Operators and KdV equation

Question

Consider

Given the linear differential operators $$ L=-\partial_x^2+u(x, t), \quad A=4 \partial_x^3-3 u \partial_x-3 \partial_x u, $$ considered as acting on a vector space of functions of $x$ verify the fact from lectures that the $\mathrm{KdV}$ equation $$ u_t+u_{x x x}-6 u u_x=0 $$ is equivalent to $L_t=[L, A]$.

This should be an algebra bash, however, I am a bit confused. In particular I know that

$$ L_t := \frac{\partial L}{\partial t}= - \partial_t \partial_x^2+\frac{\partial u}{\partial t} $$

however, according to my definition of commutor for linear operators $$ [A,B]= AB-BA. $$ The issue here is that there is no way to get a $-\partial_t \partial^2_x$ term in the right hand side. I suspect that the definition I was given does not apply int this context.

Question: What do $L_t$ and $[L,A]$ mean here?

Answer 1

It's not clear from your post on which space the operators $L$ are acting. For simplicity I'll just assume the functions are $C^\infty$ depending on only one variable $x$.

You should understand $\partial_x$ as being a linear operator (that is a linear function) which maps a function $f \in \mathcal C^\infty$ to it's partial derivative $\partial_x (f) \in C^\infty$

Similarly when we say that $L = \partial_x^2 + u(x,t)$ is a linear operator we are implicitly stating that $L : C^\infty \to C^\infty $ is a linear map which maps a function $f$ to a new function $L(f)$ defined by

$$ L(f) = \partial_x^2(f) + u f$$ where $uf$ is the product of $u$ and $f$.

We can write this more concisely as $$ L = \partial_x^2 + u.$$

Here it is important to understand that the above sum is to be understood as a sum of operators. That is, we view $u$ as an operator which maps a function $f$ to $uf$. That is,

$$ u : C^\infty \to C^\infty : f \mapsto uf$$

Another important remark is that any linear operator $T$ acting on $C^\infty$ can be multiplied by elements $g \in C^\infty$ as follows $$ (gT)(f)(x) = g(x) T(f)(x).$$

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Coming back to your specific question. You can see in the expression for $A$ the term $\partial_x u$. This is to be understood as the composition of the operators $\partial_x$ and $u$. It should not be understood as $u_x$ the partial derivative of $u$ with respect to $x$.

To see this more clearly we evaluate the operator $ \partial_x u $ at a function $f$.

$$ (\partial_x u)(f)= (\partial_x \circ u) (f) = \partial_x (uf) = u_x f + u f_x.$$ Where $u_x f + u f_x$ is exactly the operator $ u_x + u\partial_x$ evaluated in $f$.

This means that the operator $A$ is

$$ A = 4 \partial_x^3 -6 u \partial_x - 3 u_x.$$

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Similarly we evaluate in a function $f$ the operator $L$ to find $$ L(f) = \partial_x^2 (f) + uf$$ Taking the derivative with respect to $t$ we find

$$ \partial_t L(f) = u_t f$$ because the $\partial_x^2 (f)$ depends only on $x$. So in fact we have $$ L_t = u_t$$ which is a simpler expression to work with.

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Finally the meaning of the bracket $[L,A]$ is exactly the one in your post

$$ [L,A] = L \circ A - A \circ L.$$ That is $[L,A]$ is a new linear operator which can be evaluated in a function $f$ $$ [L,A](f) = L(A(f)) - A(L(f)).$$

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To prove the condition in your post I suggest you consider the operator $L_t - [L,A]$ and evaluate it in a function $f$ and make a lot of computations to prove that it is equal to 0 :).

I also suggest you make the effort to use the notation $u_x$ and $f_x$ for the partial derivaties of $u$ and $f$ and to avoid the notations $\partial_x(u)$ and $\partial_x(f)$ during the computations. In this way you will avoid the confusion between $\partial_x \circ u$ and $\partial_x(u)$.

Answer 2

The key here is that $L_t$ is $\frac{dL}{dt}$, not $\partial_t L$.

You need to see $L=L(t)$ as an operator (acting on functions of one variable $x$) that depends on time, so basically as a 1-parameter family of operators. In particular, it formally makes sense to differentiate this operator-valued function in the variable $t$.

You should be careful to distinguish between

• composition of differential operators (like $u\partial_x$ and $\partial_x u$)
• when you apply an operator to a function ($\frac{du}{dx}$, which as an operator is different from $\partial_x u$)
• when you take the time derivative of an operator ($\frac{d}{dt}L(t)$).

In this case, $L_t$ is the actual derivative of $L=L(t)$ with respect to $t$, not the composition between the operator $\partial_t$ and $L$. Since the operator $\partial_x^2$ does not depend on time, its derivative with respect to $t$ is zero, so $$ L_t=\frac{du}{dt}. $$ So, the term $\partial_t\partial_x^2$ should not be there.

To make it more precise: $\frac{du}{dt}$ is to be interpreted as the time derivative of the multiplication operator by the function $u(t)$, but it can also be seen as the multiplication operator by the function $\frac{du}{dt}$ (the two are formally the same object).