Jjjones98

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2 years, 313 days

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These are questions asked by Jjjones98

Is it possible to determine an analytic solution to the following system of two differential equations for $A$ and $B$ using Maple.  My suspicion is that trial and error would find an analytic solution in theory and so that Maple could find the solution.  M is a constant and \sigma is some arbitrary function of t and the spatial coordinates. 

\[ \Bigg( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{2} \Bigg( 1 + \frac{M}{2 \sqrt{x^2 + y^2 + z^2}} \Bigg) \Bigg( \frac{\partial \sigma}{\partial x }\frac{\partial}{\partial x} +\frac{\partial \sigma}{\partial y}\frac{\partial}{\partial y} +\frac{\partial \sigma}{\partial z}\frac{\partial}{\partial z} \Bigg) \Bigg)B=0, \]

\[\frac{d A}{dt} = AB.\]

Furthermore, the boundary conditions are 

\[B \rightarrow -1  \: \text{as}  \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty,\]

\[A \rightarrow e^{-t} \: \text{as} \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty \]

System_of_Equations.pdf

I have written a bit of code which solves a linear system for some quantities which have been Laplace and then Fourier transformed.  

e1 := -2*D*i*k*pi + A*s = 0

e2 := 2*A*i*k*pi + 2*C*i*k*pi + B*s = a

e3 := s*C + 4/5*P*w3*(2*pi*i*k*D - 2*1/3*pi*i*k*D)/w2 = -2*(C + 2*K*(2*pi*i*k*B - 2*1/3*pi*i*k*B))/(w2*K)

e4 := s*D + 2*5/4*P*t4*pi*i*k*C/(t2*K) = -5/(2*P)*D/(t2*K)

sys := {s*C + 4/5*P*w3*(2*pi*i*k*D - 2*1/3*pi*i*k*D)/w2 = -2*(C + 2*K*(2*pi*i*k*B - 2*1/3*pi*i*k*B))/(w2*K), s*D + 2*5/4*P*t4*pi*i*k*C/(t2*K) = -5/(2*P)*D/(t2*K), -2*D*i*k*pi + A*s = 0, 2*A*i*k*pi + 2*C*i*k*pi + B*s = a}

solve(sys, [A, B, C, D])

Linear_System.mw

I get at the end some fractions where everything in the fractions is a constant with some physical meaning except for k which is the only frequency as I am working in one dimension so just need one-dimensional Fourier and Laplace transforms.  s is the corresponding variable from the Laplace transform. 

I was wondering if Maple had some functionality which would enable me to inverse Laplace and then inverse Fourier transform these quantities A, B, C and D from the linear system such that I obtain an algebraic expression at the end and not a numerical result.

I have solved the following linear system for 6 variables on Maple using the following code:

sys := {w = -2*Pi*i*k_2*v + 2*Pi*i*k_2*(4*K^2)/(5*Pi)*u, z = -2*Pi*i*k_1*v + 2*Pi*i*k_1*(4*K^2)/(5*Pi)*u, p*s*x = -2*Pi*i*k_1*u - 4*Pi^2*(k_1^2 + k_2^2)*x + a_1, p*s*y = -2*Pi*i*k_2*u - 4*Pi^2*(k_1^2 + k_2^2)*y + a_2, k_1*z + k_2*w = 0, k_1*x + k_2*y = 0}

solve(sys, [x, y, z, w, v, u])

However, the solution yields z = w = 0, I was wondering if this is correct as I feel that these quantities should not be 0 in the problem which I am studying, is there a way to find a solution which does not involve setting these two to 0?

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I have a function e^(-\lambda z \sqrt(x^2 + y^2)), is it possible to use Maple to find some sequence of derivatives wrt to x and y which could be applied to this function to get

(z/(1+2*sqrt(x^2 + y^2)*lambda))*e^(-\lambda z \sqrt(x^2 + y^2))

 

 

I am trying to find 6 unknowns A, B, C, L, E, and F and have managed to write down a system of 6 equations involving these unknowns.  However, the equations are long and which I try to put the system together and solve with solve(sys, {A, B, C, E, F, L}) it says `[Length of output exceeds limit of 1000000]`.  Although the equations are long, I feel like something must be going wrong as the output cannot be that long.  The equations are:

C = (((h/(4*Pi*K)*H*e^(-k*h)*I/k - k*h^2/2 + H*A*e^(-k*h)*I/(K*k^2) - B*h - sqrt(Pi/2)*h/(2*Pi)*H*e^(-k*h)*I + sqrt(Pi/2)*K*k*h - sqrt(Pi/2)*H*A*e^(-k*h)*I/k + sqrt(Pi/2)*B*K - sqrt(Pi/2)*K*H*k*h^2*I/2) + sqrt(Pi/2)*H*A*e^(-k*h)*I/k) - sqrt(Pi/2)*H*F*h*K*I - sqrt(Pi/2)*H*A*e^(-k*h)*I/k + sqrt(Pi/2)*K*H*k*h^2/2*I) - sqrt(Pi/2)*H*F*h*K*I,

E = (((h/(4*Pi*K)*J*e^(-k*h)*I/k - k*h^2/2 + J*A*e^(-k*h)*I/(K*k^2) - L*h - sqrt(Pi/2)*h/(2*Pi)*J*e^(-k*h)*I + sqrt(Pi/2)*K*k*h - sqrt(Pi/2)*J*A*e^(-k*h)*I/k + sqrt(Pi/2)*L*K - sqrt(Pi/2)*K*J*k*h^2*I/2) + sqrt(Pi/2)*J*A*e^(-k*h)*I/k) - sqrt(Pi/2)*J*F*h*K*I - sqrt(Pi/2)*J*A*e^(-k*h)*I/k + sqrt(Pi/2)*K*J*k*h^2/2*I) - sqrt(Pi/2)*J*F*h*K*I

0 = -H*(k*z^2/2 - H*A*e^(-k*z)*I/(K*k^2) + B*z + C)*I - J*(k*z^2/2 - J*A*e^(-k*z)*I/(K*k^2) + L*z + E)*I + k*z + A*e^(-k*z)/K + F

0 = ((-H*A*e^(-k*z)*I - H*I*(-2*K*(-H*k*z^2*I/2 - H^2*A*e^(-k*z)/(K*k^2) - H*B*z*I - H*C*I)) - J*I*(-2*K*1/2*(((-J*k*z^2*I/2 - J*H*A*e^(-k*z)/(K*k^2) - J*B*z*I - J*C*I - H*k*z^2*I/2) - J*H*A*e^(-k*z)/(K*k^2)) - H*L*z*I - i*H*E)) - K*k) + H*A*e^(-k*z)*I) + K*H*z*k*I + H*A*e^(-k*z)*I + K*H*F*I

0 = ((-J*A*e^(-k*z)*I - H*(((K*J*k*z^2*I/2 + J*H*A*e^(-k*z)/k^2 + K*J*B*z*I + k*h*K*z^2*I/2) + J*H*A*e^(-k*z)/k^2) + K*H*L*z*I + K*H*E*I)*I - J*I*(-2*K*(-J*k*z^2*I/2 - J^2*A*e^(-k*z)/(K*k^2) - J*L*z*I - J*E*I)) - K*k) + J*A*e^(-k*z)*I) + K*J*z*k*I + J*A*e^(-k*z)*I + K*J*F*I

0 = (-k*A*e^(-k*z) - H*I*(-2*K*1/2*(((k*z + H*A*e^(-k*z)*I/(K*k) + B - H*k*z^2*I/2) + H*A*e^(-k*z)*I/(K*k)) - H*F*z*I - H*G*I)) - J*I*(-2*K*1/2*(((k*z + J*A*e^(-k*z)*I/(K*k) + L - J*k*z^2*I/2) + J*A*e^(-k*z)*I/(K*k)) - J*F*z*I - J*G*I)) - 2*K*k) + 2*k*A*e^(-k*z)

where h takes a constant value, H and J are constants, k is the square root of H^2 + J^2, I is the imaginary unit, and z also takes some value as a parameter.

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