Continuity of $nabla Z$ when the equation for $Z$ has a discontinuity











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I am working in a problem of fluid dynamics (diffusion flames, exactly) and I am struggling to prove the continuity of a variable. The function $Z:R^nto R$ is continuous, but not necessarily smooth. Being $H(Z-Z_0)$ the Heaviside function, the equation for $Z$ is
$$
H(Z-Z_0) nabla cdot (mathbf{u} Z) = nabla^2 Z,
$$

with $nabla cdot mathbf{u}=sum_i delta(mathbf{r}-mathbf{r}_i)$, it is, $mathbf{u}$ is divergence-free almost everywhere, except for points inside the surface $Z=Z_0$ in which there are sources of $mathbf{u}$.




What can be said about the continuity of $nabla Z$ across the surface defined by $Z=Z_0$?




(If it's relevant, the boundary condition to $Z$ is piecewise continuous).



That is my attempt, assuming that $H(Z-Z_0)nabla Z = nabla [H(Z-Z_0) Z]$: in that case, the equation can be written as
$$
nabla cdot (mathbf{u} H(Z-Z_0) (Z-Z_1)) = nabla^2 Z,
$$

in which $Z_1$ is an arbitrary constant and, supposedly, does not afect the differential operator. Integrating the equation in the volume enclosed by $Z=C$,
$$
oint_{Z=C} H(Z-Z_0) (Z-Z_1) mathbf{u}cdot mathbf{n} dS = oint_{Z=C} nabla Z cdot mathbf{n} dS.
$$

Using $C=Z_0pm epsilon$ and taking the limit $epsilon to 0$ leads to
$$
oint_{Z=Z_0^+} (Z_0-Z_1) mathbf{u} cdot mathbf{n} dS = oint_{Z=Z_0^+} nabla Z cdot mathbf{n} dS,
$$

$$
0 = oint_{Z=Z_0^-} nabla Z cdot mathbf{n} dS.
$$

If $Z_1=Z_0$, $nabla Z$ is continuous across the surface $Z=Z_0$; otherwise, it's discontinuous. However, if $Z_1$ could be chosen arbitrarily, it would make no difference to the behaviour of $nabla Z$; if $Z_1=Z_0$ cannot be chosen, it's because $nabla(Z-Z_0) neq nabla Z$ and $nabla Z$ is not continuous at $Z=Z_0$.










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  • I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
    – Calvin Khor
    Nov 16 at 20:25












  • but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
    – Calvin Khor
    Nov 16 at 20:40












  • Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
    – rafa11111
    Nov 16 at 20:50

















up vote
1
down vote

favorite
1












I am working in a problem of fluid dynamics (diffusion flames, exactly) and I am struggling to prove the continuity of a variable. The function $Z:R^nto R$ is continuous, but not necessarily smooth. Being $H(Z-Z_0)$ the Heaviside function, the equation for $Z$ is
$$
H(Z-Z_0) nabla cdot (mathbf{u} Z) = nabla^2 Z,
$$

with $nabla cdot mathbf{u}=sum_i delta(mathbf{r}-mathbf{r}_i)$, it is, $mathbf{u}$ is divergence-free almost everywhere, except for points inside the surface $Z=Z_0$ in which there are sources of $mathbf{u}$.




What can be said about the continuity of $nabla Z$ across the surface defined by $Z=Z_0$?




(If it's relevant, the boundary condition to $Z$ is piecewise continuous).



That is my attempt, assuming that $H(Z-Z_0)nabla Z = nabla [H(Z-Z_0) Z]$: in that case, the equation can be written as
$$
nabla cdot (mathbf{u} H(Z-Z_0) (Z-Z_1)) = nabla^2 Z,
$$

in which $Z_1$ is an arbitrary constant and, supposedly, does not afect the differential operator. Integrating the equation in the volume enclosed by $Z=C$,
$$
oint_{Z=C} H(Z-Z_0) (Z-Z_1) mathbf{u}cdot mathbf{n} dS = oint_{Z=C} nabla Z cdot mathbf{n} dS.
$$

Using $C=Z_0pm epsilon$ and taking the limit $epsilon to 0$ leads to
$$
oint_{Z=Z_0^+} (Z_0-Z_1) mathbf{u} cdot mathbf{n} dS = oint_{Z=Z_0^+} nabla Z cdot mathbf{n} dS,
$$

$$
0 = oint_{Z=Z_0^-} nabla Z cdot mathbf{n} dS.
$$

If $Z_1=Z_0$, $nabla Z$ is continuous across the surface $Z=Z_0$; otherwise, it's discontinuous. However, if $Z_1$ could be chosen arbitrarily, it would make no difference to the behaviour of $nabla Z$; if $Z_1=Z_0$ cannot be chosen, it's because $nabla(Z-Z_0) neq nabla Z$ and $nabla Z$ is not continuous at $Z=Z_0$.










share|cite|improve this question
























  • I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
    – Calvin Khor
    Nov 16 at 20:25












  • but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
    – Calvin Khor
    Nov 16 at 20:40












  • Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
    – rafa11111
    Nov 16 at 20:50















up vote
1
down vote

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1
down vote

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I am working in a problem of fluid dynamics (diffusion flames, exactly) and I am struggling to prove the continuity of a variable. The function $Z:R^nto R$ is continuous, but not necessarily smooth. Being $H(Z-Z_0)$ the Heaviside function, the equation for $Z$ is
$$
H(Z-Z_0) nabla cdot (mathbf{u} Z) = nabla^2 Z,
$$

with $nabla cdot mathbf{u}=sum_i delta(mathbf{r}-mathbf{r}_i)$, it is, $mathbf{u}$ is divergence-free almost everywhere, except for points inside the surface $Z=Z_0$ in which there are sources of $mathbf{u}$.




What can be said about the continuity of $nabla Z$ across the surface defined by $Z=Z_0$?




(If it's relevant, the boundary condition to $Z$ is piecewise continuous).



That is my attempt, assuming that $H(Z-Z_0)nabla Z = nabla [H(Z-Z_0) Z]$: in that case, the equation can be written as
$$
nabla cdot (mathbf{u} H(Z-Z_0) (Z-Z_1)) = nabla^2 Z,
$$

in which $Z_1$ is an arbitrary constant and, supposedly, does not afect the differential operator. Integrating the equation in the volume enclosed by $Z=C$,
$$
oint_{Z=C} H(Z-Z_0) (Z-Z_1) mathbf{u}cdot mathbf{n} dS = oint_{Z=C} nabla Z cdot mathbf{n} dS.
$$

Using $C=Z_0pm epsilon$ and taking the limit $epsilon to 0$ leads to
$$
oint_{Z=Z_0^+} (Z_0-Z_1) mathbf{u} cdot mathbf{n} dS = oint_{Z=Z_0^+} nabla Z cdot mathbf{n} dS,
$$

$$
0 = oint_{Z=Z_0^-} nabla Z cdot mathbf{n} dS.
$$

If $Z_1=Z_0$, $nabla Z$ is continuous across the surface $Z=Z_0$; otherwise, it's discontinuous. However, if $Z_1$ could be chosen arbitrarily, it would make no difference to the behaviour of $nabla Z$; if $Z_1=Z_0$ cannot be chosen, it's because $nabla(Z-Z_0) neq nabla Z$ and $nabla Z$ is not continuous at $Z=Z_0$.










share|cite|improve this question















I am working in a problem of fluid dynamics (diffusion flames, exactly) and I am struggling to prove the continuity of a variable. The function $Z:R^nto R$ is continuous, but not necessarily smooth. Being $H(Z-Z_0)$ the Heaviside function, the equation for $Z$ is
$$
H(Z-Z_0) nabla cdot (mathbf{u} Z) = nabla^2 Z,
$$

with $nabla cdot mathbf{u}=sum_i delta(mathbf{r}-mathbf{r}_i)$, it is, $mathbf{u}$ is divergence-free almost everywhere, except for points inside the surface $Z=Z_0$ in which there are sources of $mathbf{u}$.




What can be said about the continuity of $nabla Z$ across the surface defined by $Z=Z_0$?




(If it's relevant, the boundary condition to $Z$ is piecewise continuous).



That is my attempt, assuming that $H(Z-Z_0)nabla Z = nabla [H(Z-Z_0) Z]$: in that case, the equation can be written as
$$
nabla cdot (mathbf{u} H(Z-Z_0) (Z-Z_1)) = nabla^2 Z,
$$

in which $Z_1$ is an arbitrary constant and, supposedly, does not afect the differential operator. Integrating the equation in the volume enclosed by $Z=C$,
$$
oint_{Z=C} H(Z-Z_0) (Z-Z_1) mathbf{u}cdot mathbf{n} dS = oint_{Z=C} nabla Z cdot mathbf{n} dS.
$$

Using $C=Z_0pm epsilon$ and taking the limit $epsilon to 0$ leads to
$$
oint_{Z=Z_0^+} (Z_0-Z_1) mathbf{u} cdot mathbf{n} dS = oint_{Z=Z_0^+} nabla Z cdot mathbf{n} dS,
$$

$$
0 = oint_{Z=Z_0^-} nabla Z cdot mathbf{n} dS.
$$

If $Z_1=Z_0$, $nabla Z$ is continuous across the surface $Z=Z_0$; otherwise, it's discontinuous. However, if $Z_1$ could be chosen arbitrarily, it would make no difference to the behaviour of $nabla Z$; if $Z_1=Z_0$ cannot be chosen, it's because $nabla(Z-Z_0) neq nabla Z$ and $nabla Z$ is not continuous at $Z=Z_0$.







differential-equations multivariable-calculus






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edited Nov 16 at 20:52

























asked Nov 16 at 20:03









rafa11111

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  • I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
    – Calvin Khor
    Nov 16 at 20:25












  • but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
    – Calvin Khor
    Nov 16 at 20:40












  • Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
    – rafa11111
    Nov 16 at 20:50




















  • I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
    – Calvin Khor
    Nov 16 at 20:25












  • but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
    – Calvin Khor
    Nov 16 at 20:40












  • Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
    – rafa11111
    Nov 16 at 20:50


















I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
– Calvin Khor
Nov 16 at 20:25






I can't solve your original problem, but I can tell you that $ int_{partial Omega} Z_1 u cdot n dS = Z_1 int_{Omega} nabla cdot u = 0$ which recovers the independence in $Z_1$ (i.e. your proof doesn't work)
– Calvin Khor
Nov 16 at 20:25














but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
– Calvin Khor
Nov 16 at 20:40






but maybe it can be rescued if you can prove that $nabla Zin C^0$ implies $F(c) = int_{Z=c} nabla Z cdot n dS $ is continuous in $c$...?
– Calvin Khor
Nov 16 at 20:40














Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
– rafa11111
Nov 16 at 20:50






Thank you for your comment, @CalvinKhor. I think you are right with your first comment but actually $nablacdot mathbf{u} neq 0$ in my particular problem, but $nabla cdot mathbf{u} = sum_i delta (mathbf{r}-mathbf{r}_i)$, since there are sources of $mathbf{u}$ in the domain. Therefore, the net flow of $mathbf{u}$ in a closed surface is not $0$.
– rafa11111
Nov 16 at 20:50

















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