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    <p>You have a linear, coupled system of ordinary differential equations, </p> <pre><code>y' = Ay with y = [S(t); D(t); F(t)] </code></pre> <p>and you're trying to solve the <em>inverse</em> problem, </p> <pre><code>A = unknown </code></pre> <p>Interesting!</p> <h2>First line of attack</h2> <p>For given <code>A</code>, it is possible to solve such systems analytically (read <a href="http://en.wikipedia.org/wiki/Matrix_differential_equation" rel="nofollow">the wiki</a> for example). </p> <p>The general solution for 3x3 design matrices <code>A</code> take the form </p> <pre><code>[S(t) D(t) T(t)].' = c1*V1*exp(r1*t) + c2*V2*exp(r2*t) + c3*V3*exp(r3*t) </code></pre> <p>with <code>V</code> and <code>r</code> the eigenvectors and eigenvalues of <code>A</code>, respectively, and <code>c</code> scalars that are usually determined by the problem's initial values. </p> <p>Therefore, there would seem to be two steps to solve this problem: </p> <ol> <li>Find vectors <code>c*V</code> and scalars <code>r</code> that best-fit your data</li> <li>reconstruct <code>A</code> from the eigenvalues and eigenvectors.</li> </ol> <p>However, going down this road is treaturous. You'd have to solve the non-linear least-squares problem for the sum-of-exponentials equation you have (using <code>lsqcurvefit</code>, for example). That would give you vectors <code>c*V</code> and scalars <code>r</code>. You'd then have to unravel the constants <code>c</code> somehow, and reconstruct the matrix <code>A</code> with <code>V</code> and <code>r</code>. </p> <p>So, you'd have to solve for <code>c</code> (3 values), <code>V</code> (9 values), and <code>r</code> (3 values) to build the 3x3 matrix <code>A</code> (9 values) -- that seems too complicated to me.</p> <h2>Simpler method</h2> <p>There is a simpler way; use brute-force:</p> <pre><code>function test % find [A, fval] = fminsearch(@objFcn, 10*randn(3)) end function objVal = objFcn(A) % time span to be integrated over tspan = [1 2 3 4 5]; % your desired data S = [17710 18445 20298 22369 24221 ]; D = [1357.33 1431.92 1448.94 1388.33 1468.95 ]; F = [104188 104792 112097 123492 140051 ]; y_desired = [S; D; F].'; % solve the ODE y0 = y_desired(1,:); [~,y_real] = ode45(@(~,y) A*y, tspan, y0); % objective function value: sum of squared quotients objVal = sum((1 - y_real(:)./y_desired(:)).^2); end </code></pre> <p>So far so good. </p> <p>However, I tried both the complicated way and the brute-force approach above, but I found it very difficult to get the squared error anywhere near satisfyingly small. </p> <p>The best solution I could find, after numerous attempts: </p> <pre><code>A = 1.216731997197118e+000 2.298119167536851e-001 -2.050312097914556e-001 -1.357306715497143e-001 -1.395572220988427e-001 2.607184719979916e-002 5.837808840775175e+000 -2.885686207763313e+001 -6.048741083713445e-001 fval = 3.868360951628554e-004 </code></pre> <p>Which isn't bad at all :) But I would've liked a solution that was less difficult to find...</p>
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    1. POMatlab: finding coefficients of ODE system
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    1. USRody Oldenhuis
    1. USRody Oldenhuis
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    1. COAs a simple note, this is a plain example of the butterfly effect, whereas if I simply plug your A matrix into the solution, I get a very incorrect result at t=5. More sig figs are needed.
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      1. USRasman
    2. CO@Rasman: Fixed. Also using relative errors now
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      1. USRody Oldenhuis
    3. COThanks a lot @RodyOldenhuis When I run this function first time it calculates the results in a seconds, but when I try to run second time, it is running eternally and doesn't stop. I am using Matlab 7.11.0 Does it happen for you? How do you chose the best solution after numerous attempts? With lowest difference between given points?
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      1. USKarolis Valiulis
 

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