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4

The explanation in the book does not use von Neumann analysis at all but the absolute stability regions and the eigenvalues of the discrete Laplacian operator. For the result you specifically mentioned we use the fact that the maximum eigenvalues is $$\lambda_m \approx -\frac{4}{h^2}$$ from the expression given. We then want this eigenvalue to lie inside ...

3

I believe Von Neumann's stability analysis would give you the answer here. Consider the heat transfer equation: $$\frac{\partial \mathcal{T}}{\partial t} = \alpha \frac{\partial^{2} \mathcal{T}}{\partial x^{2}}$$ By using Forward Euler time integration and central difference in space discretization: $$\mathcal{T}^{t+\Delta t}_{x} = \mathcal{T}^{t}_{x} + \... 3 As Maxim Umansky mentioned in his comment if you find the k from your second equation, you would end up with this differential-integral equation:$$\frac{\partial u}{\partial t} = D \frac{\partial^{2} u}{\partial x^{2}} -5 \int_{0}^{x} u(x^{'},t) d x^{'}$$The discretized form of this equation is:$$\frac{u_{x}^{t+\Delta t} - u_{x}^{t}}{\Delta t} = D \frac{...

3

The equations you have are what is called a "differential-algebraic equation" (DAE) because you have only time derivatives for one of the variables (namely, $u$) but not for the other (namely, $k$). The prototypical case of this kind of equation is the time dependent Stokes equations, which has time derivatives for the velocity but not the pressure....

2

Assuming that you mistake a $t$ for an $x$ in the right hand side of the first equation, you have the following system of equations. \begin{align} \frac{\partial u}{\partial t} = D \frac{\partial^2u}{\partial x^2} + k\\ 0 = \frac{\partial k}{\partial x} - 5u\, . \end{align} To get a unique solution you would need 2 boundary conditions for $u$ and 1 boundary ...

2

Your nonlinear equation is a viscous Burgers' equation and could be linearized easily by using Cole-Hopf transformation. If I rewrite your equation as: $$u_{t} = 2 c \Big (\frac{\alpha}{2 c} u_{x} - \frac{u^{2}}{2}\Big)_{x}$$ And take: $$\mu = \frac{\alpha}{2c}$$ $$u = -2 \mu \frac{\phi_{x}}{\phi}$$ I would have: u_{x} = \frac{2 \mu}{\phi^{2}} (\phi_{x}^{2}...

2

You could refine your discretization and then compare the logarithm of the error ($\log |e|$) with the logarithm of the size ($\log h$) of your elements. Using a linear regression you could obtain an approximation of the order of convergence. Keep in mind that this order of convergence is asymptotic when $h \rightarrow 0$, so, for "large" $h$ you ...

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