My PDE is not obeying conservation principle
in [Matlab]
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From: Sunipa Som on 6 Jul 2010 04:40 Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <2044145382.75601.1278398289051.JavaMail.root(a)gallium.mathforum.org>... > >Hi, > >My equation is Boltzman equation. Without collision term >it is > >dN/dt+v*dN/dr+F*dN/dp=0 > >where v is velocity, F is force and p is momentum > >by rearranging few terms we are getting this > >dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > >here unit of k is 1/meter and r is also length. > >so, then my procedure of calculating total number of >particles is right or I have to do in other way? > > > >With Regards, > >Sunipa Som > > Start from the equation > dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > or > dN/dt + div(k*c1*N,-c2*r*N) = 0 > to be solved over the rectangle > V=[r_min;r_max]x[k_min,k_max]. > > Integrate over the rectangle V with boundary A to get > d/dt int_{V} N dV = int_{V} div(-k*c1*N,c2*r*N) dV. > Applying Gauss' integral theorem to the right hand side > results in > d/dt int_{V} N dV = int_{A} (-k*c1*N,c2*r*N)*n dA > where n is the unit normal pointing outwards of the > rectangle. > So if no particles enter the rectangle over the boundary, > the quantity > int_{V} N dV = > int_{r_min}^{r_max} int_{k_min}^{k_max} N(r,k) dr dk > is conserved. > This quantity can be approximated from your calculations > by > sum_{i=1}^{i=N-1} sum_{j=1}^{j=M-1} N(r_i,k_j)* > (r_(i+1)-r_(i))*(k_(j+1)-k_(j)). > > Whether this is the number of molecules in the sphere, > you must decide from the physical background of > your problem. > > Best wishes > Torsten. Hi, I am facing a typical problem to solve my Boltzmann equation. My PDE Nr(i,j)=((N(i,j)-N(i-1,j))/dr); Nk(i,j)=((N(i,j+1)-N(i,j))); Nt(i,j)=-(c1*sqrt(j).*Nr(i,j))+(c2*sqrt(j).*(i-1).*dr.*Nk(i,j)); For my equation, when I am giving Boundary condition Nr(1,j)=0;Nk(i,nk)=0; then curve is acting properly, but total number of particles decreasing with increasing time steps. But when I am giving Boundary condition Nr(1,j)=0;Nk(i,nk)=0; N(i,1)=0; then from second time steps total number of particles are constant but figure is not moving or not working properly. In my case particles are within potential trap, so it cant go out. Can you give some idea why it is happening? Thank you. With regards, Sunipa Som
From: Torsten Hennig on 6 Jul 2010 03:28 > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote > in message > <2044145382.75601.1278398289051.JavaMail.root(a)gallium. > mathforum.org>... > > >Hi, > > >My equation is Boltzman equation. Without > collision term >it is > > >dN/dt+v*dN/dr+F*dN/dp=0 > > >where v is velocity, F is force and p is momentum > > >by rearranging few terms we are getting this > > >dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > >here unit of k is 1/meter and r is also length. > > >so, then my procedure of calculating total number > of >particles is right or I have to do in other way? > > > > > >With Regards, > > >Sunipa Som > > > > Start from the equation > > dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > or > > dN/dt + div(k*c1*N,-c2*r*N) = 0 > > to be solved over the rectangle > > V=[r_min;r_max]x[k_min,k_max]. > > > > Integrate over the rectangle V with boundary A to > get > > d/dt int_{V} N dV = int_{V} div(-k*c1*N,c2*r*N) dV. > > Applying Gauss' integral theorem to the right hand > side > > results in > > d/dt int_{V} N dV = int_{A} (-k*c1*N,c2*r*N)*n dA > > where n is the unit normal pointing outwards of the > > > rectangle. > > So if no particles enter the rectangle over the > boundary, > > the quantity > > int_{V} N dV = > > int_{r_min}^{r_max} int_{k_min}^{k_max} N(r,k) dr > dk > > is conserved. > > This quantity can be approximated from your > calculations > > by > > sum_{i=1}^{i=N-1} sum_{j=1}^{j=M-1} N(r_i,k_j)* > > (r_(i+1)-r_(i))*(k_(j+1)-k_(j)). > > > > Whether this is the number of molecules in the > sphere, > > you must decide from the physical background of > > your problem. > > > > Best wishes > > Torsten. > > Hi, > I am facing a typical problem to solve my Boltzmann > equation. > My PDE > Nr(i,j)=((N(i,j)-N(i-1,j))/dr); > Nk(i,j)=((N(i,j+1)-N(i,j))); > Nt(i,j)=-(c1*sqrt(j).*Nr(i,j))+(c2*sqrt(j).*(i-1).*dr. > *Nk(i,j)); > For my equation, when I am giving Boundary condition > Nr(1,j)=0;Nk(i,nk)=0; Do you mean N(1,j) = 0; N(i,nk) = 0 here ? Otherwise your boundary conditions do not make sense since for a first-order differential equation, it is not allowed to prescribe dN/dr = 0 or dN/dk = 0 at a boundary. Or where did you read that such a boundary condition can be applied ? > then curve is acting properly, but total number of > particles decreasing with increasing time steps. If your discretization scheme is conservative, this can only happen because more particles leave the domain than enter it over the four boundaries (see my response above). > But when I am giving Boundary condition > Nr(1,j)=0;Nk(i,nk)=0; > N(i,1)=0; > then from second time steps total number of particles > are constant but figure is not moving or not working > properly. > In my case particles are within potential trap, so it > cant go out. > Can you give some idea why it is happening? > > Thank you. > With regards, > Sunipa Som Best wishes Torsten.
From: Sunipa Som on 6 Jul 2010 08:28 Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <1394417502.76713.1278415717104.JavaMail.root(a)gallium.mathforum.org>... > > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote > > in message > > <2044145382.75601.1278398289051.JavaMail.root(a)gallium. > > mathforum.org>... > > > >Hi, > > > >My equation is Boltzman equation. Without > > collision term >it is > > > >dN/dt+v*dN/dr+F*dN/dp=0 > > > >where v is velocity, F is force and p is momentum > > > >by rearranging few terms we are getting this > > > >dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > >here unit of k is 1/meter and r is also length. > > > >so, then my procedure of calculating total number > > of >particles is right or I have to do in other way? > > > > > > > >With Regards, > > > >Sunipa Som > > > > > > Start from the equation > > > dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > or > > > dN/dt + div(k*c1*N,-c2*r*N) = 0 > > > to be solved over the rectangle > > > V=[r_min;r_max]x[k_min,k_max]. > > > > > > Integrate over the rectangle V with boundary A to > > get > > > d/dt int_{V} N dV = int_{V} div(-k*c1*N,c2*r*N) dV. > > > Applying Gauss' integral theorem to the right hand > > side > > > results in > > > d/dt int_{V} N dV = int_{A} (-k*c1*N,c2*r*N)*n dA > > > where n is the unit normal pointing outwards of the > > > > > rectangle. > > > So if no particles enter the rectangle over the > > boundary, > > > the quantity > > > int_{V} N dV = > > > int_{r_min}^{r_max} int_{k_min}^{k_max} N(r,k) dr > > dk > > > is conserved. > > > This quantity can be approximated from your > > calculations > > > by > > > sum_{i=1}^{i=N-1} sum_{j=1}^{j=M-1} N(r_i,k_j)* > > > (r_(i+1)-r_(i))*(k_(j+1)-k_(j)). > > > > > > Whether this is the number of molecules in the > > sphere, > > > you must decide from the physical background of > > > your problem. > > > > > > Best wishes > > > Torsten. > > > > Hi, > > I am facing a typical problem to solve my Boltzmann > > equation. > > My PDE > > Nr(i,j)=((N(i,j)-N(i-1,j))/dr); > > Nk(i,j)=((N(i,j+1)-N(i,j))); > > Nt(i,j)=-(c1*sqrt(j).*Nr(i,j))+(c2*sqrt(j).*(i-1).*dr. > > *Nk(i,j)); > > For my equation, when I am giving Boundary condition > > Nr(1,j)=0;Nk(i,nk)=0; > > Do you mean N(1,j) = 0; N(i,nk) = 0 here ? > Otherwise your boundary conditions do not make sense > since for a first-order differential equation, > it is not allowed to prescribe dN/dr = 0 or dN/dk = 0 > at a boundary. > Or where did you read that such a boundary condition > can be applied ? > > > then curve is acting properly, but total number of > > particles decreasing with increasing time steps. > > If your discretization scheme is conservative, > this can only happen because more particles leave > the domain than enter it over the four boundaries > (see my response above). > > > But when I am giving Boundary condition > > Nr(1,j)=0;Nk(i,nk)=0; > > N(i,1)=0; > > then from second time steps total number of particles > > are constant but figure is not moving or not working > > properly. > > In my case particles are within potential trap, so it > > cant go out. > > Can you give some idea why it is happening? > > > > Thank you. > > With regards, > > Sunipa Som > > Best wishes > Torsten. Hi, Nr(1,j)=0;Nk(i,nk)=0; means dN/dr(1,j)=0 and dN/dk(i,nk)=0 which I put as boundary condition and curve is working fine. But particles numbers are decreasing. In my case particles are within potential trap, so it can not go out, so how can it decrease? and how can i resolve it? Because in my case total number of particles should be constant. Thank you. With regards, Sunipa Som
From: Torsten Hennig on 6 Jul 2010 04:50 > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote > in message > <1394417502.76713.1278415717104.JavaMail.root(a)gallium. > mathforum.org>... > > > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> > wrote > > > in message > > > > <2044145382.75601.1278398289051.JavaMail.root(a)gallium. > > > mathforum.org>... > > > > >Hi, > > > > >My equation is Boltzman equation. Without > > > collision term >it is > > > > >dN/dt+v*dN/dr+F*dN/dp=0 > > > > >where v is velocity, F is force and p is > momentum > > > > >by rearranging few terms we are getting this > > > > >dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > > >here unit of k is 1/meter and r is also > length. > > > > >so, then my procedure of calculating total > number > > > of >particles is right or I have to do in other > way? > > > > > > > > > >With Regards, > > > > >Sunipa Som > > > > > > > > Start from the equation > > > > dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > > or > > > > dN/dt + div(k*c1*N,-c2*r*N) = 0 > > > > to be solved over the rectangle > > > > V=[r_min;r_max]x[k_min,k_max]. > > > > > > > > Integrate over the rectangle V with boundary A > to > > > get > > > > d/dt int_{V} N dV = int_{V} div(-k*c1*N,c2*r*N) > dV. > > > > Applying Gauss' integral theorem to the right > hand > > > side > > > > results in > > > > d/dt int_{V} N dV = int_{A} (-k*c1*N,c2*r*N)*n > dA > > > > where n is the unit normal pointing outwards of > the > > > > > > > rectangle. > > > > So if no particles enter the rectangle over the > > > boundary, > > > > the quantity > > > > int_{V} N dV = > > > > int_{r_min}^{r_max} int_{k_min}^{k_max} N(r,k) > dr > > > dk > > > > is conserved. > > > > This quantity can be approximated from your > > > calculations > > > > by > > > > sum_{i=1}^{i=N-1} sum_{j=1}^{j=M-1} N(r_i,k_j)* > > > > (r_(i+1)-r_(i))*(k_(j+1)-k_(j)). > > > > > > > > Whether this is the number of molecules in the > > > sphere, > > > > you must decide from the physical background of > > > > > your problem. > > > > > > > > Best wishes > > > > Torsten. > > > > > > Hi, > > > I am facing a typical problem to solve my > Boltzmann > > > equation. > > > My PDE > > > Nr(i,j)=((N(i,j)-N(i-1,j))/dr); > > > Nk(i,j)=((N(i,j+1)-N(i,j))); > > > > Nt(i,j)=-(c1*sqrt(j).*Nr(i,j))+(c2*sqrt(j).*(i-1).*dr. > > > *Nk(i,j)); > > > For my equation, when I am giving Boundary > condition > > > Nr(1,j)=0;Nk(i,nk)=0; > > > > Do you mean N(1,j) = 0; N(i,nk) = 0 here ? > > Otherwise your boundary conditions do not make > sense > > since for a first-order differential equation, > > it is not allowed to prescribe dN/dr = 0 or dN/dk = > 0 > > at a boundary. > > Or where did you read that such a boundary > condition > > can be applied ? > > > > > then curve is acting properly, but total number > of > > > particles decreasing with increasing time steps. > > > > If your discretization scheme is conservative, > > this can only happen because more particles leave > > the domain than enter it over the four boundaries > > (see my response above). > > > > > But when I am giving Boundary condition > > > Nr(1,j)=0;Nk(i,nk)=0; > > > N(i,1)=0; > > > then from second time steps total number of > particles > > > are constant but figure is not moving or not > working > > > properly. > > > In my case particles are within potential trap, > so it > > > cant go out. > > > Can you give some idea why it is happening? > > > > > > Thank you. > > > With regards, > > > Sunipa Som > > > > Best wishes > > Torsten. > Hi, > Nr(1,j)=0;Nk(i,nk)=0; means dN/dr(1,j)=0 and > dN/dk(i,nk)=0 which I put as boundary condition and > curve is working fine. > But particles numbers are > decreasing. In my case particles are within potential > trap, so it can not go out, so how can it decrease? > and how can i resolve it? Because in my case total > number of particles should be constant. > > Thank you. > With regards, > Sunipa Som What is c1*sqrt(j)*N(i,j) for i=nr , -c1*sqrt(j)*N(i,j) for i=1, -c2*sqrt(j)*(i-1)*dr*N(i,j) for j=1 and c2*sqrt(j)*(i-1)*dr*N(i,j) for j=nk ? If the expressions are greater than zero, the number of particles in the volume will decrease in time. Best wishes Torsten.
From: Sunipa Som on 8 Jul 2010 05:13
Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <128269470.77069.1278420640088.JavaMail.root(a)gallium.mathforum.org>... > > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote > > in message > > <1394417502.76713.1278415717104.JavaMail.root(a)gallium. > > mathforum.org>... > > > > Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> > > wrote > > > > in message > > > > > > <2044145382.75601.1278398289051.JavaMail.root(a)gallium. > > > > mathforum.org>... > > > > > >Hi, > > > > > >My equation is Boltzman equation. Without > > > > collision term >it is > > > > > >dN/dt+v*dN/dr+F*dN/dp=0 > > > > > >where v is velocity, F is force and p is > > momentum > > > > > >by rearranging few terms we are getting this > > > > > >dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > > > >here unit of k is 1/meter and r is also > > length. > > > > > >so, then my procedure of calculating total > > number > > > > of >particles is right or I have to do in other > > way? > > > > > > > > > > > >With Regards, > > > > > >Sunipa Som > > > > > > > > > > Start from the equation > > > > > dN/dt + k*c1*dN/dr-c2*r*dN/dk=0 > > > > > or > > > > > dN/dt + div(k*c1*N,-c2*r*N) = 0 > > > > > to be solved over the rectangle > > > > > V=[r_min;r_max]x[k_min,k_max]. > > > > > > > > > > Integrate over the rectangle V with boundary A > > to > > > > get > > > > > d/dt int_{V} N dV = int_{V} div(-k*c1*N,c2*r*N) > > dV. > > > > > Applying Gauss' integral theorem to the right > > hand > > > > side > > > > > results in > > > > > d/dt int_{V} N dV = int_{A} (-k*c1*N,c2*r*N)*n > > dA > > > > > where n is the unit normal pointing outwards of > > the > > > > > > > > > rectangle. > > > > > So if no particles enter the rectangle over the > > > > boundary, > > > > > the quantity > > > > > int_{V} N dV = > > > > > int_{r_min}^{r_max} int_{k_min}^{k_max} N(r,k) > > dr > > > > dk > > > > > is conserved. > > > > > This quantity can be approximated from your > > > > calculations > > > > > by > > > > > sum_{i=1}^{i=N-1} sum_{j=1}^{j=M-1} N(r_i,k_j)* > > > > > (r_(i+1)-r_(i))*(k_(j+1)-k_(j)). > > > > > > > > > > Whether this is the number of molecules in the > > > > sphere, > > > > > you must decide from the physical background of > > > > > > > your problem. > > > > > > > > > > Best wishes > > > > > Torsten. > > > > > > > > Hi, > > > > I am facing a typical problem to solve my > > Boltzmann > > > > equation. > > > > My PDE > > > > Nr(i,j)=((N(i,j)-N(i-1,j))/dr); > > > > Nk(i,j)=((N(i,j+1)-N(i,j))); > > > > > > Nt(i,j)=-(c1*sqrt(j).*Nr(i,j))+(c2*sqrt(j).*(i-1).*dr. > > > > *Nk(i,j)); > > > > For my equation, when I am giving Boundary > > condition > > > > Nr(1,j)=0;Nk(i,nk)=0; > > > > > > Do you mean N(1,j) = 0; N(i,nk) = 0 here ? > > > Otherwise your boundary conditions do not make > > sense > > > since for a first-order differential equation, > > > it is not allowed to prescribe dN/dr = 0 or dN/dk = > > 0 > > > at a boundary. > > > Or where did you read that such a boundary > > condition > > > can be applied ? > > > > > > > then curve is acting properly, but total number > > of > > > > particles decreasing with increasing time steps. > > > > > > If your discretization scheme is conservative, > > > this can only happen because more particles leave > > > the domain than enter it over the four boundaries > > > (see my response above). > > > > > > > But when I am giving Boundary condition > > > > Nr(1,j)=0;Nk(i,nk)=0; > > > > N(i,1)=0; > > > > then from second time steps total number of > > particles > > > > are constant but figure is not moving or not > > working > > > > properly. > > > > In my case particles are within potential trap, > > so it > > > > cant go out. > > > > Can you give some idea why it is happening? > > > > > > > > Thank you. > > > > With regards, > > > > Sunipa Som > > > > > > Best wishes > > > Torsten. > > Hi, > > Nr(1,j)=0;Nk(i,nk)=0; means dN/dr(1,j)=0 and > > dN/dk(i,nk)=0 which I put as boundary condition and > > curve is working fine. > > But particles numbers are > > decreasing. In my case particles are within potential > > trap, so it can not go out, so how can it decrease? > > and how can i resolve it? Because in my case total > > number of particles should be constant. > > > > Thank you. > > With regards, > > Sunipa Som > > What is c1*sqrt(j)*N(i,j) for i=nr , > -c1*sqrt(j)*N(i,j) for i=1, > -c2*sqrt(j)*(i-1)*dr*N(i,j) for j=1 and > c2*sqrt(j)*(i-1)*dr*N(i,j) for j=nk ? > If the expressions are greater than zero, the > number of particles in the volume will decrease > in time. > > Best wishes > Torsten. Hi, In my case particle should be constant. So I want to make a reflecting wall or reflecting sphere which will reflect this particles(which moving out) inside. Do you know how can I do it? Thank you. With Regards, S Som |