integration with known boundary an 2 variables

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Abdullah Azzam 2020-2-12

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https://ww2.mathworks.cn/matlabcentral/answers/505002-integration-with-known-boundary-an-2-variables

评论： Walter Roberson 2020-2-12

Hi guys I have this equations that I have been trying to solve for different values of "n"

dz= - dh / [ (1 + v/k ) + (a * v/k * abs ( h ) ^ n ) ] where k and a are constants z range from 0 to a known value let's say Z and h range from a value h1 to h2 that are also known values.

Is there a way to write a matlab code that integrate this equations in term of h and then solve for the value of v

Thanks for the help in advance.

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Walter Roberson 2020-2-12

Are dz and dh representing derivatives? If so with what independent variable? It is especially important if dz is z(h) derivative with respect to h

What is known about n? If it is symbolic you are not going to be able to solve this in the general case. You might be able to solve for n being positive integer.

Abdullah Azzam 2020-2-12

ya dz and dh represent derivatives. Where Z represent depth and since it is goint to be integrated from 0 to desired depth I don't think it will be a problem you can consider dz as z(h). n is a constant 1,2,3,4...etc as users input.

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Answer 1

Walter Roberson 2020-2-12

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https://ww2.mathworks.cn/matlabcentral/answers/505002-integration-with-known-boundary-an-2-variables#answer_415158

"Is there a way to write a matlab code that integrate this equations in term of h and then solve for the value of v"

No. Even for n=1 there is no closed form for the integral that can be found at all easily, so you cannot solve for v except numerically given all of the constants

There is no closed form integral for this applicable to all positive integers n. Not even if you restrict h so that it does not cross 0 to avoid the question what the integral of abs() is.

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Abdullah Azzam 2020-2-12

Thanks. If possible what would be a good Numrical solution for that problem?

Walter Roberson 2020-2-12

在 MATLAB Online 中打开

balance = @(z) integral(@(h) - 1 ./ () (1 + v./k ) + (a .* v./k * abs(h).^n), h1, h2) - Z;
v0 = rand;    %or other good starting value
best_v = fsolve(balance, v0);

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