Linearity of discretized ODE
$begingroup$
I'm currently reading a section of a book on an implicit Ruge-Kutta method, and the following is written:
We start, say, with a diagonally implicit Runge-Kutta method
$$k_i = hfleft(y_0+sumlimits_{j=1}^{i-1}a_{ij}k_j + a_{ii} k_iright), i=1,dots, s$$
$$y_1 = y_0+sumlimits_{i=1}^s b_i k_i $$
applied to the autonomous differential equation
$$y'=f(y)$$
The main idea is to linearize [the formula above]. This yields
$$k_i = hf(g_i) + hf'(g_i) + hf'(g_i)a_{ii}k_i$$
$$g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$$
Maybe my question is naive but how does one know that the first equation above is not linear while the other one is?
ordinary-differential-equations computational-mathematics runge-kutta-methods
asked Jan 23 at 19:13
sequencesequence
4,25731436
$endgroup$
add a comment |
$begingroup$
I'm currently reading a section of a book on an implicit Ruge-Kutta method, and the following is written:
We start, say, with a diagonally implicit Runge-Kutta method
$$k_i = hfleft(y_0+sumlimits_{j=1}^{i-1}a_{ij}k_j + a_{ii} k_iright), i=1,dots, s$$
$$y_1 = y_0+sumlimits_{i=1}^s b_i k_i $$
applied to the autonomous differential equation
$$y'=f(y)$$
The main idea is to linearize [the formula above]. This yields
$$k_i = hf(g_i) + hf'(g_i) + hf'(g_i)a_{ii}k_i$$
$$g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$$
Maybe my question is naive but how does one know that the first equation above is not linear while the other one is?
ordinary-differential-equations computational-mathematics runge-kutta-methods
asked Jan 23 at 19:13
sequencesequence
4,25731436
$endgroup$
add a comment |
$begingroup$
I'm currently reading a section of a book on an implicit Ruge-Kutta method, and the following is written:
We start, say, with a diagonally implicit Runge-Kutta method
$$k_i = hfleft(y_0+sumlimits_{j=1}^{i-1}a_{ij}k_j + a_{ii} k_iright), i=1,dots, s$$
$$y_1 = y_0+sumlimits_{i=1}^s b_i k_i $$
applied to the autonomous differential equation
$$y'=f(y)$$
The main idea is to linearize [the formula above]. This yields
$$k_i = hf(g_i) + hf'(g_i) + hf'(g_i)a_{ii}k_i$$
$$g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$$
Maybe my question is naive but how does one know that the first equation above is not linear while the other one is?
ordinary-differential-equations computational-mathematics runge-kutta-methods
asked Jan 23 at 19:13
sequencesequence
4,25731436
$endgroup$
I'm currently reading a section of a book on an implicit Ruge-Kutta method, and the following is written:
We start, say, with a diagonally implicit Runge-Kutta method
$$k_i = hfleft(y_0+sumlimits_{j=1}^{i-1}a_{ij}k_j + a_{ii} k_iright), i=1,dots, s$$
$$y_1 = y_0+sumlimits_{i=1}^s b_i k_i $$
applied to the autonomous differential equation
$$y'=f(y)$$
The main idea is to linearize [the formula above]. This yields
$$k_i = hf(g_i) + hf'(g_i) + hf'(g_i)a_{ii}k_i$$
$$g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$$
Maybe my question is naive but how does one know that the first equation above is not linear while the other one is?
ordinary-differential-equations computational-mathematics runge-kutta-methods
ordinary-differential-equations computational-mathematics runge-kutta-methods
asked Jan 23 at 19:13
sequencesequence
4,25731436
asked Jan 23 at 19:13
sequencesequence
4,25731436
asked Jan 23 at 19:13
sequencesequence
4,25731436
asked Jan 23 at 19:13
sequencesequence
4,25731436
asked Jan 23 at 19:13
sequencesequence
4,25731436
4,25731436
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes
$$x = hf(b + mx)$$
and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols,
$$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$
You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
add a comment |
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1 Answer
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$begingroup$
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes
$$x = hf(b + mx)$$
and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols,
$$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$
You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
add a comment |
$begingroup$
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes
$$x = hf(b + mx)$$
and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols,
$$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$
You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
add a comment |
$begingroup$
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes
$$x = hf(b + mx)$$
and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols,
$$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$
You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
The first thing to ask yourself is "linear with respect to what"? In this case, the answer is $k_i$. It is still not linear with respect to $k_j$ for any $j < i$.
The second thing is, what is meant by "linear"? Here, they mean the same thing you called "linear" back in high school (or earlier), before you ever heard about linear algebra: A polynomial of degree $1$, that may or may not have a constant term.
As for why, in the first equation, $k_i$ appears in the argument of the function $f$. $f$ is an arbitrary function. Some of those functions are going to linear, and therefore the expression will indeed be linear for them. But other $f$ are not linear, so in general, we cannot consider the first equation as linear in $k_i$
Note that if all you are interested in is $k_i$, everything else inside is just a constant. So, letting $x = k_i, b = g_i = y_0 + sum_{j=1}^{i-1} a_{ij} k_j$ and $m = a_{ii}$, your equation becomes
$$x = hf(b + mx)$$
and "linearization" is just replacing the function on the right by its tangent line at $x=0$, which has the expression $y=mhf'(b)x + hf(b)$, and so the linearized version of equation is $x = hf'(b)mx + hf(b)$, or, substituting back in the original symbols,
$$k_i = hf'(g_i)a_{ii}k_i + hf(g_i)$$
You will note that the extra $hf'(g_i)$ you have is missing. I'm pretty sure you miscopied that.
This is only of first degree in $k_i$, no matter what the function $f$ is, so it is "linear" in the broader sense.
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
answered Jan 24 at 3:29
Paul SinclairPaul Sinclair
20.4k21443
20.4k21443
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
add a comment |
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
$begingroup$
Thanks. What somewhat confuses me is that $k_i$ is also constant for some $i$, but when $i$ changes, not only does $k_i$ change, but also does so $g_i$, so the analogy where we think of $g_i$ as a constant $b$ does not appear to click entirely.
$endgroup$
– sequence
Jan 24 at 21:04
1
1
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
$begingroup$
At this stage, $k_i$ is being treated as a variable, In particular, as the variable with respect to which the formula is being linearized. If you later assign a particular value to $k_i$, while you mess around with the other parameters, that does not change anything here. And similarly for all the other variables. At this point in the calculation they are linearizing with respect to $k_i$, so the other variables are not changing. They are constant as far as $k_i$ is concerned. You can change them later, This approximation holds for any value of them. It just isn't them being linearized.
$endgroup$
– Paul Sinclair
Jan 25 at 0:29
add a comment |
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