solving trascendental equations, proper setting

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

Answer 1

[Ameer Hamza]

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);

[PatrizioGraziosi]

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]

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]

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]

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]

Thank you so much!!!

It will take to me a few days to test it, I'll keep you posted.

[PatrizioGraziosi]

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]

I am glad that it worked for your case, and you got the results. Good luck with your research.