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I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions that you are looking for, in the sense that it does not imply the conservation of your quantity. To find out the correct condition, rewrite you PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right) + S(x,t) .$$

Now, the term that appear in the paranthesis, $ D\frac{\partial \phi}{\partial x} + v \phi = 0 $ is the total flux and this is the quantity that you must put to zero on the boundaries to conserve $\phi$. (I have added $S(x,t)$ for the sake of generality and for your comments). The boundary conditions that you have to impose are then (supposing your space domain is $(-10,10)$)

$$ D\frac{\partial \phi}{\partial x}(-10) + v \phi(-10) = 0 $$

for the left side and

$$ D\frac{\partial \phi}{\partial x}(10) + v \phi(10) = 0 $$

for the right side. These are the so-called Robin boundary condition (Remark that wikipedia explicitly says that these are the insulating conditions for advection diffusion equations).

If you set up these boundary conditions, you get the conservation properties that you were looking for. Indeed, integrating over the space domain, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \int \frac{\partial}{\partial x} \left( D \frac{\partial \phi}{\partial x} + v \phi \right) dx + \int S(x,t) dx$$

Using integration by parts on the right hand side, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(10) - \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(-10) + \int S(x,t) dx$$

Now, the two central terms vanish thanks to the boundary conditions. Integrating in time, we obtain

$$ \int_0^T \int \frac{\partial \phi}{\partial t} dx dt = \int_0^T \int S(x,t) dx dt$$

and if we are allowed to switch the first two integrals,

$$ \int \phi(x,T) dx - \int \phi(x,0) dx = \int_0^T \int S(x,t) dx$$

This shows that the domain is insulated from the exterior. In particular, if $S=0$, we get the conservation of $\phi$.

I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions you are looking for, in the sense that they do not imply the conservation of your quantity. To find the correct condition, rewrite your PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right) + S(x,t) .$$

Now, the term that appears in parentheses, $ D\frac{\partial \phi}{\partial x} + v \phi = 0 $ is the total flux and this is the quantity that you must put to zero on the boundaries to conserve $\phi$. (I have added $S(x,t)$ for the sake of generality and for your comments.) The boundary conditions that you have to impose are then (supposing your space domain is $(-10,10)$)

$$ D\frac{\partial \phi}{\partial x}(-10) + v \phi(-10) = 0 $$

for the left side and

$$ D\frac{\partial \phi}{\partial x}(10) + v \phi(10) = 0 $$

for the right side. These are the so-called Robin boundary condition (note that Wikipedia explicitly says these are the insulating conditions for advection-diffusion equations).

If you set up these boundary conditions, you get the conservation properties that you were looking for. Indeed, integrating over the space domain, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \int \frac{\partial}{\partial x} \left( D \frac{\partial \phi}{\partial x} + v \phi \right) dx + \int S(x,t) dx$$

Using integration by parts on the right hand side, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(10) - \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(-10) + \int S(x,t) dx$$

Now, the two central terms vanish thanks to the boundary conditions. Integrating in time, we obtain

$$ \int_0^T \int \frac{\partial \phi}{\partial t} dx dt = \int_0^T \int S(x,t) dx dt$$

and if we are allowed to switch the first two integrals,

$$ \int \phi(x,T) dx - \int \phi(x,0) dx = \int_0^T \int S(x,t) dx$$

This shows that the domain is insulated from the exterior. In particular, if $S=0$, we get the conservation of $\phi$.

I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions that you are looking for, in the sense that it does not imply the conservation of your quantity. To find out the correct condition, rewrite you PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right)$$

The quantity in paranthesis is the quantity that you have to put to zero on the boundary to get conservation of your quantity:

$$ D\frac{\partial \phi}{\partial x} + v \phi = 0 $$

To get convinced, integrate the PDE on the domain!

Remark that you end up with a Robin boundary condition.

Hope it helps!

I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions that you are looking for, in the sense that it does not imply the conservation of your quantity. To find out the correct condition, rewrite you PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right) + S(x,t) .$$

Now, the term that appear in the paranthesis, $ D\frac{\partial \phi}{\partial x} + v \phi = 0 $ is the total flux and this is the quantity that you must put to zero on the boundaries to conserve $\phi$. (I have added $S(x,t)$ for the sake of generality and for your comments). The boundary conditions that you have to impose are then (supposing your space domain is $(-10,10)$)

$$ D\frac{\partial \phi}{\partial x}(-10) + v \phi(-10) = 0 $$

for the left side and

$$ D\frac{\partial \phi}{\partial x}(10) + v \phi(10) = 0 $$

for the right side. These are the so-called Robin boundary condition (Remark that wikipedia explicitly says that these are the insulating conditions for advection diffusion equations).

If you set up these boundary conditions, you get the conservation properties that you were looking for. Indeed, integrating over the space domain, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \int \frac{\partial}{\partial x} \left( D \frac{\partial \phi}{\partial x} + v \phi \right) dx + \int S(x,t) dx$$

Using integration by parts on the right hand side, we have

$$ \int \frac{\partial \phi}{\partial t} dx = \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(10) - \left( D \frac{\partial \phi}{\partial x} + v \phi \right)(-10) + \int S(x,t) dx$$

Now, the two central terms vanish thanks to the boundary conditions. Integrating in time, we obtain

$$ \int_0^T \int \frac{\partial \phi}{\partial t} dx dt = \int_0^T \int S(x,t) dx dt$$

and if we are allowed to switch the first two integrals,

$$ \int \phi(x,T) dx - \int \phi(x,0) dx = \int_0^T \int S(x,t) dx$$

This shows that the domain is insulated from the exterior. In particular, if $S=0$, we get the conservation of $\phi$.

I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions that you are looking for, in the sense that it does not imply the conservation of your quantity. To find out the correct condition, rewrite you PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right)$$

The quantity in paranthesis is the quantity that you have to put to zero on the boundary to get conservation of your quantity:

$$ D\frac{\partial \phi}{\partial x} + v \phi = 0 $$

To get convinced, integrate the PDE on the domain, on the left hand side you should get a quantity that quantifies the conservation of $\phi$, on the right hand side just a boundary term.

Remark that you end up with a Robin boundary condition.

Hope it helps!

I think that one of your problems is that (as you observed in your comments) Neumann conditions are not the conditions that you are looking for, in the sense that it does not imply the conservation of your quantity. To find out the correct condition, rewrite you PDE as

$$ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial x}\left( D\frac{\partial \phi}{\partial x} + v \phi \right)$$

The quantity in paranthesis is the quantity that you have to put to zero on the boundary to get conservation of your quantity:

$$ D\frac{\partial \phi}{\partial x} + v \phi = 0 $$

To get convinced, integrate the PDE on the domain!

Remark that you end up with a Robin boundary condition.

Hope it helps!