potential() compute scalar potential from gradient vector field

I have 2 [m, n] matrices dR_dx, dR_dy and 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).

I have a density gradient field:

The total differential reads

with dx and dy being the size of the pixels in the image .

Now I would like to integrate this total differential to get the actual density. I think it should be doable with the Matlab function potential()

X and Y are my cartesian coordinate vectors (in mm) along the dimensions of the image:

[m, n] = size(dRho); [1, n] = size(X); [m, 1] = size(Y);

dRho = (dR_dx + dR_dy) * dpix

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.

dR_sym = sym(dRho, 'f');

dRdx_sym = sym(dRdx, 'f');

dRdy_sym = sym(dRdy, 'f');

And I create two symbolic vectors x and y:

x = sym('x', [1 n]);

y = sym('y', [1 m]);

Then I call the potential() function inside 2 for loops:

for i = 1:m

for j = 1:n

Rho_sym(i, j) = potential([dRdx_sym(i, j), dRdy_sym(i, j)], [x(j) y(i)], [x(1) y(1)]); % compute the potential of this value as analytical function

rxX = subs(Rho_sym(i, j), x(j), X(j)); % substitute all x symbolic values with the numeric one from X

ryY = subs(rxX, y(i), Y(i)); % substitute all y symbolic values with the numeric one from Y

Rho_pot(i, j) = double(ryY); % convert it to double precision

end

end

I don't know how to call exactly the potential() function/which matrices I should use as input: Can I give as input directly dRho_sym or should I use the the single partial derivatives dR_dx, dR_dy?

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.

Answer 1

[Sachin Lodhi]

Hello Philipp,

The ‘potential()’ function in MATLAB is used to compute the scalar potential of a vector field. Given a vector field ‘V’ and its coordinates ‘X’, the function integrates the vector field to find the scalar potential field.

For your scenario, the partial derivatives ‘dR_dx’ and ‘dR_dy’ represent the ‘x’ and ‘y’ components of the density gradient vector field. These should be input to ‘potential()’ function to find the scalar field Rho you seek.

Here's a modified version of your code.

% Assuming dR_dx and dR_dy are the m-by-n matrices of partial derivatives
% X and Y are the vectors representing the coordinates in the image
[m, n] = size(dR_dx); % Assuming dR_dx and dR_dy have the same size
X = linspace(x_start, x_end, n); % Define the X coordinates
Y = linspace(y_start, y_end, m); % Define the Y coordinates
% Convert the matrices to symbolic form for the potential function
dRdx_sym = sym(dR_dx, 'f');
dRdy_sym = sym(dR_dy, 'f');
% Create symbolic variables for x and y
syms x y
% Initialize the matrix to hold the potential values
Rho_pot = zeros(m, n);
% Compute the potential
for i = 1:m
        for j = 1:n
            % Compute the potential at each point
            Rho_sym = potential([dRdx_sym(i, j), dRdy_sym(i, j)], [x, y], [X(1), Y(1)]);
            % Substitute the symbolic variables with the actual coordinates
            Rho_pot(i, j) = double(subs(Rho_sym, {x, y}, {X(j), Y(i)}));
        end
end

Please replace ‘x_start’, ‘x_end’, ‘y_start’, and ‘y_end’ with the actual coordinate limits of x and y components respectively. Also, the code presumes that gradient field is conservative; otherwise, the ‘potential' function may not apply.

Hope this helps.

Best Regards,

Sachin