Heat (or diffusion) equation in semi infinite region

[shultz]

Homework Statement

In a one dimensional problem, there is a liquid where x>0 and a wall in x=0. In time t=0 an ideal mass M is injected to the liquid at the point x=x0. The diffusion coefficient of the molecules injected is D.

a. Find the concentration of the material as a function of time and space C(x,t) with the boundry condition that the material cannot pass the wall.
b. Now, let's assume that the material particles get glued to the wall such that the concentration of the material in the liquid is zero on the wall. Find the flow of the molecules on the wall as a function of time and calculate the total mass of the molecules glued to the wall from time t=0 until t=t0.

Clue: In order to meet the right boundery conditions, put an imaginary point mass at x=-x0. The sign of this mass needs to be positive or negative according to the boundary condition.

Homework Equations

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

The Attempt at a Solution

First of all, I am not sure how to translate the given data to exact mathematical boundary conditions.
Are the boundary condition in part a are:
$$\[ \left\{ \begin{array}{l} c\left( {x,0} \right) = \delta \left( {x - x_0 } \right) \\ c_x \left( {0,t} \right) = 0 \\ \end{array} \right. \]$$
If it is correct, how do I continue from here?
I know I can solve the equation for infinite region (from both sides) using Fourier transform but I am not sure how to do it on semi infinite region using the given clue.

Thanks a lot!

P.S. I am new to this forum and I am not sure where to put this question (math homework, physics homework, maybe math differential equations etc.) so please forgive me and transfer the question to its more appropriate place.

 

[mighty2000]

Hello,

Thank you for your question. I would approach this problem by first understanding the physical system and the given boundary conditions. From the given information, it seems that we are dealing with a one-dimensional problem in which a mass M is injected into a liquid at a point x=x0 at time t=0. The molecules injected have a diffusion coefficient D, and there is a wall at x=0 that prevents the material from passing through.

Based on this information, we can state the following boundary conditions:

a. At time t=0, the concentration of the material is a delta function at x=x0, indicating that all of the injected mass is located at this point.

b. The concentration of the material cannot pass through the wall at x=0, meaning that the derivative of the concentration with respect to x is equal to 0 at this point.

Using this information, we can now write the following boundary conditions:

\[
\left\{ \begin{array}{l}
c\left( {x,0} \right) = \delta \left( {x - x_0 } \right) \\
\frac{\partial c}{\partial x} \left( {0,t} \right) = 0 \\
\end{array} \right.
\]

To solve for the concentration of the material as a function of time and space, we can use the diffusion equation:

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

Since we are dealing with a one-dimensional problem, we can use the method of separation of variables to solve this equation. Let's assume that the concentration can be written as a product of two functions, one depending only on x and the other only on t:

\[
c\left( {x,t} \right) = X\left( x \right)T\left( t \right)
\]

Substituting this into the diffusion equation, we get:

\[
X\left( x \right)\frac{dT}{dt} = D\frac{d^2X}{dx^2}T\left( t \right)
\]

Dividing both sides by DT, we get:

\[
\frac{1}{T