# Show integral is independent of time

1. Dec 10, 2014

### motherh

1. The problem statement, all variables and given/known data

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 ($\frac{\partial I_1}{\partial t}=0$).

2. Relevant equations

-

3. 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?

2. Dec 10, 2014

### ShayanJ

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.

3. Dec 10, 2014

### Ray Vickson

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

4. Dec 10, 2014

### motherh

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?

5. Dec 10, 2014

### Ray Vickson

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.

6. Dec 11, 2014

### motherh

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!

7. Dec 11, 2014

### ShayanJ

Integrate by parts!

8. Dec 11, 2014

### motherh

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?

9. Dec 11, 2014

### motherh

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?

10. Dec 11, 2014

### ShayanJ

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!

11. Dec 11, 2014

### motherh

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?

12. Dec 11, 2014

### ShayanJ

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

13. Dec 11, 2014

### PeroK

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:

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

14. Dec 11, 2014

### motherh

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.

15. Dec 11, 2014

### ShayanJ

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.

16. Dec 11, 2014

### motherh

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