I have this Non-linear PDE $$ \frac{\partial C}{\partial t}=\left(\frac{\partial C}{\partial x}\right)^2+C\frac{\partial^2 C}{\partial x^2} $$

Where C is a function of (x,t) It comes from the diffusion equation where D is concentration depending, and has the linear form $D=k \cdot C$. The PDE made dimensionless for simplicity.

I have tried to find a solution with finite difference methods but without luck, The PDE can be linearized but this will make the numerical solution to inaccurate so no luck there either.

So how can I get a proper numerical solution?

  • $\begingroup$ Are there boundary/initial conditions? What kind of problems did you run into when solving? $\endgroup$ – Dylan May 17 at 13:43
  • $\begingroup$ In my particular case the BC's is C(0,t)=14 and C(xi,t)=2, the IC is C(x,0)=2. And the PDE is the diffusion equation if D linear dependent of C, so D=C*k where k is some constant and C is the concentration of some contaminant. $\endgroup$ – Peter Panduro Jørgensen May 17 at 20:09
  • $\begingroup$ That doesn't make sense. If $k$ is just a constant, then $D_t = D_{xx} \implies kC_t = kC_{xx} $ which is just the same diffusion equation in $C$ $\endgroup$ – Dylan May 17 at 20:32
  • $\begingroup$ The Diffusion equation with a concentration depending diffusion coefficient is $$ \frac{\partial C}{\partial t}=\frac{\partial }{\partial x} \left(D \frac{\partial C}{\partial x}\right) $$ The k disappears because it is made dimensionless $\endgroup$ – Peter Panduro Jørgensen May 18 at 5:44
  • $\begingroup$ You should definitely specify that in the question. I was assuming $D$ was the unknown variable in the diffusion equation. $\endgroup$ – Dylan May 18 at 9:53

Assuming a lattice

$$ x_j = j h,\ \ j = 0,1,\cdots, J,\ \ J = \frac Xh\\ t_n = n k,\ \ n = 0,1,\cdots, N,\ \ N = \frac{T}{k} $$

to cope with the boundary conditions $\Omega = \{0\le x \le X, 0\le t\le T\}$ with $C_x = 0$ for $x = 0, x = X$ and $C(x,0) = f(x)$ and defining

$$ \begin{array}{rcll} \Delta C_j^n & = & C_{j+1}^n-C_j^n & \mbox{forward difference}\\ \nabla C_j^n & = & C_{j}^n-C_{j-1}^n & \mbox{backward difference}\\ \delta^2C_j^n & = & C_{j+1}^n-2C_j^n+C_{j-1}^n & \mbox{central difference} \end{array} $$

we have the difference representation

$$ \frac{C_j^{n+1}-C_j^n}{k} = C_j^n\frac{\delta^2C_j^n}{h^2}+\left(\frac{\Delta C_j^n}{h}\right)\left(\frac{\nabla C_j^n}{h}\right) $$


$$ C_j^{n+1}=C_j^n+\frac{k}{h^2} \left(C_j^n (C_{j-1}^n-2 C_j^n+C_{j+1}^n)+(C_j^n-C_{j-1}^n) (C_{j+1}^n-C_j^n)\right) $$

This process lead to truncation errors of order $k+h^2$

  • $\begingroup$ How can I implement this in a numerical solution, only have experience with FDM $\endgroup$ – Peter Panduro Jørgensen May 17 at 8:59
  • $\begingroup$ This proves that you will be able to find one specific solution to the given problem. How does this help to find a solution with an arbitrary initial condition? $\endgroup$ – Artem May 17 at 13:35
  • $\begingroup$ It does not matter. Assume that you have a boundary (say, Dirichlet) condition and some initial condition. Your suggestion in this answer is basically worthless in this situation. $\endgroup$ – Artem May 17 at 14:51
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    $\begingroup$ These are not "eigenfunctions", this is my main point: for a nonlinear equation the superposition principle does not work and knowing your specific solutions will not help to solve the original problem. $\endgroup$ – Artem May 17 at 17:13

You can look at this article

The way you obtain the result that is shown in the article is to look for a scaling function : $$ C(t,r) = t^{-a s} F(r t^{-a} )$$

You obtain an equation and you impose that only the variable : $x=r t^{-a} $ remains, sinceyou want a separation of variables. You obtain a relation between $s$ and $n$ that are defined in the article : $a=1/(sn+2)$. And you finish the work.

Tell me if you want me to write the whole calculation.


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