potential() compute scalar potential from gradient vector field; Error using sym/potential Second argument must be a vector of variables.

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Philipp Bisignano
Philipp Bisignano 2022-5-31
评论： Philipp Bisignano 2022-5-31
I have a matrix dRho which is a (density) gradient vector field. I want to compute the scalar potential (the density) Rho by means of the matlab-function potential(V,X,Y).
X and Y are my cartesian coordinate vectors:
[m, n] = size(dRho); [1, n] = size(X); [m, 1] = size(Y);
As the function potential() deals with symbolic variables, I convert my quantities to symbolic matrix/vectors with the sym() function, 'f' for floating point precision. (According to numeric to symbolic conversion https://ch.mathworks.com/help/symbolic/numeric-to-symbolic-conversion-1.html#buyp69j-1)
dRho_sym = sym(dRho, 'f');
X_sym = sym(X, 'f');
Y_sym = sym(Y, 'f');
Then I'm trying to compute the potential for each row of my matrix wrt to X_sym, using the respective y value (Y_sym(j)) as base point for the integration, as suggested in the description:
"potential(V,X,Y) computes the potential of vector field V with respect to X using Y as base point for the integration".
for j = 1:m
Rho = potential( dRho_sym(j,:), X_sym, Y_sym(j));
end
I'm getting this error:
Error using sym/potential
Second argument must be a vector of variables.
Are the symbolic numbers not understood as symbolic variables?
Or do I mess up something with the dimensions?
This is a more detailed description in the potential() function:
% P = POTENTIAL(V,X,Y) computes the potential of vector field V with
% respect to X using Y as base point for the integration. If Y is not a
% scalar, then Y must be of the same dimension as V and X. If Y is
% scalar, then Y is expanded into a vector of the same size as X with
% all components equal to Y.
%
% Examples:
% syms x y z; potential([x y z*exp(z)], [x y z])
% returns x^2/2 + y^2/2 + exp(z)*(z - 1).
%
% syms x0 y0 z0; potential([x y z*exp(z)], [x y z], [x0 y0 z0])
% returns x^2/2 - x0^2/2 + y^2/2 - y0^2/2 + exp(z)*(z - 1) -
% exp(z0)*(z0 - 1)
%
% potential([x y z*exp(z)], [x y z], [1,1,1])
% returns x^2/2 + y^2/2 + exp(z)*(z - 1) - 1
%
% potential([x y z*exp(z)], [x y z], 1)
% returns x^2/2 + y^2/2 + exp(z)*(z - 1) - 1
%
% potential([y -x], [x y])
% returns NaN since the potential does not exist.
I will be very gratefull for any help! Also suggestions about how to compute the scalar potential of a gradient vector field by other means are welcome.

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采纳的回答

Walter Roberson
Walter Roberson 2022-5-31
No, the second parameter must be a vector of symbolic names.
When you do not have a starting point, each term in V is integrated with respect to the corresponding variable in the second parameter, and all of the integrals are added.
But you cannot integrate anything with respect to a number, so what you are trying to do cannot work.
  1 个评论
Philipp Bisignano
Philipp Bisignano 2022-5-31
x = sym('x', [1 n]);
y = sym('y', [1 m]);
for j = 1:m
Rho_sym(j) = potential(dRho_sym(j,:), x, y(j) );
end
Like this it is compiling. But I'm getting one single value per iteration,
which is function of single entries x1, x2, ..., y1, y2, .. of x and y.
And it takes plenty of computation time.
Could I match the vector x = [x1, x2, ...] and y to my vectors X and Y?
Somehow x1 = X(1), ... ? Would that make any sense to you?

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