Scaling Invariant, Non-Linear PDE

[BrainHurts]

Homework Statement

Consider the nonlinnear diffusion problem

$u_t - (u_x)^2 + uu_{xx} = 0, x \in \mathbb{R} , t >0$

with the constraint and boundary conditions

$\int_{\mathbb{R}} u(x,t)=1, u(\pm \inf, t)=0$

Investigate the existence of scaling invariant solutions for the equation and then use the constraint condition to show that

$u=t^{- \frac{1}{3}}y(z)$ and $z=xt^{- \frac{1}{3}}$

Homework Equations

The Attempt at a Solution

So this is what I did
Let
$x = AX$ and $t = BT$ and $u=CU$

$\frac{\partial}{\partial x} = \frac{1}{A} \frac{\partial}{\partial X}$

$\frac{\partial^2}{\partial x^2} = \frac{1}{A^2} \frac{\partial^2}{\partial X^2}$

$\frac{\partial}{\partial t} = \frac{1}{B} \frac{\partial}{\partial T}$

so

$u_t = \frac{1}{B} \frac{\partial u}{\partial T}$

$u_x = \frac{1}{A} \frac{\partial u}{\partial X}$

$u_{xx} = \frac{1}{A^2} \frac{\partial^2 u}{\partial X^2}$

and

$u_t - (u_x)^2 + uu_{xx} = 0$ becomes

$u_T - \frac{B}{A^2} (u_X)^2 - \frac{BC}{A^2}uu_{XX} = 0$

so if $\frac{B}{A^2} = 1$ and $A = \lambda, then B = \lambda^2$

and I want $\frac{BC}{A^2} = 1$ means $C = 1$

so a scaling invariant solution is $u(\lambda x, \lambda^2 t)$

but I'm getting that $z = \frac{x}{t^{\frac{1}{2}}}$

so I'm not really sure where the

$z=xt^{- \frac{1}{3}}$

 

[haruspex]

$u_T - \frac{B}{A^2} (u_X)^2 - \frac{BC}{A^2}uu_{XX} = 0$

Do you mean $U_T - \frac{B}{A^2} (U_X)^2 - \frac{BC}{A^2}UU_{XX} = 0$? If so, I think you dropped a C or two, and reversed a sign.

 

[BrainHurts]

So I redid everything from the point

$\frac{\partial}{\partial x} u = \frac{C}{A} \frac{\partial U}{\partial X}$

$\frac{\partial^2}{\partial x^2} u = \frac{C^2}{A^2} \frac{\partial^2 U}{\partial X^2}$

$\frac{\partial}{\partial t} u = \frac{C}{B} \frac{\partial U}{\partial T}$

so I get

$\frac{\partial U}{\partial T} - \frac{CB}{A^2} (\frac{\partial U}{\partial X})^2 - \frac{C^2 B}{A^2}U \frac{\partial^2 U}{\partial X^2} = 0$

so if I let $A=\lambda$ It'll follow that $C=1, B = \lambda^2$

so if I take $\frac{x}{t^{\frac{1}{2}}} = \frac{\lambda X}{(\lambda^2 T)^\frac{1}{2}} = \frac{X}{T^{\frac{1}{2}}}$

So essentially I'm not seeing that $z=xt^{\frac{-1}{3}}$

I see $z=xt^{\frac{-1}{2}}$

 

[haruspex]

Sorry, I'm not familiar with this type of problem and I don't know what principles you're applying to find y and z. The OP says to use the constraint conditions for that part, but you don't appear to have done so.
If you like, I'll post on the Homework Helpers forum to see if someone else can help you.

 

[BrainHurts]

oh yes please do so, that'll be excellent, yes i haven't used the constraint conditions quite yet and to be quite honest I'm not quite sure on how to do so.

 

[TSny]

$\frac{\partial^2}{\partial x^2} u = \frac{C^2}{A^2} \frac{\partial^2 U}{\partial X^2}$

Shouldn't C just be raised to the first power in the expression above?

Also, I think the constraint will yield a relation between A and C.

 

[haruspex]

Shouldn't C just be raised to the first power in the expression above?

Good catch - I missed that.

 

[BrainHurts]

Shouldn't C just be raised to the first power in the expression above?

Also, I think the constraint will yield a relation between A and C.

yup you're right, now let me see what I can do with this

 

[fzero]

You seem to be on the right track, but it would be simpler to be a little clearer about how the scaling works. I would consider the scaling transformation

$$ t \rightarrow \lambda t , ~~ x \rightarrow \lambda^a x, ~~ u \rightarrow \lambda^b u.$$

You can use this to define your $T,X,U$ and it should be a bit clearer since you will get a pair of linear relationships between $a$ and $b$.

 

[BrainHurts]

You seem to be on the right track, but it would be simpler to be a little clearer about how the scaling works. I would consider the scaling transformation

$$ t \rightarrow \lambda t , ~~ x \rightarrow \lambda^a x, ~~ u \rightarrow \lambda^b u.$$

You can use this to define your $T,X,U$ and it should be a bit clearer since you will get a pair of linear relationships between $a$ and $b$.

hmm, I'm getting this

$\lambda^{b-1}u_t - \lambda^{2b-2a} (u_x)^2 -\lambda^{b-2a}u_{xx} = 0$

so we'll get a scaling invariant solution if

$b-1 = 2b-2a = b-2a$ correct?


I'm getting that $a = - \frac{1}{2}, b = -2$

so if i define $z = \frac{\lambda^{\frac{-1}{2}} x}{(\lambda^{-2}t)^{(\frac{-1}{4})}}$
and we get that $z = xt^\frac{-1}{4}$

I feel like I should use the constraint based on what TSny said

I mean if we have that $\int_{\mathbb{R}} u(x,t) dx = 1$ I feel like it has something to do with

$\int_{\mathbb{R}} CU(AX,BT) A dX = 1$

any other thoughts?

 

[haruspex]

$b-1 = 2b-2a = b-2a$ correct?
I'm getting that $a = - \frac{1}{2}, b = -2$

I get $a = \frac{1}{2}, b = 0$

 

[fzero]

hmm, I'm getting this

$\lambda^{b-1}u_t - \lambda^{2b-2a} (u_x)^2 -\lambda^{b-2a}u_{xx} = 0$

You missed a factor of $u$ in the last term.

I mean if we have that $\int_{\mathbb{R}} u(x,t) dx = 1$ I feel like it has something to do with

$\int_{\mathbb{R}} CU(AX,BT) A dX = 1$

any other thoughts?

Yes, this gives a 2nd linear constraint on the scaling exponents.

 

[TSny]

I mean if we have that $\int_{\mathbb{R}} u(x,t) dx = 1$ I feel like it has something to do with

$\int_{\mathbb{R}} CU(AX,BT) A dX = 1$

any other thoughts?

I could be wrong, but here is how I interpret $u = CU$:

$u(x,t) = CU(X,T)$ where $x = AX$ and $t = CT$.

Then $\int_{\mathbb{R}} u(x,t) dx =C\int_{\mathbb{R}} U(X,T) dx$

Now make the appropriate substitution for $dx$ and see what you get.

 

[BrainHurts]

ALRIGHT,

I wrestled with this problem but I did it both ways!

so sticking with it haruspex and TSny I got that

$\frac{C}{B}U_T - \frac{C^2}{A^2}U_X - \frac{C^2}{A^2}UU_XX = 0$

so if we divide everything by $\frac{C}{B}$ we'll get the equation

$\frac{CB}{A} = 1$

using the constraint

$\int_{\mathbb{R}} u(x,t)dx = \int_{\mathbb{R}} CU(AX,BT) AdX = 1 \Rightarrow CA = 1 \Rightarrow C = \frac{1}{A}$

plugging this back into

$\frac{CB}{A} = 1 \Rightarrow \frac{B}{A^3} = 1 \Rightarrow B=A^3$

letting $A = \lambda$ means $B = \lambda^3$

so if we let $z = \frac{x}{t^{\frac{1}{3}}} = \frac{X}{T^{\frac{1}{3}}}$

similarly doing it your way fzero

i came up with

$\lambda^{b-1} = \lambda^{2b-a}$

so $b - 2a + 1 = 0$

using the constraint I get that

$\int_{\mathbb{R}} u(x,t)dx = \int_{\mathbb{R}} \lambda^b u(\lambda^a x, \lambda t) \lambda^a dX = 1 \Rightarrow \lambda^{b+a}= \lambda^0 \Rightarrow b = -a$

so

$-a -2a +1 = 0 \Rightarrow 3a = 1 \Rightarrow a = \frac{1}{3}$ and it follows that $b = - \frac{1}{3}$

so using the same idea as above, I get the same result!

 

[TSny]

so if we divide everything by $\frac{C}{B}$ we'll get the equation

$\frac{CB}{A} = 1$
...

plugging this back into

$\frac{CB}{A} = 1 ...$

I think you just forgot to type the square on A: $\frac{CB}{A^2} = 1$

Otherwise, it all looks good to me.

 

[BrainHurts]

I think you just forgot to type the square on A: $\frac{CB}{A^2} = 1$

Otherwise, it all looks good to me.

yup it should be A²

nm i figured that part out and that was garbage.

 

[fzero]

yup it should be A²

so we have that

$\frac{1}{A}U(AX,BT)=u(x,t)$

and this is o as $x \rightarrow \pm ∞$

not sure how this is going to help me get

$u = t^{-1}{3} y(z)$.

You found that the equation is homogenous with respect to the transformation

$$ t \rightarrow \lambda t , ~~ x \rightarrow \lambda^{1/3} x, ~~ u \rightarrow \lambda^{-1/3} u.$$

The combination $z = x/t^{1/3}$ is invariant, while $u$ scales like $t^{-1/3}$. To understand what this means for the form of $u$, imagine that we knew $u(t,x)$ and that it had a well-defined Taylor series at some point. This Taylor series would have to be of the form

$$ u(t,x) = \sum_{\alpha,\beta} c_{\alpha\beta} t^\alpha x^\beta.$$

However, $u$ has a definite scaling property, so each term in this series must also have the same scaling property. Can you work out what sort of constraint is placed on $\alpha$ and $\beta$? This will lead you to an expression for the Taylor series of $y(z)$ and establish the above relation between $u$ and $y$.