Numerically solve of integral equation with known Integrand

Question

AHMAD TALAEI 2019-3-1

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https://ww2.mathworks.cn/matlabcentral/answers/447617-numerically-solve-of-integral-equation-with-known-integrand

评论： AHMAD TALAEI 2019-3-4

采纳的回答： Torsten

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I'm trying to solve the attached integral equation (Eq. 5), numerically. The value of particle acceleration

and fluid acceleration

are known up to the time step n where t = n

and the integral is from time 0 to t. I'm currently using the basic definition of integral to solve it (namely, Eq. 4).

The integral can be evaluated numerically except at a region where τ is approaching t. There, the integrand becomes infinite and needs to be treated separately. Therefore, I split the integral into two terms: Z1 and Z2 where Z1 is the series term expressed in Eq. 5 and Z2 is the last term in Eq. 5. Here is the code that I wrote:

if n > 2
    Z1= 0;
    for m = 2:n-1
         Z1 = Z1+ ((d2z_p(n-m) - d2z_f(n-m))/sqrt(m)) * dt^(1/2)
    end
    Z2 = (d2z_p(n-1) - d2z_f(n-1)) * dt^(1/2);
    Z_1 = Z1 + Z2;
end

Z_1 is supposed to be the solution of the integral, but I know that it doesn't give the right answer either because Eq. 4 is not the right one to use, or my code is wrong. Anyone can help me on how to solve the given integral?

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Answer 1

Torsten 2019-3-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/447617-numerically-solve-of-integral-equation-with-known-integrand#answer_363302

编辑：Torsten 2019-3-1

If d2z_p(i) = dv_p / dtau at tau = i * t / n = i * dt and d2z_f(i) = dv_f / dtau at tau = i * t / n = i * dt in your code from above, it is correct.

Note that you can also use

Z= 0;
for m = 1:n-1
   Z = Z+ ((d2z_p(n-m) - d2z_f(n-m))/sqrt(m)) * dt^(1/2)
end

to approximate the integral without splitting in Z1 and Z2.

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Torsten 2019-3-4

编辑：Torsten 2019-3-4

t = ...;      % vector of t-values where dvpdt and dvfdt are available
dvpdt = ...;  % dvpdt vector with the same length as t vector 
dvfdt = ...;  % dvfdt vector with the same length as t vector
t1 = ...;     % upper limit of the integral
t0 = 0.0;     % lower limit of the integral
eps = 1.0e-6;
fun=@(tau)1./sqrt(t1-tau).*(interp1(t,dvpdt,tau)-interp1(t,dvfdt,tau));
sol = integral(fun,t0,t1-eps)

AHMAD TALAEI 2019-3-4

Thanka a lot!

Both give the same answer, although the "integral" function slows down the code. I just noticed that I have multiplied the integral by a wrong coefficient and that is the reason it didn't give the correct solution.

Ahmad

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