Question

Partial differential equation (PDE)

Answer 1

Need some help to solve the following PDE:
\[\frac{ \partial C _{2} }{ \partial t }=D_{2}\frac{ \partial ^{2} C _{2} }{ \partial x ^{2} }\]

I know the following relations is right for the equation:
\[J _{2}=-\frac{ RT }{ R_{r} f _{2}}\frac{ \partial C _{2} }{ \partial x }=-D _{2}\frac{ \partial C _{2} }{ \partial x }\]

Answer 2

When I try solve it I always get a dummy veriable I can't get rid off.

Answer 3

What have you tried?

Answer 4

I toke a minor cheat way (as I work with science :P ) and assumed the following conditions:
C=0 (x=infinity), C=C0 (x=-infinity)
then I toke and make the guess the solution will take the form C=C(u), where u=x/t^(1/2) and then just use substitution, y=dC/du

Answer 5

Doing so I get to the following result:
\[C_{2}=\frac{ C _{0} }{ 2 }(1.\frac{ 2 }{ \sqrt{2} }\int\limits_{0}^{\frac{ x }{ (2Dt)^{\frac{ 1 }{ 2 }} }}e ^{-y ^{2}}dy)\]
And where is y the annoying dummy variable.

Answer 6

typo: 1-2/sqrt(2)

Answer 7

you want t o use fourier method ?
or some Numerical method like crank nicolsan ?

Answer 8

Think a numerical method could be fine. I have heard that Crank–Nicolson method should be good for this kind of work. However I don't know how to use it.

Answer 9

actually what u did is the method called Combination of Variables which is perfect for this kind of PDE :)

Answer 10

Yes but the result is not that easy to work with :/

Answer 11

the result is found out to be error function which is already exists in tables

Answer 12

Well I just thought if I were able to make a simple function.

Answer 13

solve it using the method of separation of variables.
Let the C(t,x) be C=u(x)v(t)
now substitute this into the equation and separate variables.
Later notice that for bouth sides to be equal they must be constant. With this in mind solve two ODE for u and v. And you got it

Answer 14

What if it takes more than one dimension and we rewrite to:
\[\frac{ \partial C }{ \partial t }=D \nabla ^{2}C\]
Does it have any particular solution?

Answer 15

What do you mean be "What if it takes more than one dimension ". The method that I described you is 2 dimentional. It is, because the PDE you have is dependent on 2 varaiables

Answer 16

Yea sorry meant from a physical perspective. The equation before take account for the time t and the distance x

Answer 17

The new one should take account for the time t distance x and wide y fx.

Answer 18

Your 1º PDE looks like heat equation. 2º one not really sure, but similar to Poisson

Answer 19

Right. Think I need to work a bit more with it.

Answer 20

yes it is heat equation with the effect of time (not steady state ) . Poisson equation does not count time effect .
we used to solve them in my Heat and Mass transfer course . we used Finite difference approximation .