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Lax pair and compatibility for KdV

dwsmith

Well-known member
Feb 1, 2012
1,673
\begin{align}
u_t + u_{xxx} + 6uu_x &= 0\\
L &= \partial_{xx} + u_t\\
M &= -4\partial_{xxx} - 3(2u\partial_x + u_x)
\end{align}
For the \(L\) operator, should that be \(u\) not \(u_t\)?
I ask because I found
\begin{align}
\partial_{xx}\psi &= (\lambda - u)\psi\\
\partial_t\psi &= (-4\partial_{xxx} - 6u\partial_x - 3u_x)\psi
\end{align}
Also, where did \(\lambda\) come from?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Yes, typically $L$ is chosen to be a Schrödinger operator; in your case, since you're using $u$ as the variable in KdV, you would usually write $L= \partial_{xx}+v$. So, just to be clear (and it's usually a good idea to write operators with the test function, since operators are such slippery characters), we have that $Lu=(\partial_{xx}+v)u=u_{xx}+uv$. I say this is typical. However, if you find a different $L$ and $M$ that, when plugged into the Lax equation, reproduce your pde, then those are fine.

The $\lambda$ comes from one of the two Lax equations: $Lv=\lambda v$. It's called the eigenvalue.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Yes, typically $L$ is chosen to be a Schrödinger operator; in your case, since you're using $u$ as the variable in KdV, you would usually write $L= \partial_{xx}+v$. So, just to be clear (and it's usually a good idea to write operators with the test function, since operators are such slippery characters), we have that $Lu=(\partial_{xx}+v)u=u_{xx}+uv$. I say this is typical. However, if you find a different $L$ and $M$ that, when plugged into the Lax equation, reproduce your pde, then those are fine.

The $\lambda$ comes from one of the two Lax equations: $Lv=\lambda v$. It's called the eigenvalue.
So are my first L and M ok then?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
So are my first L and M ok then?
To find out if they work, you have a significant computation ahead of you. As a guide, in my Ph.D. dissertation, I compute whether the standard Schrödinger operator and its mate produce the KdV equation when plugged into Lax's equation. You can follow that program and see if those operators work. I'd be curious to know, so if you could please post the results, I'd be most grateful.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
To find out if they work, you have a significant computation ahead of you. As a guide, in my Ph.D. dissertation, I compute whether the standard Schrödinger operator and its mate produce the KdV equation when plugged into Lax's equation. You can follow that program and see if those operators work. I'd be curious to know, so if you could please post the results, I'd be most grateful.
So I found \(LM\), \(ML\), and \(L_t\) but I don't think I did something right or there is a simplification I don't see.
\begin{align}
LMv &= -4v_{xxxxx} - 6u_{xx}v_x - 12u_xv_{xx} - 6uv_{xxx} - 3u_{xxx} - 4u_tv_{xxx} - 6u_tuv_x - 3u_tu_xv\\
-MLv &= 4v_{xxxxx} + 4u_{txxx}v + 4u_tv_{xxx} + 6uu_{tx}v + 6uu_tv_x + 3u_xv_{xx} + 3u_xu_tv\\
[LM - ML]v &= 4u_{txxx}v + 6uu_{tx}v - 6u_{xx}v_x - 9u_xv_{xx} - 3u_{xxx}\\
L_tv &= v_{xxt} + u_{tt}v\\
[L_t + [L,M]]v &= v_{xxt} + u_{tt}v + 4u_{txxx}v + 6uu_{tx}v - 6u_{xx}v_x - 9u_xv_{xx} - 3u_{xxx}
\end{align}
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Well, one thing right away I can see. If you look at the bottom of page 6 of my dissertation, you will see that when you compute $L_{t}$, everywhere you see just a $\partial_{x}$, it vanishes. That is, if $L=-\partial_{x}^{2}+u$, then $L_{t}=u_{t}$. It is not true that $L_{t}=-\partial_{t} \partial_{x}^{2}+u_{t}$.

Therefore, by analogy in your case, if you have $L=\partial_{x}^{2}+u_{t}$, then you should have $L_{t}=u_{tt}$.

Checking one or two other computations (you would make me do that, wouldn't you? ;) ). You must assemble $LM$, which in your case is
$$LMv= \left( \partial_{x}^{2}+u_{t} \right) \left( -4 \partial_{x}^{3}-6u \partial_{x}-3u_{x} \right)v= \left[ -4 \partial_{x}^{5}-6\partial_{x}^{2}u \partial_{x} -3 \partial_{x}^{2}u_{x}-4u_{t} \partial_{x}^{3}-6u_{t}u \partial_{x}-3u_{t}u_{x}\right]v.$$
The second and third terms require more computation (you want the $\partial_{x}^{n}$ parts to be at the far right of the expressions). So, you must have the following:
\begin{align*}
\partial_{x}^{2}u \partial_{x}&=u_{xx}\partial_{x}+2u_{x} \partial_{x}^{2}+u \partial_{x}^{3} \quad \text{(Equation 1.1.15 in the dissertation)}\\
\partial_{x}^{2}u&=u_{xx}+2u_{x} \partial_{x}+ \partial_{x}^{2}, \quad \text{so by analogy, we have}\\
\partial_{x}^{2}u_{x}&=u_{xxx}+2u_{xx} \partial_{x}+ \partial_{x}^{2}.
\end{align*}
Dropping these into the previous $LM$ expression yields
\begin{align*}
LMv&=\left[ -4 \partial_{x}^{5}-6(u_{xx}\partial_{x}+2u_{x} \partial_{x}^{2}+u \partial_{x}^{3}) -3 (u_{xxx}+2u_{xx} \partial_{x}+ \partial_{x}^{2})-4u_{t} \partial_{x}^{3}-6u_{t}u \partial_{x}-3u_{t}u_{x}\right]v \\
&=\left[ -4 \partial_{x}^{5}-6u_{xx}\partial_{x}-12u_{x} \partial_{x}^{2}-6u \partial_{x}^{3} -3 u_{xxx}-6u_{xx} \partial_{x}-3 \partial_{x}^{2}-4u_{t} \partial_{x}^{3}-6u_{t}u \partial_{x}-3u_{t}u_{x}\right]v \\
&=\left[ -4 \partial_{x}^{5}-12u_{xx}\partial_{x}-12u_{x} \partial_{x}^{2}-6u \partial_{x}^{3} -3 u_{xxx}-3 \partial_{x}^{2}-4u_{t} \partial_{x}^{3}-6u_{t}u \partial_{x}-3u_{t}u_{x}\right]v \\
&=-4v_{xxxxx}-12u_{xx}v_{x}-12u_{x}v_{xx}-6uv_{xxx}-3u_{xxx}v-3v_{xx}-4u_{t}v_{xxx}-6u_{t}uv_{x}-3u_{t}u_{x}v.
\end{align*}
You can see there are a couple of differences between this expression and yours. Double-check your $MLv$ equation also.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
For $MLv$, I get
\begin{align*}
MLv&= \left( -4 \partial_{x}^{3}-6u \partial_{x}-3u_{x} \right) \left( \partial_{x}^{2}+u_{t} \right)v \\
&= \left( -4 \partial_{x}^{5}-4 \partial_{x}^{3}u_{t}-6u \partial_{x}^{3}-6u \partial_{x} u_{t}-3u_{x} \partial_{x}^{2}-3u_{x}u_{t} \right) v.
\end{align*}
We need two of these finished out:
\begin{align*}
\partial_{x}^{3} u&=u_{xxx}+3u_{xx} \partial_{x}+3u_{x} \partial_{x}^{2}+u \partial_{x}^{3}, \quad \text{so by analogy}\\
\partial_{x}^{3} u_{t}&=u_{txxx}+3u_{txx} \partial_{x}+3u_{tx} \partial_{x}^{2}+u_{t} \partial_{x}^{3}.
\end{align*}
I don't have an analogy for $u \partial_{x} u_{t}$, so here goes:
$$(u \partial_{x} u_{t})v=u \partial_{x}(u_{t}v)=u(u_{tx}v+u_{t}v_{x})=
(uu_{tx}+uu_{t} \partial_{x})v,$$
and hence
$$u \partial_{x} u_{t}=uu_{tx}+uu_{t} \partial_{x}.$$
Dropping these into $MLv$ yields
\begin{align*}
MLv&=\left( -4 \partial_{x}^{5}-4 (u_{txxx}+3u_{txx} \partial_{x}+3u_{tx} \partial_{x}^{2}+u_{t} \partial_{x}^{3})-6u \partial_{x}^{3}-6(uu_{tx}+uu_{t} \partial_{x})-3u_{x} \partial_{x}^{2}-3u_{x}u_{t} \right) v \\
&=\left( -4 \partial_{x}^{5}-4 u_{txxx}-12u_{txx} \partial_{x}-12u_{tx} \partial_{x}^{2}-4u_{t} \partial_{x}^{3}-6u \partial_{x}^{3}-6uu_{tx}-6uu_{t} \partial_{x}-3u_{x} \partial_{x}^{2}-3u_{x}u_{t} \right) v \\
&=-4v_{xxxxx}-4u_{txxx}v-12u_{txx}v_{x}-12u_{tx}v_{xx}-4u_{t}v_{xxx}-6uv_{xxx}-6uu_{tx}v-6uu_{t}v_{x}-3u_{x}v_{xx}-3u_{x}u_{t}v.
\end{align*}
How does that compare?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
So, I think that $L= \partial_{x}^{2}+u_{t}$ is doomed, and here's why: there's no way you're going to get the resulting $u_{tt}$ to cancel from anything in the commutator. And the fact is, KdV is first-order in time. Therefore, that operator is incorrect.

You can see from my dissertation that the operators
\begin{align*}
L&=-\partial_{x}^{2}+u \\
M&=-4 \partial_{x}^{3}+3( \partial_{x} u+u \partial_{x})
\end{align*}
will work.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
When I was doing some reading, I noticed the operator you used that is why I asked the question. I wasn't sure if I wrote it down wrong or if it would work.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Even when I use
\begin{align}
L &= \partial_{xx} + u\\
M &= -4\partial_{xxx} - 6u\partial_x - 3\partial_xu
\end{align}
the equations don't work out.

For \([L,M]\), I obtain:
\[
-6\partial_{xx}u\partial_x + u_{xxx} + 2u\partial_{xxx} - 6u^2\partial_x + 3u\partial_xu + 3\partial_xu\partial_{xx} + 3\partial_xu^2
\]
which simplifies to
\[
u_{xxx} - 6u_{xx}\partial_x - 9u_x\partial_{xx} + 9uu_x - u\partial_{xxx}
\]
Therefore, \(- 6u_{xx}\partial_x - 9u_x\partial_{xx} + 9uu_x - u\partial_{xxx}\) has to equal \(6uu_x\) some how.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Using these identities:
\begin{align}
\partial_{xxx}u &= u_{xxx} + 3u_{xx}\partial_x + 3u_x\partial_{xx} + u\partial_{xxx}\\
\partial_{xx}u\partial_x &= u_{xx}\partial_x + 2u_x\partial_{xx} + u\partial_{xxx}\\
u\partial_xu &= uu_x + u^2\partial_x\\
\partial_xu\partial_{xx} &= u_x\partial_{xx} + u\partial_{xxx}\\
\partial_xu^2 &= 2uu_x + u^2\partial_x
\end{align}
I re-wrote \(M\) to see if it would help as
\begin{align}
L &= \partial_{xx} + u\\
M &= -4\partial_{xxx} - 6u\partial_x - 3\partial_xu
\end{align}
Then for \(LM\) and \(-ML\), I obtain:
\begin{align}
LM &= -4\partial_{xxxxx} - 6\partial_{xx}u\partial_x - 3\partial_{xxx}u - 4u\partial_{xxx} - 6u^2\partial_x - 3u\partial_xu\\
&= -4\partial_{xxxxx} - 15u_{xx}\partial_x - 21u_x\partial_{xx} - 6u\partial_{xxx} - 3u_{xxx} - 7u\partial_{xxx} - 9u^2\partial_x - 3uu_x\\
-ML &= 4\partial_{xxxxx} + 4\partial_{xxx}u + 6u\partial_{xxx} + 6u\partial_xu + 3\partial_xu\partial_{xx} + 3\partial_xu^2\\
&= 4\partial_{xxxxx} + 4u_{xxx} + 12u_{xx}\partial_x + 15u_x\partial_{xx} + 13u\partial_{xxx} + 12uu_x + 9u^2\partial_x\\
[L,M] &= u_{xxx} - 3u_{xx}\partial_x - 6u_x\partial_{xx} + 9uu_x
\end{align}
Where have I gone wrong? I have also done this with out writing \(3\partial_xu\) but instead using \(3u_x\) but leads no where as well.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Using these identities:
\begin{align}
\partial_{xxx}u &= u_{xxx} + 3u_{xx}\partial_x + 3u_x\partial_{xx} + u\partial_{xxx}\\
\partial_{xx}u\partial_x &= u_{xx}\partial_x + 2u_x\partial_{xx} + u\partial_{xxx}\\
u\partial_xu &= uu_x + u^2\partial_x\\
\partial_xu\partial_{xx} &= u_x\partial_{xx} + u\partial_{xxx}\\
\partial_xu^2 &= 2uu_x + u^2\partial_x
\end{align}
I re-wrote \(M\) to see if it would help as
\begin{align}
L &= \partial_{xx} + u\\
M &= -4\partial_{xxx} - 6u\partial_x - 3\partial_xu
\end{align}
Then for \(LM\) and \(-ML\), I obtain:
\begin{align}
LM &= -4\partial_{xxxxx} - 6\partial_{xx}u\partial_x - 3\partial_{xxx}u - 4u\partial_{xxx} - 6u^2\partial_x - 3u\partial_xu\\
&= -4\partial_{xxxxx} - 15u_{xx}\partial_x - 21u_x\partial_{xx} - 6u\partial_{xxx} - 3u_{xxx} - 7u\partial_{xxx} - 9u^2\partial_x - 3uu_x\\
-ML &= 4\partial_{xxxxx} + 4\partial_{xxx}u + 6u\partial_{xxx} + 6u\partial_xu + 3\partial_xu\partial_{xx} + 3\partial_xu^2\\
&= 4\partial_{xxxxx} + 4u_{xxx} + 12u_{xx}\partial_x + 15u_x\partial_{xx} + 13u\partial_{xxx} + 12uu_x + 9u^2\partial_x\\
[L,M] &= u_{xxx} - 3u_{xx}\partial_x - 6u_x\partial_{xx} + 9uu_x
\end{align}
Where have I gone wrong? I have also done this with out writing \(3\partial_xu\) but instead using \(3u_x\) but leads no where as well.
I am not to sure about these derivative identities. On page number 10 or page 30 of the pdf, we have that
\begin{align}
LM &= -4\partial_{xxxxx} - \frac{5}{3}\alpha u\partial_{xxx} - \frac{5}{2}\alpha u_x\partial_{xx} - \frac{1}{6}\alpha^2u^2\partial_x - 2\alpha u_{xx}\partial_x - \frac{1}{12}\alpha^2uu_x - \frac{1}{2}\alpha u_{xxx}\\
-ML &= 4\partial_{xxxxx} + \frac{5}{3}\alpha u\partial_{xxx} + \frac{5}{2}\alpha u\partial_{xx} + \frac{1}{6}\alpha^2u^2\partial_x + 2\alpha u_{xx}\partial_x + \frac{1}{4}\alpha^2uu_x + \frac{2}{3}\alpha u_{xxx}
\end{align}
In my case, \(\alpha = 6\). Using these identities, I obtain the desired results. I then compared the terms with using the previous method. The identities for \(\partial_{xxx}u\) and \(u\partial_xu\) add too many terms. The identity \(\partial_{xx}u\partial_x\) is needed though. So those two identities either don't work for all \(\alpha\) or there is an issue with them not sure what it is though.

Here is the link to the other document:
http://inside.mines.edu/~whereman/papers/Larue-MS-Thesis-2011.pdf
 
Last edited: