Conservation Laws for the KdV: Difference between revisions

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}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
</math></center>
The second integral will be zero if <math>u\rightarrow 0</math> as <math>x\rightarrow \pm
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore  
\infty .</math> Therefore  
<center><math>
<center><math>
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\int_{-\infty }^{\infty }u\mathrm{d} x
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
</math></center>
is conserved. This corresponds to conservation of momentum. We can also
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as  
write the KdV equation as  
<center><math>
<center><math>
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\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
</math></center>
must be conserved. This corresponds to conservation of energy. It turns out
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
that there is an infinite number of conserved quantities and we give here
the proof of this.
the proof of this.
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^{n}w_{n}\left( x,t\right)  
^{n}w_{n}\left( x,t\right)  
</math></center>
</math></center>
Since the equation is in conservation form then  
Since we can rewrite the equation into the conservation form <math>\partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0</math>, then  
<center><math>
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
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<center><math>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>
</math></center>


We can solve recurrsively to obtain
We can solve recursively to obtain
<center><math>
<center><math>
w_{0}=u
w_{0}=u
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</math></center>
</math></center>
<center><math>
<center><math>
w_{3} = \partial _{x}^3 u  + 4 u \partial_{x} u
w_{3} = \partial _{x}^3 u  + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u  + 2 u^2)
</math></center>
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just the derivative (with some modification) of the previous law and  
<center><math>
therefore does not actually provide a new conservation law.
w_{4} = \partial _{x}^4 u  + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u  + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just an <math>\partial _{x}</math> of some <math>X(u)</math> and therefore does not actually provide a conservation law.
 
As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>u</math> (mass), <math>u^2</math> (momentum), and <math>2 u^3 - (\partial_{x} u)^2</math> (energy).


== Lecture Videos ==  
== Lecture Videos ==  

Latest revision as of 19:17, 24 September 2025

Nonlinear PDE's Course
Current Topic Conservation Laws for the KdV
Next Topic Introduction to the Inverse Scattering Transform
Previous Topic Numerical Solution of the KdV


One of the most interesting freatures of the KdV is the existence of infinitely many conservation laws. Lets begin with some basics of conservation laws. If we can write our equation of the form

[math]\displaystyle{ \partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0 }[/math]

Then we can integrate this equation from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \infty }[/math] to obtain

[math]\displaystyle{ \int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty }^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x }[/math]

The second integral will be zero if [math]\displaystyle{ X(u)\rightarrow 0 }[/math] as [math]\displaystyle{ x\rightarrow \pm \infty . }[/math] Therefore

[math]\displaystyle{ \partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0 }[/math]

so that the quantity

[math]\displaystyle{ \int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x }[/math]

must be conserved by the solution of the equation. For the KdV we can write

[math]\displaystyle{ \partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0. }[/math]

so that we immediately see that the quantity

[math]\displaystyle{ \int_{-\infty }^{\infty }u\mathrm{d} x }[/math]

is conserved. This corresponds to conservation of mass. We can also write the KdV equation as

[math]\displaystyle{ \partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial _{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0 }[/math]

so that the quantity

[math]\displaystyle{ \int_{-\infty }^{\infty }u^{2}\mathrm{d} x }[/math]

must be conserved. This corresponds to conservation of momentum. It turns out that there is an infinite number of conserved quantities and we give here the proof of this.

Modified KdV

The modified KdV is

[math]\displaystyle{ \partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0 }[/math]

It is connected to the KdV by Miura's transformation

[math]\displaystyle{ u=-\left( v^{2}+\partial _{x}v\right) }[/math]

If we substitute this into the KdV we obtain

[math]\displaystyle{ \partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v\right) }[/math]

Note that this shows that every solution of the mKdV is a solution of the KdV but not vice versa.

Proof of an Infinite Number of Conservation Laws

An ingenious proof of the exisitence of an infinite number of conservation laws can be obtain from a generalization of Miura's transformation

[math]\displaystyle{ u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2} }[/math]

If we substitute this into the KdV we obtain

[math]\displaystyle{ \partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left( 1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial _{x}^{3}w\right) }[/math]

Therefore [math]\displaystyle{ u }[/math] solves the KdV equation provided that

[math]\displaystyle{ \partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial _{x}^{3}w=0 }[/math]

We write the solution to this equation as a formal power series

[math]\displaystyle{ w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) }[/math]

Since we can rewrite the equation into the conservation form [math]\displaystyle{ \partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0 }[/math], then

[math]\displaystyle{ \int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant} }[/math]

and since this is true for all [math]\displaystyle{ \varepsilon }[/math] this implies that

[math]\displaystyle{ \int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant} }[/math]

We then consider the expression

[math]\displaystyle{ u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2} }[/math]

which implies that

[math]\displaystyle{ u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon \partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) \right) ^{2} }[/math]

It we equate powers of [math]\displaystyle{ \varepsilon }[/math] we obtain

[math]\displaystyle{ u=w_{0} }[/math]
[math]\displaystyle{ 0=w_{1}-\partial _{x}w_{0} }[/math]
[math]\displaystyle{ 0=w_{2}-\partial _{x}w_{1}-w_{0}^{2} }[/math]
[math]\displaystyle{ 0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1} }[/math]
[math]\displaystyle{ 0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2 }[/math]

We can solve recursively to obtain

[math]\displaystyle{ w_{0}=u }[/math]
[math]\displaystyle{ w_{1}=\partial _{x}u }[/math]
[math]\displaystyle{ w_{2} = \partial_{x}^2 u + u^{2} }[/math]
[math]\displaystyle{ w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u + 2 u^2) }[/math]
[math]\displaystyle{ w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2 }[/math]

Note that each of the odd conservation laws ([math]\displaystyle{ w_1, w_3 }[/math] etc.) are just an [math]\displaystyle{ \partial _{x} }[/math] of some [math]\displaystyle{ X(u) }[/math] and therefore does not actually provide a conservation law.

As [math]\displaystyle{ \int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0 }[/math], [math]\displaystyle{ w_0 }[/math], [math]\displaystyle{ w_2 }[/math], and [math]\displaystyle{ w_4 }[/math] correspond to conservation of [math]\displaystyle{ u }[/math] (mass), [math]\displaystyle{ u^2 }[/math] (momentum), and [math]\displaystyle{ 2 u^3 - (\partial_{x} u)^2 }[/math] (energy).

Lecture Videos

Part 1

Part 2