Stationary Solution to Reaction-Diffusion Eq w/ Boundary Conditions

[Signifier]

Homework Statement

What is the stationary (steady state) solution to the following reaction diffusion equation:

<br /> \frac{\partial C}{\partial t}= \nabla^2C - kC<br />

Subject to the boundary conditions C(x, y=0) = 1, C(x = 0, y) = C(x = L, y) (IE, periodic boundary conditions along the x-axis, the value at x=0 is the same as at x=L). Also, at y = 0 and y = L, \frac{\partial C}{\partial x} = \frac{\partial C}{\partial y} = 0.

Homework Equations

With

\frac{\partial C}{\partial t} = 0,

rearrange to:

<br /> \nabla^2C = kC<br />
...

The Attempt at a Solution

I believe I can solve this PDE without the boundary conditions, at least the one equation is satisified by a sum of hyberbolic sine or cosine functions. I have absolutely no idea how to incorporate the boundary conditions though. That they are periodic across x tells me that the solution should be symmetric about x = L / 2, but I have no mathematical reasons for this. I have never taking a PDE class before so I am a bit out of my element... any help would be very useful. I know that there IS an analytic solution with these constraints, but I haven't a clue what it is.

 

[Kummer]

Homework Statement

What is the stationary (steady state) solution to the following reaction diffusion equation:

<br /> \frac{\partial C}{\partial t}= \nabla^2C - kC<br />

Subject to the boundary conditions C(x, y=0) = 1, C(x = 0, y) = C(x = L, y) (IE, periodic boundary conditions along the x-axis, the value at x=0 is the same as at x=L). Also, at y = 0 and y = L, \frac{\partial C}{\partial x} = \frac{\partial C}{\partial y} = 0.

Steady state implies that \frac{\partial C}{\partial t} = 0
And so,
\frac{\partial ^2 C}{\partial x^2} + \frac{\partial ^2 C}{\partial y^2} = kC

I do not understand the Dirichlet problem. Is this on a rectangle? Can you be more implicit with the boundary conditions?

 

[HallsofIvy]

As Kummer said, your "stationary solution" implies
\frac{\partial ^2 C}{\partial x^2} + \frac{\partial ^2 C}{\partial y^2} = kC
Now "separate variables"- Let C= X(x)Y(y) so that the equation becomes
Y\frac{d^2X}{dx^2}+ X\frac{d^2Y}{dy^2}= kXY
Divide by XY to get
\frac{1}{X}\frac{d^2X}{dx^2}+ \frac{1}{Y}\frac{d^2Y}{dy^2}= k
In order that that be true for all x the two parts involving only X and only Y must be constants (other wise, by changing x but not y, we could change the first term but not the second- their sum could not remain the same constant, k). That is, we must have
\frac{1}{X}\frac{d^2X}{dx^2}= \lambda
or
\frac{d^2X}{dx^2}= \lambda X
and
\frac{1}{Y}\frac{dY^2}{dy^2}= k- \lambda
or
\frac{d^2Y}{dx^2}= (k- \lambda )Y

The general solution will be a sum of X(x,\lambda)Y(y,\lambda) summed over all possible values of \lambda.

Can you see that, in order to satisfy periodic boundary conditions on the x-axis, \lambda must be -2n\pi for some integer n?

 

[Kummer]

The general solution will be a sum of X(x,\lambda)Y(y,\lambda) summed over all possible values of \lambda.

Can you see that, in order to satisfy periodic boundary conditions on the x-axis, \lambda must be -2n\pi for some integer n?

1)What are the boundary conditions? I tried reading the post several times, I do not understand what they are?

2)Is it on a rectangle?

 

[Signifier]

I am sorry to have not been more descriptive. I have not yet had time to digest HallsofIvy's response, which seems to be the most complete. To respond otherwise, though, the equation is being solved for the stationary state on a square of length/width L. The boundary conditions are: no flux at y = 0 or y = L (that is, top and bottom of square = no flux); periodic boundary conditions along x (that is, N(x = 0) = N(x = L)), and N = 1 along y = 0 (at the top of the square).

Thank you all; HallsofIvy, I will now proceed to consider what you've posted.

 

[Kummer]

and N = 1 along y = 0 (at the top of the square).

What is that supposed to mean?

Anyway, it seems to me this is a partial differential equations with a non-homogenous boundary value problem.

Which means you will have to solve for u_1(x,y) so to satisfy:
\frac{\partial^2 u_1}{\partial x^2}+\frac{\partial^2 u_1}{\partial y^2} = 0 \mbox{ with }\left\{ \begin{array}{c}u_1(x,0)=u_1(L,y)=u_1(x,L)=0\\ u_1(0,y)=f(y) \end{array} \right.

And then you need to solve for u_2(x,y) so to satify:
\frac{\partial^2 u_2}{\partial x^2}+\frac{\partial^2 u_2}{\partial ^2 u_2}{\partial y^2}=0 \mbox{ with }\left\{ \begin{array}{c}u_2(x,0)=u_2(L,y)=u_2(x,0)=0\\u_2(L,y)=f(y) \end{array} \right.

Then, u(x,y)=u_1(x,y)+u_2(x,y) will be the solution to this equation.

But to solve for those two individually use the methods of separation of variables.

 

[Signifier]

Kummer: N = 1 along y = 0 means: N(x, 0) = 1 (all of the points along the line at the top of the square, at y = 0, have unit concentration).

I will now consider your response... thank you.