Mixing
Calculating the jacobian matrix for a time-dependent system
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For testing purposes I would like to calculate the Jacobian matrix of the time-dependent heat equation, i.e.
$$partial_tu=nabla^2u+f$$
As far as I understood, the Jacobian can be calculated using
$$F(u)=partial_tu-nabla^2u-f$$
and (with $delta u$ a small step)
$$begin{split}F'(u)(delta u)&=limlimits_{hrightarrow0}frac{F(u+hdelta u)-F(u)}{h}\
&=limlimits_{hrightarrow0}frac{partial_t(u+hdelta u)-nabla^2(u+hdelta u)-f-partial_tu+nabla^2u+f}{h}\
&=limlimits_{hrightarrow0}frac{hpartial_t(delta u)-hnabla^2(delta u)}{h}\
&=partial_t(delta u)-nabla^2(delta u)forall hneq0end{split}$$
That result can be rewritten into
$$F'(u)(delta u)=frac{delta u_1-delta u_0}{dt}-nabla^2(delta u)$$
Is that correct? If yes, what would $delta u_1$ and $delta u_0$ be? Or is there a mistake in the last step?
numerical-methods jacobian
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
$endgroup$
add a comment |
$begingroup$
For testing purposes I would like to calculate the Jacobian matrix of the time-dependent heat equation, i.e.
$$partial_tu=nabla^2u+f$$
As far as I understood, the Jacobian can be calculated using
$$F(u)=partial_tu-nabla^2u-f$$
and (with $delta u$ a small step)
$$begin{split}F'(u)(delta u)&=limlimits_{hrightarrow0}frac{F(u+hdelta u)-F(u)}{h}\
&=limlimits_{hrightarrow0}frac{partial_t(u+hdelta u)-nabla^2(u+hdelta u)-f-partial_tu+nabla^2u+f}{h}\
&=limlimits_{hrightarrow0}frac{hpartial_t(delta u)-hnabla^2(delta u)}{h}\
&=partial_t(delta u)-nabla^2(delta u)forall hneq0end{split}$$
That result can be rewritten into
$$F'(u)(delta u)=frac{delta u_1-delta u_0}{dt}-nabla^2(delta u)$$
Is that correct? If yes, what would $delta u_1$ and $delta u_0$ be? Or is there a mistake in the last step?
numerical-methods jacobian
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
$endgroup$
$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33
add a comment |
$begingroup$
For testing purposes I would like to calculate the Jacobian matrix of the time-dependent heat equation, i.e.
$$partial_tu=nabla^2u+f$$
As far as I understood, the Jacobian can be calculated using
$$F(u)=partial_tu-nabla^2u-f$$
and (with $delta u$ a small step)
$$begin{split}F'(u)(delta u)&=limlimits_{hrightarrow0}frac{F(u+hdelta u)-F(u)}{h}\
&=limlimits_{hrightarrow0}frac{partial_t(u+hdelta u)-nabla^2(u+hdelta u)-f-partial_tu+nabla^2u+f}{h}\
&=limlimits_{hrightarrow0}frac{hpartial_t(delta u)-hnabla^2(delta u)}{h}\
&=partial_t(delta u)-nabla^2(delta u)forall hneq0end{split}$$
That result can be rewritten into
$$F'(u)(delta u)=frac{delta u_1-delta u_0}{dt}-nabla^2(delta u)$$
Is that correct? If yes, what would $delta u_1$ and $delta u_0$ be? Or is there a mistake in the last step?
numerical-methods jacobian
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
$endgroup$
For testing purposes I would like to calculate the Jacobian matrix of the time-dependent heat equation, i.e.
$$partial_tu=nabla^2u+f$$
As far as I understood, the Jacobian can be calculated using
$$F(u)=partial_tu-nabla^2u-f$$
and (with $delta u$ a small step)
$$begin{split}F'(u)(delta u)&=limlimits_{hrightarrow0}frac{F(u+hdelta u)-F(u)}{h}\
&=limlimits_{hrightarrow0}frac{partial_t(u+hdelta u)-nabla^2(u+hdelta u)-f-partial_tu+nabla^2u+f}{h}\
&=limlimits_{hrightarrow0}frac{hpartial_t(delta u)-hnabla^2(delta u)}{h}\
&=partial_t(delta u)-nabla^2(delta u)forall hneq0end{split}$$
That result can be rewritten into
$$F'(u)(delta u)=frac{delta u_1-delta u_0}{dt}-nabla^2(delta u)$$
Is that correct? If yes, what would $delta u_1$ and $delta u_0$ be? Or is there a mistake in the last step?
numerical-methods jacobian
numerical-methods jacobian
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
asked Jan 31 at 16:55
arc_lupusarc_lupus
232218
232218
$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33
add a comment |
$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33
$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33
add a comment |
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$begingroup$
Why do you have $delta u_0$? The first term should be $partial_t(delta u)$.
$endgroup$
– xpaul
Jan 31 at 17:10
$begingroup$
I intended to use the approximation $partial_tu=frac{u_1-u_0}{dt}$, if that is possible here?
$endgroup$
– arc_lupus
Jan 31 at 17:12
$begingroup$
If you want, you can write it as $frac{partial delta u}{partial t}$. What are meanings of $u_1$ and $u_0$?
$endgroup$
– xpaul
Jan 31 at 17:20
$begingroup$
When comparing it to the usual approach for time-dependent systems, $u_1$ is the new solution, and $u_0$ the old/current solution. But I do not know if that applies here, too.
$endgroup$
– arc_lupus
Jan 31 at 17:23
$begingroup$
What are the old/current solutions? I doubt that they can be used here.
$endgroup$
– xpaul
Jan 31 at 17:33