- #1
vladimir69
- 130
- 0
hi
i have been trying to solve the diffusion equation using separation of variables. i know the answer should turn out something like the normal probability density function but its just turns into a mess when i try it.
i am given the following information:
[tex]\frac{\partial p}{\partial t}=D\frac{\partial^2 p}{\partial x^2}[/tex]
where [tex]p=p(x,t)[/tex]
[tex]p(0,0)=1[/tex]
[tex]p(L,t)=0[/tex]
[tex]p(-L,t)=0[/tex]
where D is a constant (its not specified whether its negative or positive). it would be safe to say the sign for D which gives the easiest solution is the right one.
now this is what i did
[tex]p(x,t)=T(t)X(x)[/tex]
[tex]X\frac{dT}{dt}=DT\frac{d^2 X}{dx^2}[/tex]
which gives 2 equations, namely
[tex]\frac{1}{DT}\frac{dT}{dt}=\lambda[/tex] (1) and
[tex]\frac{1}{X}\frac{d^2X}{dx^2}=\lambda[/tex] (2)
solving (1) and applying the initial conditions and boundary conditions i get
[tex]T(t)=\exp(D\lambda t)[/tex]
for (2) i assumed [tex]\lambda>0[/tex] to get
[tex]X(x)=a \exp(\sqrt{\lambda}x)+b \exp(-\sqrt{\lambda}x)[/tex]
then finally you plug all back into [tex]p(x,t)=T(t)X(x)[/tex]
now I'm not sure where to go here, or how to apply the initial and boundary conditions and nor can i see how this would end up being something like a normal pdf (with an exp(-x^2) in there somwhere)
thanks in advance
i have been trying to solve the diffusion equation using separation of variables. i know the answer should turn out something like the normal probability density function but its just turns into a mess when i try it.
i am given the following information:
[tex]\frac{\partial p}{\partial t}=D\frac{\partial^2 p}{\partial x^2}[/tex]
where [tex]p=p(x,t)[/tex]
[tex]p(0,0)=1[/tex]
[tex]p(L,t)=0[/tex]
[tex]p(-L,t)=0[/tex]
where D is a constant (its not specified whether its negative or positive). it would be safe to say the sign for D which gives the easiest solution is the right one.
now this is what i did
[tex]p(x,t)=T(t)X(x)[/tex]
[tex]X\frac{dT}{dt}=DT\frac{d^2 X}{dx^2}[/tex]
which gives 2 equations, namely
[tex]\frac{1}{DT}\frac{dT}{dt}=\lambda[/tex] (1) and
[tex]\frac{1}{X}\frac{d^2X}{dx^2}=\lambda[/tex] (2)
solving (1) and applying the initial conditions and boundary conditions i get
[tex]T(t)=\exp(D\lambda t)[/tex]
for (2) i assumed [tex]\lambda>0[/tex] to get
[tex]X(x)=a \exp(\sqrt{\lambda}x)+b \exp(-\sqrt{\lambda}x)[/tex]
then finally you plug all back into [tex]p(x,t)=T(t)X(x)[/tex]
now I'm not sure where to go here, or how to apply the initial and boundary conditions and nor can i see how this would end up being something like a normal pdf (with an exp(-x^2) in there somwhere)
thanks in advance
Last edited: