Post by matthorr on Sept 30, 2015 17:49:47 GMT
I'm not sure I did the analysis correctly - I kind of reach the right conclusion (from the plot you can see the stability limit is 1/4), but the roots are complex and only the real part was 1/4... Has anyone come up with anything different?
\[ \tilde{u}_j^{n+1} \equiv \tilde{\sigma}u_j^n \quad \text{and} \quad u_j^{n+1} \equiv \sigma u_j^n \]
For the predictor
\[ \tilde{\sigma} = 1 + \frac{\nu\Delta t}{x^2}\Big( e^{ik\Delta x} + e^{-ik\Delta x} -2 \Big) \]
\[ \tilde{\sigma} = 1 + 2\frac{\nu\Delta t}{x^2}\Big( \cos(k\Delta x) - 1 \Big) \]
and for the corrector
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}\Big(\frac{\tilde{\sigma}e^{ik\Delta x}-2\tilde{\sigma}+\tilde{\sigma}e^{-ik\Delta x}}{2} + \frac{e^{ik\Delta x}-2+e^{-ik\Delta x}}{2}\Big) \]
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}(\tilde{\sigma} + 1)\Big( \frac{e^{ik\Delta x}+e^{-ik\Delta x}}{2}-1\Big) \]
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}(\tilde{\sigma} + 1)\Big( \cos(k\Delta x)-1\Big) \]
The predictor stability (\(|\tilde{\sigma}| \leq 1\)) for all \(k\Delta x\) is shown below
\[ -1 \leq 1 + 2\frac{\nu\Delta t}{\Delta x^2}\Big(\cos(k\Delta x)-1\Big) \leq 1 \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2}\Big(1-\cos(k\Delta x)\Big) \leq 1 \quad \text{where} \quad \Big(1-\cos(k\Delta x)\Big) \in [0,2] \quad \forall \quad k\Delta x \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2} \leq \frac{1}{2} \]
Looking at the corrector (and overall) stability (\(|\sigma| \leq 1\))
\[ -1 \leq 1 + \frac{\nu\Delta t}{\Delta x^2}\Big(\tilde{\sigma}+1\Big)\Big(\cos(k\Delta x)-1\Big) \leq 1 \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2}\Big(\tilde{\sigma}+1\Big)\Big(1-\cos(k\Delta x)\Big) \leq 2 \]
let the diffusion number \(d \doteq \frac{\nu\Delta t}{\Delta x^2}\) and substitute the predictor equation for \(\tilde{\sigma}\)
\[ 0 \leq d\Big(2-2d(1-cos(k\Delta x) \Big)\Big(1-\cos(k\Delta x)\Big) \leq 2 \]
given that \(\Big(1-\cos(k\Delta x)\Big) \in [0,2] \quad \forall \quad k\Delta x\)
\[ 0 \leq -
^2 + 4d \leq 2 \]
solving the quadratic \(-^2 + 4d = 2\) gives the real part of the root \(d = \frac{1}{4}\). Being more restrictive than the requirement for the predictor, overall stability is ensured by
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2} \leq \frac{1}{4} \quad \forall \quad k\Delta x \]
HW4 copy.pdf (199.52 KB)
\[ \tilde{u}_j^{n+1} \equiv \tilde{\sigma}u_j^n \quad \text{and} \quad u_j^{n+1} \equiv \sigma u_j^n \]
For the predictor
\[ \tilde{\sigma} = 1 + \frac{\nu\Delta t}{x^2}\Big( e^{ik\Delta x} + e^{-ik\Delta x} -2 \Big) \]
\[ \tilde{\sigma} = 1 + 2\frac{\nu\Delta t}{x^2}\Big( \cos(k\Delta x) - 1 \Big) \]
and for the corrector
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}\Big(\frac{\tilde{\sigma}e^{ik\Delta x}-2\tilde{\sigma}+\tilde{\sigma}e^{-ik\Delta x}}{2} + \frac{e^{ik\Delta x}-2+e^{-ik\Delta x}}{2}\Big) \]
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}(\tilde{\sigma} + 1)\Big( \frac{e^{ik\Delta x}+e^{-ik\Delta x}}{2}-1\Big) \]
\[ \sigma = 1 + \frac{\nu\Delta t}{x^2}(\tilde{\sigma} + 1)\Big( \cos(k\Delta x)-1\Big) \]
The predictor stability (\(|\tilde{\sigma}| \leq 1\)) for all \(k\Delta x\) is shown below
\[ -1 \leq 1 + 2\frac{\nu\Delta t}{\Delta x^2}\Big(\cos(k\Delta x)-1\Big) \leq 1 \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2}\Big(1-\cos(k\Delta x)\Big) \leq 1 \quad \text{where} \quad \Big(1-\cos(k\Delta x)\Big) \in [0,2] \quad \forall \quad k\Delta x \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2} \leq \frac{1}{2} \]
Looking at the corrector (and overall) stability (\(|\sigma| \leq 1\))
\[ -1 \leq 1 + \frac{\nu\Delta t}{\Delta x^2}\Big(\tilde{\sigma}+1\Big)\Big(\cos(k\Delta x)-1\Big) \leq 1 \]
\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2}\Big(\tilde{\sigma}+1\Big)\Big(1-\cos(k\Delta x)\Big) \leq 2 \]
let the diffusion number \(d \doteq \frac{\nu\Delta t}{\Delta x^2}\) and substitute the predictor equation for \(\tilde{\sigma}\)
\[ 0 \leq d\Big(2-2d(1-cos(k\Delta x) \Big)\Big(1-\cos(k\Delta x)\Big) \leq 2 \]
given that \(\Big(1-\cos(k\Delta x)\Big) \in [0,2] \quad \forall \quad k\Delta x\)
\[ 0 \leq -
^2 + 4d \leq 2 \]solving the quadratic \(-
^2 + 4d = 2\) gives the real part of the root \(d = \frac{1}{4}\). Being more restrictive than the requirement for the predictor, overall stability is ensured by\[ 0 \leq \frac{\nu\Delta t}{\Delta x^2} \leq \frac{1}{4} \quad \forall \quad k\Delta x \]
HW4 copy.pdf (199.52 KB)
