Solving a differential equation using matlab

Hi!

I have the following differential equation where x is unknown and p_dot is the derivative of p(p is given):

x * p_dot = qa -qb.

The vectors qa, qb and p are given by samples of 1 sec, and n samples. In this case n = 3597.

To find x, I've tried several procedures, none that actually works. One challenge for me is to find the derivative of p. Using both matlab and simulink has given me no logical answers. I've also tried the other way around, by integrating both sides of the equal sign. This also gives faulty answers.

Anyone who can provide som hint/tips, especialy some coding examples, would be appreciated!

Cheers

Answer 1

[Star Strider]

I am not certain that I understand what you want to do, so see if this approximates it:

p = linspace(0,50,25) + rand(1,25)*0.01;    % Create Data
qa = randi(10,1,25);                        % Create Data
qb = randi(20,1,25);                        % Create Data

p_dot = gradient(p);                        % Create ‘p_dot’

x = (qa - qb)./p_dot;                       % Solve For ‘x’

[Andreas Volden]

Thanks for replying!

I have tried your procedure without any luck. I may have been unclear in my formulation, let me rephrase: I know the x in this case is a constant while p, qa and qb varies.

When i calculate x straight forward like you suggest, x varies a lot over the elements in the [nx1] vector it becomes.

I have tried to solve it only when values in the terms (qa-qb) and p_dot is nonzero. The vectors then becomes shorter, like: [n-jx1]. Even then, my x varies a lot. Any further tips?

[Star Strider]

My pleasure!

Without knowing more about your data and what you want to do, I cannot be specific. (A sample of your data would be helpful.) If x is supposed to be a constant, one possibility might be to do a linear regression, for instance:

x = [ones(size(p))' p_dot']\[qa' - qb'];    % Solve For ‘x’

This replaces my previous line calculating ‘x’. It gives two scalar values, a slope ‘x(1)’ and an intercept ‘x(2)’, but it might be what you want. Note that this assumes all your original vectors are row vectors, so they need to be transposed to column vectors to work in the regression equation (thus the transpose (') operators).

Does this do what you want?