How to integrate this function numerically?

14 Dec 2019
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Updated 15 Dec 2019
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Hello everyone,
I'm stuck at this equation and I want to integrate it numerically
where
and r = 695510, a = -0.0021, b = 1.34, Wo = 438.1
I need to integrate the first equation numerically to get V as a function of R
Numerical integration from R = 10, where it is assumed V = U, to 215 should give V(R).
I appreciate your help.
Thank you,

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Walter Roberson
Walter Roberson on 14 Dec 2019
Is e the base of the natural logarithms? And is it acting as a constant multiplier or is e being raised to the part after it?
Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Yes that's right, it's the base of the natural logarithm and it's being raised to the part after it.

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 Accepted Answer

Chuguang Pan
Chuguang Pan on 14 Dec 2019

1 vote

You can use Euler formula. V(n+1)=V(n)+h*f(R,V(n)), which f(R,V) is the right side of differential equation.
But you need to know the initial value V(10).
r=695510;a=-0.0021;b=1.34;Wo=438.1;
h=0.1;%Integral step size, you can change this value
N=100;%It means that you want to integral from 10 to 10+h*N
V=zeros(1,N+1);%Initialization V array
V(1)=?;%Need to know the initial value V(10)
for n=1:N
R=10+(n-1)*h;
W=Wo*sqrt(1-exp(1)*(2.8-R)/8.1);
V(n+1)=V(n)+h*r*a*R^(-b)*(1-W/V(n));
end
X=10+(0:N)*h;
Y=V;
plot(X,Y);

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Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Edited: Mohamed Nedal on 14 Dec 2019
Thanks for your response. The initial value of V(10) is known (693).
I think the 3rd line of the for loop should be like this
W=Wo*sqrt(1-exp((2.8-R)/8.1));
Could you please elaporate on h and N?
What are the consequences of changing the integral step size (h) and N?
Chuguang Pan
Chuguang Pan on 14 Dec 2019
r=695510;a=-0.0021;b=1.34;Wo=438.1;
h=0.1;%Integral step size, you can change this value
N=(215-10)/h;%It means that you want to integral from 10 to 215
V=zeros(1,N+1);%Initialization V array
V(1)=693;%Need to know the initial value V(10)
for n=1:N
R=10+(n-1)*h;
W=Wo*sqrt(1-exp((2.8-R)/8.1));
V(n+1)=V(n)+h*r*a*R^(-b)*(1-W/V(n));
end
X=10+(0:N)*h;
Y=V;
plot(X,Y);
you can decrease h for increasing presision
Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Thank you so much, that's very helpful.
Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Please I have another question. Knowing the above parameters, How do we get t from the folowing equation?
Chuguang Pan
Chuguang Pan on 14 Dec 2019
If t is independent variable, it is similar to R. You need to decide for youself.
Such as t=t0+(0:N)*h.
Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Edited: Mohamed Nedal on 14 Dec 2019
t is the time (in seconds) the object takes to cross the distance R at speed V.
to = 0, I need to find t(R=215).
Mohamed Nedal
Mohamed Nedal on 14 Dec 2019
Could you check this please.
r = 695510; a = -0.0021; b = 1.34; Wo = 438.1;
h = 1; % Integral step size, you can change this value
N = (215 - 10)/h; % It means that you want to integral from 10 to 215
V = zeros(1,N+1); % Initialization V array
t = zeros(1,N+1);
t(1) = 0;
V(1) = 693; % Need to know the initial value V(10)
for n = 1:N
R = 10 + (n-1)*h;
W = Wo * sqrt(1-exp((2.8-R)/8.1));
V(n+1) = V(n) + h*r*a*R^(-b)*(1-W/V(n));
t(n+1) = t(n) + h*a*R^(-b)*(V(n)-W);
end
X = 10 + (0:N)*h;
t = (0:N)*h; % to = 0
plot(X,V);
t here is monotonically increasing (t=0,1,2,...) and when I removed this line
t = (0:N)*h;
it gives negative fractions and it's not correct.
Chuguang Pan
Chuguang Pan on 14 Dec 2019
V=dR/dt. dt=dR/V.
R=10+(0:N)*h;
dR=diff(R);
V=(V(2:end)+V(1:end-1))*0.5;%average speed
dt=dR./V;
t=[0 cumsum(dt)];
Mohamed Nedal
Mohamed Nedal on 15 Dec 2019
Excuse me another question..
I would like to apply this equation on several objects and I tried to replace V(1), which's the initial value V(10), with a vector column instead of a single value like this one for instance
Vtest = [136
490
359
361
693
1374
585
542
1118
910
1188];
so that each value is assigned for one object. But I got this error
In an assignment A(:) = B, the number of elements
in A and B must be the same.
Error in numeric_integ (line 22)
V(1) = Vtest; % Need to know the initial value V(10)
Could you help me out?
and if I want to find only the value of V(215) and t(215), which are the last values of V and t at distance 215, how can I do that?
Chuguang Pan
Chuguang Pan on 15 Dec 2019
result_V=zeros(1,length(Vtest));% save last V for every Vtest
result_t=zeros(1,length(Vtest));% save last t for every Vtest
for i=1:length(Vtest) %for every Vtest
V=zeros(1,N+1);
V(1)=Vtest(i);
%do the computing
result_V=V(end); %using end to index last value of array V
result_t=t(end); %using end to index last value of array t
end

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