solving trascendental equations, proper setting

2020 6 14
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更新时间：2020 6 17
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Hello everybody,
I'd like to solve for y = y(x) the following equation
d log( y ) / d x + y = 1 + f
with f = f(x).
f is a 1D numerically known array, I don't know its nalytical form.
I cannot set properly solve or fzero.
Can you help me, please?
Patrizio

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Ameer Hamza
Ameer Hamza 2020-6-14
编辑：Ameer Hamza 2020-6-14

2 个投票

This is a differential equation and you can use symbolic toolbox to find an anayltical solution
syms y(x) f
eq = diff(log(y), x) + y == 1 + f;
sol = dsolve(eq);
Result
>> sol
sol =
(exp((C1 + x)*(f + 1))*(f + 1))/(exp((C1 + x)*(f + 1)) + 1)
f + 1
Following shows how to get a numerical solution using ode45
syms y(x) f
eq = diff(log(y), x) + y == 1 + f;
sol = dsolve(eq);
odeFun = matlabFunction(odeToVectorField(eq), 'Vars', {'t', 'Y', 'f'});
tspan = [0 10]; % time span for numerical solution
ic = 1; % initial condition: y(0)==1
fv = 1; % numerical solution for f=1
[t, y] = ode45(@(t, y) odeFun(t, y, fv), tspan, ic);
plot(t, y);

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PatrizioGraziosi
PatrizioGraziosi 2020-6-14
编辑：PatrizioGraziosi 2020-6-14
Thank you Ameer!
In the system I'm stuying, for each 'x' value there is one value for 'f' and one value for 'y'.
I mean, there is a f(x) that I know and a y(x) I don't know but for each 'x' value there are only one 'f' and one 'y' . I know f(x) and the relation between f(x) and y(x), and should find y(x).
As far as I understand, the tspan plays the role of the 'x' variable and I get a y(x) for each 'fv' but this not actually the case: for each 'x' value I have also a different 'fv' value.
Hope this makes my problem clearer. If I have misunderstood your answer, sorry.
About the symbolic toolbok lines you wrote, it looks to be the case for a scalar f.
if I input
syms y(x) f(x)
I get this
sol =
exp(int(f(x) + 1, x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true))/(C6 - int(-exp(int(f(x) + 1, x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true)), x, 'IgnoreSpecialCases', true, 'IgnoreAnalyticConstraints', true))
that I don't understand clearly. I wonder if you can hel me again, please.
Thanks
Patrizio
Ameer Hamza
Ameer Hamza 2020-6-14
I get your point. Yes, tspan plays the role of 'x'. Even if f is a function of 'x', you can consider that in the above code, f(x)=1, i.e., it is a constant function. Although, I agree with your point that in the general case, f and y need to vary with 'x'. Do you have a specific expression for f(x)? Solving it using dsolve() for general case f(x) might not give any useful results.
PatrizioGraziosi
PatrizioGraziosi 2020-6-14
Hi Ameer, thanks!
Unfortunately f(x) is an experimental measurements and I don't have a precise analytical expression for it. I would need of y(x) to fit it and disentangle between two physical possibilities...
Anyway, I'l keep thinking about.
Ameer Hamza
Ameer Hamza 2020-6-14
编辑：Ameer Hamza 2020-6-14
If f is a function of 'x' but obtained experimentally, then you can use the solution in my answer by considering 'f' as a symbolic variable. Following shows how it can be done
syms y(x) f
eq = diff(log(y), x) + y == 1 + f;
sol = dsolve(eq);
odeFun = matlabFunction(odeToVectorField(eq), 'Vars', {'t', 'Y', 'f'});
xspan = 0:0.1:10; % time span for numerical solution
ic = 1; % initial condition: y(0)==1
fv = rand(size(xspan));
ffun = @(x) interp1(xspan, fv, x);
[x, y] = ode45(@(x, y) odeFun(t, y, ffun(x)), xspan, ic);
plot(x, y);
Here I used xspan as a vector, and the values of function f (fv) are also available for the values in xspan. I generated fv randomly in this code.
PatrizioGraziosi
PatrizioGraziosi 2020-6-15
Thank you so much!!!
It will take to me a few days to test it, I'll keep you posted.
PatrizioGraziosi
PatrizioGraziosi 2020-6-16
Hi Ameer,
you're solution is brilliant!
Sorry that I couldn't test it before... Your support has been excellent!
In the case we publish the data analysis done thanks to your solution, we'll aknowledge your support.
Thanks
Patrizio
Ameer Hamza
Ameer Hamza 2020-6-17
I am glad that it worked for your case, and you got the results. Good luck with your research.

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