# Homework Help: Stationary solution to reaction-diffusion equation with certain boundary conditions

1. Aug 4, 2007

### Signifier

1. The problem statement, all variables and given/known data
What is the stationary (steady state) solution to the following reaction diffusion equation:

$$\frac{\partial C}{\partial t}= \nabla^2C - kC$$

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, $$\frac{\partial C}{\partial x} = \frac{\partial C}{\partial y} = 0$$.

2. Relevant equations

With

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

rearrange to:

$$\nabla^2C = kC$$
...

3. 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.

2. Aug 4, 2007

### Kummer

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?

3. Aug 4, 2007

### 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?

Last edited by a moderator: Aug 4, 2007
4. Aug 4, 2007

### Kummer

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?

5. Aug 4, 2007

### 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.

6. Aug 5, 2007

### Kummer

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 seperation of variables.

7. Aug 5, 2007

### 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.