# Derive on a specific border for a particular calculus

**URL:** <https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051>\
**Category:** General Discussion\
**Created:** [June 21, 2021, 2:51pm UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051 "2021-06-21T14:51:50Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![Antonin\_a](https://avatars.discourse-cdn.com/v4/letter/a/bc8723/32.png) [@Antonin\_a](https://community.freefem.org/u/Antonin_a)\
**Post date:** [June 21, 2021, 2:51pm UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051/1 "2021-06-21T14:51:50Z")

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Dear users,

In order to obtain a partial reflective B.C. condition for a Helmholtz equation, I’d like to calculate the angle of incidence of a waterwave on a border.  
I call u the solution of the equation, and the researched angle is $\gamma$.

Here is the formula (Steward - Panchang, 1999) :

$f(\gamma) = tan(\gamma) - \frac{\frac{1}{k} \frac{\partial arg(u)}{\partial x\_s}}{ \frac{1}{k} \frac{\partial arg(u)}{\partial x\_n} + \frac{2K\_r(cos(k\beta)+K\_r)}{1+2K\_rcos(k\beta)+K\_r^2}cos(\gamma)} $

$u$ is obtained by solving the equation a first time, with $\gamma= 0$.  
I use a P1 triangulation.  
Here I’d like to determine $\gamma$, knowing k, K\_r, $\beta$, and $u$, buy solving $f(\gamma) = 0$ (with bracketing and disection).  
x\_n is the normal, and x\_s is the tangent to the border, and arg(u) is the argument of u.

My problem is to calculate the derivatives of u.

I saw that I could obtain the tangent to the border by rotating the normal by pi/2.

Then on another topic, I read that I could collect the values of u on the border : [getting data of specific points](https://ljll.math.upmc.fr/pipermail/freefempp/2010/000557.html)

So I tried to collect those values of u on the border, put them in a FE variable, take the argument of it, and apply the “dx” and “dy” operators to it. But I had an error.

Then my question is :  
**Do you have any idea on how to obtain $darg(u)/dxs$ or $darg(u)/dxn$ on a specific border ?**

Here is a simple code I made to test the method :

//param  
real k = 0.002; // wavenumber

real Kr = 0.4; //reflection coeff

real theta = 11_pi/8 ;  
real k1 = k_cos(theta);  
real k2 = k\*sin(theta);

real internRadius = 2000;  
real externRadius = 5000;

// interior disc  
border Cint(t=0,2_pi){x= internRadius_cos(t) ; y = internRadius\*sin(t) ;}

// exte disc

border CF(t=5/4._pi, 3/2.pi){x = externRadiuscos(t) ; y = externRadius_sin(t);}  
border Cext(t=-pi/2, 5/4._pi){x = externRadius_cos(t) ; y = externRadius\*sin(t);}

mesh Th = buildmesh(Cint(-90) + CF(45) + Cext(90));

//Finite element space

fespace Vh(Th, P1);  
Vh uh, vh, uinc = 0.5_exp(-1i_(k1_x+k2_y)); // uinc = incoming wave

Vh u,ur, uim; (for the real/imag solution)

Vh argu; // dNargu; // argu = argu of uh, dNargu = normal derivative ?

// Helmoltz  
solve Helmoltz(uh,vh) = -int2d(Th)(dx(uh)_dx(vh) + dy(uh)dy(vh)  
-k^2uh_vh)  
-int1d(Th,Cint)(1i_k_((1-Kr)/(1+Kr))_uh_vh)  
+on(CF, uh = -uinc);

//solution

// Obtain uh on the interior circle

// boundary dof  
int ndof;  
{ int1d(Th,Cint,qfe=qf1pE)((ndof++)\*1.); }

// indexes on the boundary  
int[int] i1(ndof);  
{  
int[int] a(2_ndof);  
Vh index=0;  
for(int i=0;i\<Vh.ndof;i++) index[][i]=i;  
int n=0;  
int1d(Th,Cint,qfe=qf1pElump)((a(n++)=floor(index+0.5))1.);  
Vh next=0;  
for(int i=0;i\<ndof;i++) next[][a(2i+1)]=a(2_i);  
i1(0)=a(1);  
for(int i=1;i\<ndof;i++) i1(i)=next[][i1(i-1)];  
}

//display the value of uh on the boundary Cint  
//for (int i=0;i\<ndof;i++){  
//cout\<\<i1(i)\<\<" uh("\<\<Th(i1(i)).x\<\<","\<\<Th(i1(i)).y\<\<")="\<\<uh[][i1(i)]\<\<endl;  
//;  
//}

//How to get darg(uh)/dn on Cint??

// plot the solution  
u = real(uh);

plot(u);

Thank you all for reading,

Best

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**Author:** ![frederichecht](https://yyz2.discourse-cdn.com/flex030/user_avatar/community.freefem.org/frederichecht/32/15_2.png) [@frederichecht](https://community.freefem.org/u/frederichecht)\
**Post date:** [June 23, 2021, 9:03am UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051/2 "2021-06-23T09:03:34Z")

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I have a problem the function argument is only defined with complex number ur+ ui _i  
and it is atan2(ui, ur) = imag( log( (ur + ui_i)/sqrt(ur^2+ui^2)) )  
then you can derive with maple in function of ui and ur , so not difficulty.

otherwise what is function argument.

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**Author:** ![Antonin\_a](https://avatars.discourse-cdn.com/v4/letter/a/bc8723/32.png) [@Antonin\_a](https://community.freefem.org/u/Antonin_a)\
**Post date:** [June 23, 2021, 10:17am UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051/3 "2021-06-23T10:17:32Z")

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Hello,

Thank you for your answer.

Indeed, the solution u will be a complex number so I will be able to take its argument easily.  
But my problem is that I can’t find how to get all the values of `grad(argu(u))*n`, where `n` is the normal to the border on only a single border of my domain.

I’d like to get those values because I’ll have to do calculations with them.

I’m not used to maple, i’ll check this right away.

Thanks again,

Regards

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<div class="post-metadata">

**Author:** ![frederichecht](https://yyz2.discourse-cdn.com/flex030/user_avatar/community.freefem.org/frederichecht/32/15_2.png) [@frederichecht](https://community.freefem.org/u/frederichecht)\
**Post date:** [June 23, 2021, 11:40am UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051/4 "2021-06-23T11:40:03Z")

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you can compute formerly

func dxargu = dx(argut(u))  
func dyargu =dy(argu(u)),

so you can do [dxargu, dyargu]’ \*[N.x,N.y]

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<div class="post-metadata">

**Author:** ![Antonin\_a](https://avatars.discourse-cdn.com/v4/letter/a/bc8723/32.png) [@Antonin\_a](https://community.freefem.org/u/Antonin_a)\
**Post date:** [June 23, 2021, 12:00pm UTC](https://community.freefem.org/t/derive-on-a-specific-border-for-a-particular-calculus/1051/5 "2021-06-23T12:00:48Z")

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I will try using this, thanks a lot !