# No idea where to start

by stunner5000pt
Tags: start
 P: 1,443 define $$L[u] = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0$$ Show that if L[u] 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!
 HW Helper P: 2,079 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[u] in terms of the new variables and show that $$k L[u]=v\frac{{\partial}^2u}{\partial x'^2}-w\frac{{\partial}^2u}{\partial t'^2}$$ where v,w>0 k some constant
 P: 1,443 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?
P: 1,443

## No idea where to start

 P: 41 what class is this for?
 P: 1,443 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[u] itself?
HW Helper
P: 2,079
 Quote by 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[u] itself?
Use the relations between old and new partials I gave above like so
$$L[u] = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0$$

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x'}$$
and
$$\frac{\partial}{\partial t}=-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}$$
so
$$L[u] = a\left(-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}\right)^2 u+ B\left(\frac{\partial}{\partial x'}\right) \left(-\frac{B}{2A} \ \frac{\partial}{\partial x'}+\frac{\partial}{\partial t'}\right)u+ C \left(\frac{\partial}{\partial x'}\right)^2 u = 0$$
 P: 1,443 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?
HW Helper
P: 2,079
 Quote by 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?
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
 HW Helper Sci Advisor P: 1,593 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.
 P: 1,443 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[u] 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 ! http://physicsforums.com/showthread.php?t=92515
 HW Helper Sci Advisor P: 1,593 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.
 HW Helper Sci Advisor P: 1,593 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.
HW Helper
P: 2,079
 Quote by 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[u] 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 ! http://physicsforums.com/showthread.php?t=92515
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[u] 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[u] is a multiple of the wave operator that is
$$L[u]=w{\square}^2u=w\left[{\nabla}^2-v^2\frac{\partial^2}{{\partial t'}^2}\right]u=w\left[\frac{\partial^2}{{\partial x'}^2}-v^2\frac{\partial^2}{{\partial t'}^2}\right]u=w\left[\frac{\partial^2 u}{{\partial x'}^2}-v^2\frac{\partial^2 u}{{\partial t'}^2}\right]$$
for some real numbers w and v that depend on A,B,C

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