Can MATLAB solve a integrodifferential equation?

Hi, I have checked some posts on the forum, however to no avail.

I have the given system:

{
syms h g x C  
h = 2
g = 4
C = 1/sqrt(2.*pi)
y = ((exp(2.*g.*i.*x./h)).*(((h.^2)./2) + (g.*x)/2.*h.*i));
z = ((exp(-2.*g.*i.*x./h)).*(((h.^2)./2) + (g.*x)/2.*h.*-i));
[int(y.*z*.u, x], 0, 2.*pi)]==1

} which should solve for u. Is it feasible with some modifications?

Thanks

Answer 1

[Torsten]

u = 1/int(y*z,x,0,2*pi)

Best wishes

Torsten.

[Sergio Manzetti]

Hi Torsten, maybe I haven't outline the problem correctly:

it should be:

[int(y.*z*.u )*, x], 0, 2.*pi)]==1

where u is inside the integrator. It is not possible to extract u from the integrator by dividing by int(y,z,x,o,2i), that is why it is an integrodifferential eqn. u is inside the integration operation.

[Torsten]

If u is a scalar, you can take it out of the int-operator.

What is u if it's not a scalar value ?

Best wishes

Torsten.

[Sergio Manzetti]

It's a function (unknown).

[Torsten]

A scalar is also a function ...

Best wishes

Torsten.

[Sergio Manzetti]

u is a function which is multiplied with y and z, which are functions. The three of them are under then dx symbol, and thus cannot be divided per se in our out of the integral sign.

[Torsten]

Of course u can be taken out of the integral if it does not depend on x.

Your equation is solved if you set u to be equal to the scalar value

u = 1/int(y*z,x,0,2*pi)

Maybe there are other requirements on u which make this impossible, but up to now, you didn't state them.

Best wishes

Torsten.

[Sergio Manzetti]

Yes, it depends on x.

[Torsten]

As long as you can't set up your equation such that MATLAB doesn't treat u as a scalar, you will get the above result for u when using MATLAB's "solve".

Or how do you want to tell MATLAB that u explicitly depends on x ?

Best wishes

Torsten.

[Sergio Manzetti]

This is precisely the reason I ask here. In fact, u can even be an operator, such as d/dx

[Torsten]

I wonder why you call your equation an "integro-differential equation". Where is the differential operator ?

Best wishes

Torsten.

[Sergio Manzetti]

u can be the differential operator, or it can be a simple function, such as "x"

[Torsten]

I think you should first take a look at the definition of an integro-differential equation:

https://en.wikipedia.org/wiki/Integro-differential_equation

There is no MATLAB program that directly solves this type of equations.

But what exactly is the equation you are trying to solve ?

If it's the one you stated and u depends on x, you must explicily specify how u depends on x, e.g. u(x)=a*cos(x) with unknown a. Then - as a next step - you can determine "a" via the "solve" command.

Best wishes

Torsten.

[Sergio Manzetti]

Thanks Torsten! I will give that a try.

[Sergio Manzetti]

Tried it, and got "a" not defined

end
syms h g x C  
h = 2
g = 4
C = 1/sqrt(2.*pi)
y = ((exp(2.*g.*i.*x./h)).*(((h.^2)./2) + (g.*x)/2.*h.*i));
z = ((exp(-2.*g.*i.*x./h)).*(((h.^2)./2) + (g.*x)/2.*h.*-i));
u = 1/[int(y.*z*.*a*cos(x), x], 0, 2.*pi)]
solve

[Sergio Manzetti]

Dear Torsten, I tried this new variant:

if true
  % code
end
syms h g x C
h = 1
g = 5
y=@(x)(C - (exp(2.*g.*1i.*x./h));
z=@(x)(C - (exp(-2.*g.*1i.*x./h));
prod=@(x)y(x).*z(x)*(1+x^2);
W= integral(prod,0,2*pi)==1;

and I should get the result for "C", however that is simply 0. It looks incorrect, or is the W part correctly written?

Thanks!

[Torsten]

syms C x
h = 1;
g = 5;
y = C-exp(2*g*1i*x/h);
z = C-exp(-2*g*1i*x/h);
prod = y*z*(1+x^2);
Csol = solve(int(prod,x,0,2*pi)-1==0,C);

Best wishes

Torsten.

[Sergio Manzetti]

All ok! Worked Thanks!

[Sergio Manzetti]

PS: Torsten, I get TWO values if I use double(Csol), how does that figure?

[Torsten]

From your Ansatz, you get a quadratic equation in C which usually has two solutions.

Let MATLAB show "int(prod,x,0,2*pi)" and you'll see.

Best wishes

Torsten.

[Sergio Manzetti]

Thanks Torsten!

[GNANASEKARAN S]

Dear Torsten,

I have the source term of the form u(1-int u) in two dimension. How to write the FDM code? Could you please tell.