Show integral is independent of time

[motherh]

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).

Homework Equations

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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?

 

[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.

 

[Ray Vickson]

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).

Homework Equations

-

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 $dI_1/dt = 0$.

 

[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?

 

[Ray Vickson]

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.

 

[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!

 

[ShayanJ]

Integrate by parts!

 

[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?

 

[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?

 

[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!

 

[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?

 

[ShayanJ]

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

 

[PeroK]

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

 

[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.

 

[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.

 

[motherh]

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