K*k)*z*exp(k*z) - H*A*I/(2*K*k)*h*exp(k*z))*I - J*(C*exp(-k*z) + J*A*I/(2*K*k)*z*exp(k*z) - J*A*I/(2*K*k)*h*exp(k*z))*I + A/(2*K)*exp(-k*z) - k*A/(2*K)*z*exp(-k*z) + k*A/(2*K)*h*exp(-k*z).

I then create the system and solve it with solve(sys, {A, B, C}) but the value of A is very long.  I assume that the solution which Maple finds is the unique solution to the system of equations?

 

@Kitonum When I try When I try Determinant(`<|>`(seq(`if`(k=2,b,A(..,k)),k=1..6)))/Determinant(A); with the linear algebra package, it says Error, (in LinearAlgebra:-Determinant) matrix must be square.

Also although the result cannot be simplifed, is there at least a command which collects like coefficients, for example, so you could have all the coefficients which multiply a z collected together in one set of brackets multiplying z.

@Kitonum Ah right I see, so the result for C=Sol[3] would be Determinant(`<|>`(seq(`if`(k=3,b,A(..,k)),k=1..6)))/Determinant(A);, for example.  When I try Determinant(`<|>`(seq(`if`(k=2,b,A(..,k)),k=1..6)))/Determinant(A); with the linear algebra package, it says Error, (in LinearAlgebra:-Determinant) matrix must be square

It's a shame that these expressions are so long.  I feel like Maple sometimes makes expressions much longer than they need to be, perhaps I could these expressions much shorter if I solved the system by hand by removing the unknown coefficients one by one?

@Kitonum So does Sol[1] correspond to the unknown A and asking for Sol[2] correspond to B for example?  Is there any way of simplifying the final result as this is extremely bulky as you say.

@Christian Wolinski No sorry, I should have clarified, K is another separate constant with a physical meaning which I won't go into.  I is meant to be the imaginary unit i, I thought that the imaginary unit had to be written as I.

I did not realise that the exponential function was written in this way, I will change that.  I am wondering if I might have to try and do this by hand as it feels like Maple is trying to solve it by brute force which could result in an extremely long expression.

@Rouben Rostamian  Apologies, what I said did not make sense.  I actually mean the derivatives to be with respect to the Cartesian coordinate z.

1 2 3 4 Page 1 of 4