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Is there any numerical method (like the method of splitting of variables for the equation of 2D-diffusion) for solving the following boundary value problem

$$\left\{\begin{aligned}&\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x \partial y}+\frac{\partial^2 u}{\partial x^2};\\ &(x,y,t)\in\Omega=[0,a]\times[0,b];\\ &u(x,y,0)=u_{0}(x,y); \\ &u(\partial\Omega)=0? \end{aligned}\right.$$

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Here is one suggestion.

First notice your equation is: $$ \frac{\partial u}{\partial t} = \nabla \cdot (A\nabla u), $$ where $$ A = \begin{pmatrix}1& 0 \\ 1 &0\end{pmatrix}. $$ $A$ is not positive definite here, the right side is not an elliptic operator, traditional numerical methods like finite difference or finite element is good for positive definite $A$.

Now you can make a little perturbation on $A$, say $$ A_{\epsilon} = \begin{pmatrix}1& 0 \\ 1 &\epsilon\end{pmatrix}, $$ for $\epsilon >0$ but small, now the eigenvalues of $A$ are $1$ and $\epsilon$, which gives you a well-posed diffusion equation with diffusion matrix $A_{\epsilon}$: $$ \frac{\partial u_{\epsilon}}{\partial t} = \nabla \cdot (A_{\epsilon}\nabla u_{\epsilon}). $$ Use any of your favorite numerical methods to solve above equation and let $\epsilon \to 0$ see what happened on $u_{\epsilon}$.


About finite differences: $$ \frac{\partial^2 u}{\partial x^2}\Bigg|_{(i,j);t=t_n} \approx \frac{u^n(i+1,j) -2u^n({i,j})+ u^n({i-1,j})}{(\Delta x)^2} \\ =\frac{1}{\Delta x}\left(\frac{u^n(i+1,j) -u^n(i,j)}{\Delta x}-\frac{u^n(i,j)-u^n({i-1,j})}{\Delta x}\right), $$ this is how you approximate the second partial derivative in $x$. As you can see, this is nothing but approximating $\partial/\partial x$twice. Mimicing above you could use the following to approximate the mixed partial derivative $$ \frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right)\Bigg|_{(i,j);t=t_n} \approx \frac{1}{2\Delta y} \left( \frac{\partial u}{\partial x} \Bigg|_{(i,j+1);t=t_n} - \frac{\partial u}{\partial x} \Bigg|_{(i,j-1);t=t_n}\right) \\ \approx \frac{1}{2\Delta y}\left(\frac{u^n(i+1,j+1) -u^n(i,j+1)}{\Delta x}-\frac{u^n(i,j- 1)-u^n({i-1,j- 1})}{\Delta x}\right), $$ if you use the central difference to approximate the partial derivative in $y$.

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  • $\begingroup$ Perhaps, I should have asked this question more correct: how exactly works the method of finite difference here? For example, if the equation was $$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial y^2}+\frac{\partial^2 u}{\partial x^2} $$ than we would have written the scheme $$\frac{u^{n+1/2}_{i,j}-u^{n}_{i,j}}{\Delta t}=\frac{u^{n+1}_{i+1,j}-2u^{n+1}{i,j}+u^{n+1}_{i-1,j}}{\Delta x_{1}};\\ $$ $$\frac{u^{n+1}_{i,j}-u^{n+1/2}_{i,j}}{\Delta t}=\frac{u^{n+1}_{i,j+1}-2u^{n+1}{i,j}+u^{n+1}_{i,j-1}}{\Delta x_{1}};$$ and used the algorythm of triagonal matrix. Is in this case the same? $\endgroup$
    – cool
    Jul 3, 2013 at 19:34
  • $\begingroup$ @Paul Your formula given here is not correct, second order partial numerical differentiation should have a factor of $(\Delta x)^2$ on the bottom. $\endgroup$
    – Shuhao Cao
    Jul 3, 2013 at 19:38
  • $\begingroup$ Sure, I'm editing now $\endgroup$
    – cool
    Jul 3, 2013 at 19:40
  • $\begingroup$ I can't now edit, but I see my mistake. Anyway, would this method work in the same way in my scheme? I'm not sure that I can use tridiagonal method,because when we approximate $u_{xy}$, there are 4 unknown values, and here there are only 3 $\endgroup$
    – cool
    Jul 3, 2013 at 19:46
  • $\begingroup$ @Paul Please see my edit. $\endgroup$
    – Shuhao Cao
    Jul 3, 2013 at 19:52

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