# Thread: Solve Diffusion Equation with Laplace Transform

1. ## Solve Diffusion Equation with Laplace Transform

Solve the diffusion equation from this differential equation (Fick's Second Law). Note that there are two different independent variables t (time) and x (distance).

$\frac {\partial c}{\partial t} = D \frac {\partial ^2 c}{\partial x^2}$

With these boundary conditions:

$c(x, t) = \begin{cases} 0, & \mbox{for } 0 < x < \infty; t = 0 \\ 0, & \mbox{for } x \to -\infty \\ 0, & \mbox{for } x \to +\infty \\ c_0, & \mbox{for } x = 0 ; t > 0 \\ \end{cases}$

The professor said to use Laplace Transform. I'm familiar with Laplace transforms and initial value problems, but I'm not sure how to use them here since there are two independent variables and there seems to be a missing limit condition:

$\frac {\partial c}{\partial x}(x=0)$

Any help is appreciated. Thanks!

2. Outline of procedure: Use the Laplace Transform (LT) in the time domain only. Result: second-order ODE in C(x,s). Solve using standard methods. Result: C(x,s). Take the inverse LT of this expression to obtain c(x,t).

Warning: the inverse LT can easily become quite hairy here, since you may not be able to find the inverse LT in a table. You might have to go back to the definition of the inverse LT in terms of a complex line integral, compute residues, etc. It's entirely possible that your solution will turn out to be an infinite series solution. That's typical.

Does this make sense?

3. Is D a constant?

4. @Prove It, yes D is a constant.

@Ackbeet, That sounds like the correct approach, but I'm stumbling on the first step. How do I take the time domain Laplace transform of the right side of the equation?

$s \cdot L\{c(x,t)\} = ?$

5. $\mathcal{L}\left\{\frac{\partial c(x,t)}{\partial t} =D\frac{\partial^{2}c(x,t) }{\partial x^{2}} \right\}\to s C(x,s)-c(x,0)=D\frac{\mathrm{d}^{2} C(x,s)}{\mathrm{d} x^{2}}.$

My capital letters and lowercase letters are very intentional. Lowercase c represents the original concentration function. Capital C represents the LT in time of the concentration function.

6. Thank you. That makes perfect sense thus far. I hate to ask again, but how do you solve that for C(x, s)? Most straight-forward, analytical techniques for solving a second order ODE require constant coefficients. The coefficient s is clearly not a constant.

Since $c(x,0) = 0$

How do you solve this:

$s C(x,s) = D \frac{\mathrm{d}^{2} C(x,s)}{\mathrm{d} x^{2}}$

7. s is a constant with respect to the differential equation at hand. You have

$C(x,s)=A\sinh\left(\sqrt{\frac{s}{D}}\,x\right)+B \cosh \left(\sqrt{\frac{s}{D}}\,x\right).$

You could use exponentials if you want, but when I worked a similar problem, I found the hyperbolic trig functions a little cleaner.

8. Thank you so much Ackbeet. I finished the problem from that point. BTW, this is a textbook derivation, although the texts refer to this as the heat equation rather than the diffusion equation (they are basically the same).

9. You're very welcome. Two books of particular worth on this subject are the following: Crank's The Mathematics of Diffusion, and Carslaw and Jaeger's Conduction of Heat in Solids. They are both definitive. Loads of Laplace Transform solutions, as well as other types of solutions.

10. ## Re: Solve Diffusion Equation with Laplace Transform

Originally Posted by Ackbeet
$\mathcal{L}\left\{\frac{\partial c(x,t)}{\partial t} =D\frac{\partial^{2}c(x,t) }{\partial x^{2}} \right\}\to s C(x,s)-c(x,0)=D\frac{\mathrm{d}^{2} C(x,s)}{\mathrm{d} x^{2}}.$

My capital letters and lowercase letters are very intentional. Lowercase c represents the original concentration function. Capital C represents the LT in time of the concentration function.
How is the Laplace transform for two-dimensional diffusion as $\frac{\partial c}{\partial t} =D\left\{\frac{\partial^{2}c(x,t) }{\partial x^{2}} + \frac{\partial^{2}c(y,t) }{\partial y^{2}}\right\}$?

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# How to solve dimensional diffusion equations using laplace transform

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