# No idea where to start

1. Oct 1, 2005

### stunner5000pt

define $$L = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0$$

Show that if L is hyperbolic and A is not zero, the transofrmation to moving coordinates
$$x' = x- \frac{B}{2A} t$$
$$t' =t$$
takes L into a mutliple of the wave operator

Now the moving coordiantes looks very much like the galilean transforamtions i did in a relaitivity class a while ago.
Hyperbolic means the B^2 - 4AC > 0. But what does the transformation to moving coordiantes mean?
Also the solution to the second order PDE was solved by using
$$\xi = \alpha x + \beta t$$ and
$$\eta = \delta x + \gamma t$$

Please give me a hint on how to connect the two together... i am quite lost!

2. Oct 2, 2005

### lurflurf

Change variables with the chain rule so that
$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x'}$$
and
$$\frac{\partial}{\partial t}=-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}$$
express L in terms of the new variables and show that
$$k L=v\frac{{\partial}^2u}{\partial x'^2}-w\frac{{\partial}^2u}{\partial t'^2}$$
where v,w>0
k some constant

3. Oct 2, 2005

### stunner5000pt

am i supposed to use the transformations that i listed in post 1?

and use them to find something like $$\frac{\partial \xi}{\partial t} = \frac{-B}{2A} \frac{\partial \eta}{\partial x} + \frac{\partial \eta}{\partial t'}$$

is this in the right direction?

4. Oct 3, 2005

### stunner5000pt

5. Oct 4, 2005

what class is this for?

6. Oct 4, 2005

### stunner5000pt

this is for my Partial Differential Equations class...

i just said that it reminds me of that physics formula

anyway how would i relate the transformation coordinates to the equation L itself?

7. Oct 5, 2005

### lurflurf

Use the relations between old and new partials I gave above like so
$$L = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0$$

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x'}$$
and
$$\frac{\partial}{\partial t}=-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}$$
so
$$L = a\left(-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}\right)^2 u+ B\left(\frac{\partial}{\partial x'}\right) \left(-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}\right)u+ C \left(\frac{\partial}{\partial x'}\right)^2 u = 0$$

8. Oct 5, 2005

### stunner5000pt

one thing lurf, you write $$\frac{\partial}{\partial t'}$$... what is the variable/function that is being differentiated? and also is it tru that
$$(\frac{\partial}{\partial x'})^2 = \frac{\partial^2 u}{\partial x^2}$$
how are you figuring out this for t and t' likewise?

Last edited: Oct 5, 2005
9. Oct 6, 2005

### lurflurf

by the chain rule
$$\frac{\partial}{\partial x}=\frac{\partial x'}{\partial x} \ \frac{\partial}{\partial x'}+\frac{\partial t'}{\partial x} \ \frac{\partial}{\partial t'}=\frac{\partial}{\partial x'}$$
also
$$\frac{\partial}{\partial t}=\frac{\partial x'}{\partial t} \ \frac{\partial}{\partial x'}+\frac{\partial t'}{\partial t} \ \frac{\partial}{\partial t'}=-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}$$
Thus the relations of the partial derivatives are easily found by using the chain rule and partial derivatives of the relations of the variables you gave in the first post

The derivatives act on functions they are right multiplied by in this case u and depending on your intermediate steps partial derivatives of u
$$\frac{\partial}{\partial x} \ u=\frac{\partial u}{\partial x}$$
$$\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)u=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}$$
$$\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)\frac{\partial u}{\partial x}=\frac{\partial^2 u}{{\partial x}^2}+\frac{\partial^2 u}{\partial t\partial x}$$
Writing derivatives this way is sometimes convient, but no new concept is in use.
as far as the u it depents how technical one wants to get I am considering u as a variable so that u can be determined given x and t or x' and t' (also other variables or combinations) thus there are functions relating the variables
u=f(x,t)
u=g(x',t')
u=f(x'+(B/(2A))t',t')
u=g(x+(-B/(2A))t,t)
so the actual function form of u depents which variables are in use,
one could use different letters if confusion were likely.
$$\left(\frac{\partial}{\partial x'}\right)^2u=\frac{\partial^2 u}{{\partial x'}^2}=\frac{\partial^2 u}{{\partial x}^2}$$
since x and x' partials of u are equal

10. Oct 6, 2005

### saltydog

Nice. Actually I'd like to see a real . . . well nevermind. I had trouble with this too Stunner. May I add some points which may help:

This approach is best viewed in terms of "operators". That is what Lurflurf is doing, I believe. When you see an expression like:

$$\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)^2$$

That really means "operate on the operator" and not multiply or raise to a power, that is:

$$\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)^2=\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)$$

That's not multiplying ok, but rather one operator "operating" on another operator. Now, its similar to the distributive law with algebraic equations but instead of adding and multiplying, we "operate". Now, look at the RHS. It has two operators, each themselves having two operators. Now, break them apart and remember these are "operators" ok:

\begin{align*} \left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right) &= \left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)\left(\frac{\partial}{\partial t'}\right) \\ &- \left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right) \left(\frac{b}{2a}\frac{\partial}{\partial x'}\right) \end{align}

Ok, now lets do the first one:

$$\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)\left(\frac{\partial}{\partial t'}\right)=\frac{\partial^2 }{\partial t^{'2}}-\frac{b}{2a}\frac{\partial^2}{\partial x'\partial t'}$$

Now you do the second one.

Edit: Oh yea, same dif for the other ones. Can you post the mixed-partial one? Remember, its just one operator operating on another.

Last edited: Oct 6, 2005
11. Oct 6, 2005

### stunner5000pt

so would the otehr temr look like this:

$$\frac{-b}{2A} \frac{\partial x}{\partial x' \partial t'} + \frac{b^2}{4A^2} \frac{\partial^2}{\partial x'^2}$$

ok now taht we got that settled once i expand that L that was formed by all of this, am i supposed to find equalities such that the the parritals with the x' and t' go away? I won't type this all out here yet but is this the right way to go from there?

so far thank you for the help... it had me quite perplexed before.

ALso can you have a look at these problems, please? I would greatly appreciate it ! https://www.physicsforums.com/showthread.php?t=92515

12. Oct 6, 2005

### saltydog

Good for you Stunner! And I don't wish to take away from Lurflurf's input as I was unable to solve this until he posted his results.

For the mixed partial, wouldn't that just be:

$$\frac{\partial}{\partial x}\left(\frac{\partial}{\partial t}\right)= \frac{\partial}{\partial x}\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)$$

But:

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x'}$$

so:

$$\frac{\partial}{\partial x}\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)= \frac{\partial}{\partial x'}\left(\frac{\partial}{\partial t'}-\frac{b}{2a}\frac{\partial}{\partial x'}\right)$$

Can you finish this one? Remember:

$$\frac{\partial}{\partial x'}$$

is "operating" on the quantity in paranthesis (which is also an operator).

Make all the substitutions in the original equation and get one in terms of the prime variables. The mixed-partials drop out if your algebra is correct.

Last edited: Oct 6, 2005
13. Oct 6, 2005

### saltydog

Hey Stunner, you know something, I think Lurflurf has been sleeping all this time. I think he may be in Europe. Anyway, I hope when he wakes up and looks at this thread, everything I've said will meet with his requirements (other people too you know). If not, well, I guess I'll be corrected then.

14. Oct 6, 2005

### lurflurf

The partials will not go away, but the mixed partials will.
here is the outline
1)Use the chain rule to write the derivatives in terms of the new variables x' and y'
2)Use the derivative relations to write L in terms of the new variables x' and y'
3)use the facts B^2 - 4AC > 0 and A is not zero to show that L is a multiple of the wave operator that is
$$L=w{\square}^2u=w\left[{\nabla}^2-v^2\frac{\partial^2}{{\partial t'}^2}\right]u=w\left[\frac{\partial^2}{{\partial x'}^2}-v^2\frac{\partial^2}{{\partial t'}^2}\right]u=w\left[\frac{\partial^2 u}{{\partial x'}^2}-v^2\frac{\partial^2 u}{{\partial t'}^2}\right]$$
for some real numbers w and v that depend on A,B,C