Question

The diffusion equation and a solution.

Answer 1

@Hunus
Problem is found as attachment

Answer 2

In addition to that I need to determine what N0 is.
and I know the following relations can help me by looking at it:
\[\int\limits_{0}^{\infty}e ^{-a ^{2}x ^{2}}dx=\frac{ 1 }{ 2a }\sqrt{\pi}\]
and
\[\int\limits_{-\infty}^{\infty}c(x,t)dx=N0\]

Answer 3

\[\Large \frac{\partial c(x,t)}{\partial t} = \frac{D N e^{\frac{-x^2}{4Dt}} (x^2-2 D t)}{8 \sqrt{\pi} (D t)^{5/2}}\]\[\Large \frac{\partial^2 c(x,t)}{\partial x^2} = \frac{ N e^{\frac{-x^2}{4Dt}} (x^2-2 D t)}{8 \sqrt{\pi} (D t)^{5/2}}\]

Answer 4

\[\Large \frac{\partial c(x,t)}{\partial t} = D \frac{\partial^2 c(x,t)}{\partial x^2}\]

Answer 5

So we simply take the partial derivative to the function with respect to the correct variable and see they equal each other?

Answer 6

Or it is a solution to the PDE?

Answer 7

yes

Answer 8

No idea why we where given the expressions I wrote up then.

Answer 9

Perhaps they want you to do it another way?

Answer 10

It is that other way I simply don't understand.

Answer 11

If I had to guess for another way would be to look at the generalized diffusion equation\[\frac{ \partial c }{ \partial t }=D \frac{ \partial ^{2}c }{ \partial x ^{2} }-v \frac{ \partial c }{ \partial x }\]
and use that:
\[\frac{ \partial c }{ \partial t }=\left\{ c-\left[ c+\left( \frac{ \partial c }{ \partial x } \right)l \right] \right\} \frac{ v }{ l }=-v \frac{ \partial c }{ \partial x }\]
Then assume that x=x0 at t=0.
But the expression only become longer... god my brain is frying.

Answer 12

Think I go with your way, it show what we want and most be fine :)

Answer 13

Ones again, thank you :)

Answer 14

Do you still have to find N0?

Answer 15

Yup :)

Answer 16

I know it is N0 =n0 Na where n0 is the amount of molecules and Na Avogadro's constant :)