integration with known boundary an 2 variables

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Abdullah Azzam
Abdullah Azzam 2020 年 2 月 12 日
コメント済み: Walter Roberson 2020 年 2 月 12 日
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
Walter Roberson 2020 年 2 月 12 日
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
Abdullah Azzam 2020 年 2 月 12 日
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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Walter Roberson
Walter Roberson 2020 年 2 月 12 日
"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
Abdullah Azzam 2020 年 2 月 12 日
Thanks. If possible what would be a good Numrical solution for that problem?
Walter Roberson
Walter Roberson 2020 年 2 月 12 日
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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