How do i discretize this Temperature flow equation with the Fnite Difference method, when the conductivity K, is not constant?: $$ \frac{\partial}{\partial x}\left(K\frac{\partial T}{\partial x}\right) + \frac{\partial}{\partial y}\left(K\frac{\partial T}{\partial y}\right) = 0 $$

My attempt at this problem was to do the following:

$$ K(x)\frac{\partial^2 T}{\partial x^2} + K(y)\frac{\partial^2 T}{\partial y^2} = 0 $$ then discretize the second order partial equations: $$ K(x) \left(\frac{T_{i+1,j} -2T_{i,j} + T_{i-1,j} }{\Delta x}\right) + K(y) \left(\frac{T_{i,j+1} -2T_{i,j} + T_{i,j-1} }{\Delta y}\right) = 0 $$

However, i have been informed that this is an incorrect method as this working assumes a constant conductivity, where that is not the case in the problem.

would someone be able to help me with this?


Let us consider the 1D case for simplicity. The second-order centered finite-difference may be viewed as the composition of two first-order centered finite-differences: $$ \left.\frac{\text d^2 T}{\text d x^2} \right|_{x=x_i}\simeq \frac{T_{i+1}-2T_{i}+T_{i-1}}{{\Delta x}^2} = \frac{1}{\Delta x}\left(\frac{T_{i+1}-T_{i}}{{\Delta x}} - \frac{T_{i}-T_{i-1}}{{\Delta x}}\right) . $$ In the spatially-varying conductivity case, the latter finite-difference writes as $$ \left.\frac{\text d}{\text d x}\left(K\frac{\text d T}{\text d x}\right) \right|_{x=x_i}\simeq K_{i+1/2}\frac{T_{i+1}-T_{i}}{{\Delta x}^2} - K_{i-1/2}\frac{T_{i}-T_{i-1}}{{\Delta x}^2}\, , $$ where $K_{i\pm 1/2} = K\big(x_i\pm\frac{1}{2}\Delta x\big)$. It is incorrect to take $K$ out of the derivative as proposed in OP, since by the product rule, $(K T')' = K T'' + K'T'$ where prime $'$ denotes differentiation w.r.t. $x$. This expression suggests to compute the finite-difference \begin{aligned} \left.\frac{\text d}{\text d x}\left(K\frac{\text d T}{\text d x}\right) \right|_{x=x_i} &\simeq \frac{K_{i+1/2}+K_{i-1/2}}{2}\frac{T_{i+1}-2T_{i}+T_{i-1}}{{\Delta x}^2}\\ &\phantom{ \simeq } + \frac{K_{i+1/2}-K_{i-1/2}}{{\Delta x}}\frac{T_{i+1}-T_{i-1}}{2\,{\Delta x}}\, \end{aligned} which leads to the same method as before. Other suggestions and the same one can be found here. In 2D, we have $\partial_x(K \partial_x T) + \partial_y(K \partial_y T)$ with conductivity $K(x,y)$ instead, but the same principle applies (see e.g. this post).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.