# Show integral is independent of time

## Homework Statement

Given that $u(x,t)$ satisfies $u_t-6uu_x+u_{xxx}=0$ (*) and $u, u_x, u_{xx} \to 0$ as $|x| \to \infty$ show that

$I_1 = \int_{-\infty}^{\infty}u^2dx$

is independent of time ($\frac{\partial I_1}{\partial t}=0$).

-

## The Attempt at a Solution

I guess it would suffice to show that $I_1$ is equal to some function of (only) $x$. I can't think of a way to do this which involves the equation (*). My attempt so far has been to do some kind of substitution so that I integrate with respect to $u$ rather than $x$. Something like

$I_1 = \int_{-\infty}^{\infty}u^2dx = \int_{u(-\infty,t)}^{u(\infty, t)}u^2 \frac{dx}{du} du = \int_{0}^{0} \frac{u^2}{u_x} dx = 0$

Am I on the right track?

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ShayanJ
Gold Member
That doesn't seem right to me. Use Leibniz integral rule for differentiating the derivative w.r.t. time and then use the differential equation for substitution.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Given that $u(x,t)$ satisfies $u_t-6uu_x+u_{xxx}=0$ (*) and $u, u_x, u_{xx} \to 0$ as $|x| \to \infty$ show that

$I_1 = \int_{-\infty}^{\infty}u^2dx$

is independent of time ($\frac{\partial I_1}{\partial t}=0$).

-

## The Attempt at a Solution

I guess it would suffice to show that $I_1$ is equal to some function of (only) $x$. I can't think of a way to do this which involves the equation (*). My attempt so far has been to do some kind of substitution so that I integrate with respect to $u$ rather than $x$. Something like

$I_1 = \int_{-\infty}^{\infty}u^2dx = \int_{u(-\infty,t)}^{u(\infty, t)}u^2 \frac{dx}{du} du = \int_{0}^{0} \frac{u^2}{u_x} dx = 0$

Am I on the right track?

As "Shyan" has already suggested, you can show that ##I_1## is a constant by showing that ##dI_1/dt = 0##.

Thanks guys.

I've given your advice a go. So far I have used Leibniz integral rule and $f=u^2$:

$\frac{\partial I_1}{\partial t} = \frac{\partial}{\partial t}(\int_{-\infty}^{\infty} f(x,t)dx) = \int_{-\infty}^{\infty} \frac{\partial f}{\partial t} dx = \int_{-\infty}^{\infty} uu_tdx$.

Where do I go from here?

Ray Vickson
Homework Helper
Dearly Missed
Thanks guys.

I've given your advice a go. So far I have used Leibniz integral rule and $f=u^2$:

$\frac{\partial I_1}{\partial t} = \frac{\partial}{\partial t}(\int_{-\infty}^{\infty} f(x,t)dx) = \int_{-\infty}^{\infty} \frac{\partial f}{\partial t} dx = \int_{-\infty}^{\infty} uu_tdx$.

Where do I go from here?
Are you forgetting that ##u## satisfies a partial differential equation? You have all the information you need; it is just a matter of applying it.

I thought about subbing in $u_t=6uu_x-u_{xxx}$ to get

$6\int_{-\infty}^{\infty}u^2u_xdx-\int_{-\infty}^{\infty}uu_{xxx}dx$.

I can compute the first integral to get $\frac{1}{3}u^3$ evaluated at $-\infty$ to $\infty$ (so that integral is equal to 0) but I can't do anything with the second integral. I had this problem last night but decided to sleep on it - unfortunately that didn't help!

ShayanJ
Gold Member
Integrate by parts!

Right, I got it thanks!

There's another part to the question, it's pretty much the same as part one but now for

$I_2=\int_{-\infty)^{\infty}(u^3+\frac{1}{2}u_x^2)dx$.

I got it down to

$\frac{\partial I_2}{\partial t}=3\int_{-\infty)^{\infty}u^2u_tdx + \int_{-\infty)^{\infty}u_xu_{xt}$.

Now subbing in $u_t=6uu_x-u_{xxx}$ and $u_{tx}=6u_x^2+6u_{xx}-u_{xxxx}$ gives a few difficult integrals to solve. I can do three of them but the other two I can't get past are

$\int_{-\infty)^{\infty}u^2u_{xxx}dx$ and $\int_{-\infty)^{\infty}u_x^3dx$.

Have I made a mistake to get to this point? If not, are these integrals possible?

Right, I got it thanks!

There's another part to the question, it's pretty much the same as part one but now for

$I_2 = \int_{-\infty}^{\infty}(u^3+\frac{1}{2}u_x^2)dx$.

I got it down to

$\frac{\partial I_2}{\partial t}=3\int_{-\infty}^{\infty}u^2u_tdx + \int_{-\infty}^{\infty}u_xu_{xt}$.

Now subbing in $u_t=6uu_x-u_{xxx}$ and $u_{tx}=6u_x^2+6u_{xx}-u_{xxxx}$ gives a few difficult integrals to solve. I can do three of them but the other two I can't get past are

$\int_{-\infty}^{\infty}u^2u_{xxx}dx$ and $\int_{-\infty}^{\infty}u_x^3dx$.

Have I made a mistake to get to this point? If not, are these integrals possible?

ShayanJ
Gold Member
That seems right. But you can also use $u_{xt}=u_{tx}$ and integrate by parts to eliminate the x derivative. Then substitute using the differential equation. But I'm stuck at the end too! My problem is I can't deal with $\int_{-\infty}^{\infty} u u_x u_{xx} dx$. If you prove this is zero, then its finished.

EDIT: I found it. There are two such integrals which cancel each other and give zero. This was nice!

I'm really sorry, I don't follow what you're saying. How is $u_{xt} = u_{tx}$ useful and what should be integrated by parts?

ShayanJ
Gold Member
$\int_{-\infty}^{\infty} u_x u_{tx}dx=-\int_{-\infty}^{\infty} u_{xx} u_t dx$.

PeroK
Homework Helper
Gold Member
2020 Award
Given that $u(x,t)$ satisfies $u_t-6uu_x+u_{xxx}=0$ (*) and $u, u_x, u_{xx} \to 0$ as $|x| \to \infty$ show that

$I_1 = \int_{-\infty}^{\infty}u^2dx$

is independent of time ($\frac{\partial I_1}{\partial t}=0$).
I just wanted to point out that ##I_1## is actually a function of ##t## only. So, you shouldn't be using the partial derivative wrt ##t##.

You need to be careful to check what sort of function you have - especially when you're using condensed notation. It's often better to put the variable(s) in so that can see what you're doing:

$I_1(t) = \int_{-\infty}^{\infty}u(x, t)^2dx$

So far I have
$\frac{\partial I_2}{\partial t} = \frac{\partial}{\partial t}\int_{-\infty}^{\infty}(u^3+\frac{1}{2}u_x^2)dx = 3\int_{-\infty}^{\infty}u^2u_tdx + \int_{-\infty}^{\infty}u_xu_{xt} = 3\int_{-\infty}^{\infty}u^2u_tdx - \int_{-\infty}^{\infty}u_{xx}u_tdx$

$\int_{-\infty}^{\infty}u^2u_tdx = 18\int_{-\infty}^{\infty}u^3u_xdx-3\int_{-\infty}^{\infty}u^2u_{xxx}dx$.

$\int_{-\infty}^{\infty}u_{xx}u_tdx= -6 \int_{-\infty}^{\infty}u_{xx}u_xdx + \int_{-\infty}^{\infty}u_{xx}u_{xxx}dx$.

I can compute all of these integrals (to be 0) but not $\int_{-\infty}^{\infty}u^2u_{xxx}dx$. If I can compute that integral then I'm done.

I didn't come across $\int_{-\infty}^{\infty}uu_xu_{xx}dx$ at all.

ShayanJ
Gold Member
As PeroK noted, you should write $\frac{dI_2}{dt}$ because a definite integral on x of a function of x and t, is no more a function of x.
Anyway, you did it wrong! $u_{xx}u_x$ should be $u u_{xx} u_x$. And the term you can't evaluate, can be integrated by parts to give a similar term.

I've got it! Thanks so much for all of the help, I certainly needed it!