Trying to get this PDE in terms of 'y'

[jaketodd]

I will love forever whoever can show me the steps of how to get the following equation in terms of y=[...] This is not a homework question. I have a calculus book that has given me some progress, such as expanding the equation to a mixture of terms and first order partial derivatives, and I know to hold all other variables constant when getting a partial derivative, but I clearly don't understand how to do it right because my work doesn't agree with the answers in the textbook, and it seems like there might be more than one answer from what the textbook seems to say. I am really confused. My mom is a math tutor and she is having trouble with it. I really hope someone comes through for me.

Thanks,

Jake

 

[TylerH]

I don't know how to do it, but it will help those who do if you post your attempted solution.

Also, in general, this would still be considered a "homework" question, becuase it is in a textbook question format.

 

[jaketodd]

I don't know how to do it, but it will help those who do if you post your attempted solution.

Also, in general, this would still be considered a "homework" question, becuase it is in a textbook question format.

Well it's not from a textbook. I am just using a textbook to try to figure it out. Also, the furthest I can get it is to first order partial derivatives, then I am lost. Anyone who can solve this would know that step.

Jake

 

[Hootenanny]

You mean you want to solve the equation? If that's the case, then it is pretty straight forward. Or do you want to write in PDE in the form,

$$y = f\left(x,t,\frac{\partial y}{\partial x}, \ldots, \frac{\partial^n y}{\partial x^n}, \frac{\partial y}{\partial t}, \ldots, \frac{\partial^n y}{\partial t^n}\right)$$

 

[hunt_mat]

First method is separation of variables, so you write $y=X(x)T(t)$, what does this say now?

 

[HallsofIvy]

That is a "wave equation". Another method is to make the change of variable [itex]u= x- \nu t[itex], $v= x+ \nu t$.

$$\frac{\partial y}{\partial x}= \frac{\partial y}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial y}{\partial v}{\partial v}{\partial x}$$
$$= \frac{\partial y}{\partial u}+ \frac{\partial y}{\partial v}$$

$$\frac{\partial^2 y}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial u}+ \frac{\partial y}{\partial v}\right)$$
$$= \frac{\partial }{\partial x\left(\frac{\partial y}{\partial u}\right)+ \frac{\partial}\partial x}\left(\frac{\partial y}{\partial v}\right)$$
$$= \frac{\partial v}{\partial x}\frac{\partial^2 y}{\partial u^2}+ 2\frac{\partial u}{\partial x}\frac{\partial^2 y}{\partial u\partial v}+ \frac{\partial^2 y}{\partial v^2}$$
$$= \frac{\partial^2 y}{\partial x^2}+ 2\frac{\partial^2 y}{\partial u\partial v}+ \frac{\partial^2 y}{\partial v^2}$$

Do the same with $\partial^2y/\partial x^2$ and put them into the original equation. It simplifies remarkably.

This is called the "method of characteristics". $x- \nu t= const$ and $x+ \nu t= const$ are the "characteristic lines" for the equation.

 

[jaketodd]

I read all you guys' replies and I thank you, but I am still lost. I tried applying what you guys said to the equation, but I don't get it. I am just looking for y= ...I am just looking for the equation solved for y. I don't know how, and I did try.

Please help,

Jake

Oh, and Hootenanny, yes I just want to solve the equation for y. If it's pretty straightforward as you say, could you show me how? Many thanks! But anyone, please help.

 

[hunt_mat]

Jake, using my method, inserting y into the PDE shows:
$$X''(x)T(t)=\frac{1}{\nu^{2}}X(x)T''(t)$$
Divide both sides through by X(x)T(t) to obtain:
$$\frac{X''(x)}{X(x)}=\frac{1}{\nu^{2}}\frac{T''(t)}{T(t)}$$
Now one side is totally a function of x and the other side is a function of t, this is only possible is both sides are constant, call this constant k say, so this should leave you with two ODEs, what are they?

 

[jaketodd]

Jake, using my method, inserting y into the PDE shows:
$$X''(x)T(t)=\frac{1}{\nu^{2}}X(x)T''(t)$$
Divide both sides through by X(x)T(t) to obtain:
$$\frac{X''(x)}{X(x)}=\frac{1}{\nu^{2}}\frac{T''(t)}{T(t)}$$
Now one side is totally a function of x and the other side is a function of t, this is only possible is both sides are constant, call this constant k say, so this should leave you with two ODEs, what are they?

First of all, thank you. However, two things: Firstly, y is already in the PDE as in the image of my original post. Secondly, I have no idea how to get two ODEs from what you provided, and I don't even know what an ODE is. I looked it up and I think it means "Ordinary Differential Equation," but ...please, anyone, I am lost here.

 

[hunt_mat]

If you don't know what an ODE is then attacking a PDE may be a little beyond you.
the differential equations are:
$$\frac{X''(x)}{X(x)}=k,\quad\frac{1}{\nu^{2}}\frac{T''(t)}{T(t)}=k$$
which gives two odes:
$$X''(x)-kX(x)=0,\quad T''(t)-\nu^{2}kT(t)=0$$

 

[TylerH]

He's saying to let y=X(x)T(t). $\frac{\partial^2y}{\partial x^2}=X''(x)T(t) \mbox{ and } \frac{\partial^2y}{\partial t^2}=X(x)T''(t)$

From there, I'm as lost as you are. :)

EDIT: Note: This was typed before the above post.

 

[jaketodd]

I am still lost (for instance, v seems to disappear and then reappear, and also the variable definitions don't seem to match the original post even when I take into account that some variables are being expressed differently). I am trying to get it. Can all the variables in the original post simply be expressed as y=? That's what I really need.

 

[TylerH]

I am still lost (for instance, v seems to disappear and then reappear, and also the variable definitions don't seem to match the original post even when I take into account that some variables are being expressed differently). I am trying to get it. Can all the variables in the original post simply be expressed as y=? That's what I really need.

Well, you know y=X(x)T(t) and you have 2 ODEs describing X(x) and T(t), respectively. So solve them for X(x) and T(t) and multiply them together.

 

[TylerH]

Oh... I just noticed you said you didn't know what an ODE is.

$$X(x)=C_1e^2+C_2e^-x$$
$$T(t)=C_3e^{vt}+C_4e^{-vt}$$
$$y(x,t)=(C_1e^x+C_2e^{-x})(C_3e^{vt}+C_4e^{-vt})$$

where all the Cs are arbitrary constants.

 

[TylerH]

How far are you in the calculus sequence? This stuff comes after that, and ODEs are usually introduced in calc II. To give you an idea, the only PDE class my school offers is a second year graduate class ie, people getting master's degrees don't get to them until their second year, here.

I STRONGLY recommend you learn ODEs first. PDEs are an extension of ODEs to functions of many variables, so it should be clear that since the latter underlies the former, the latter is required knowledge for even beginning to grasp the former.

Anyway, MIT's OCW is a good place for calculus and ODEs. Paul's Online Notes has an entire class worth of notes on ODEs and a small section on PDEs.

 

[jaketodd]

So there are arbitrary constants in the y=? Isn't there a way to get it just in terms of the original variables? If there must be constants in there, can I solve for them by plugging in arbitrary values for the other variables? Can you please show me how the previous post to this one fits into the one before it? And in the most recent post, did you intend to put x in the exponent of the first line there?

Thanks,

Jake

 

[hunt_mat]

You can't just solve a PDE in that way, you need boundary conditions, initial conditions and what not.

Have you done any differential equations before? What have you done?

 

[jaketodd]

You can't just solve a PDE in that way, you need boundary conditions, initial conditions and what not.

Have you done any differential equations before? What have you done?

I know how to take the derivative of a function, and I know how limits work. That's about it.

Earlier in this thread, Hootenanny said that solving for y is "pretty straight forward," so that gives me the impression that it can be done, somehow.

 

[hunt_mat]

So if I asked you to find y when:
$$\frac{dy}{dx}+2xy=x$$
Could you tell me how to calculate y?
Solving PDEs such as this is straightforward, if you have the right experience. It looks to me as if you're jumping way ahead of your experience. I would humbly suggest that you start with how to solve the equation I gave you before you start on the wave equation.

On an aside, I would also suggest you look how to solve 1st order partial differential equations before you look into 2nd order PDEs.

 

[jaketodd]

So if I asked you to find y when:
$$\frac{dy}{dx}+2xy=x$$
Could you tell me how to calculate y?
Solving PDEs such as this is straightforward, if you have the right experience. It looks to me as if you're jumping way ahead of your experience. I would humbly suggest that you start with how to solve the equation I gave you before you start on the wave equation.

On an aside, I would also suggest you look how to solve 1st order partial differential equations before you look into 2nd order PDEs.

Well you see, that's the whole reason I am coming to physics forums; I don't have the experience to figure it out on my own.

 

[hunt_mat]

My advice would be to get a book on differential equations, and start reading it. If you get stuck on some concepts then by all means ask and we will help you.

 

[jaketodd]

I wish someone could just show me the steps involved... I'll transfer $50 with Paypal to the first person who does this for me; anyone who provides the steps of solving for y from the original post with all variables definable (for example, no unsolvable constants). It's for a physics paper I'm working on; I don't need to understand all the mathematical concepts, I just need the equation solved for y please.

Jake

 

[jaketodd]

Please help guys; it's for a physics paper and I really don't need to understand all the math.

 

[TylerH]

The arbitrary constants are arbitrary. That means the equation is true no matter their value. Think about what taking a derivative does to a constant, it's like that.

For example, if y'=k, where k is constant, then y=kx+C, where C is an arbitrary constant. As you can see, no matter what C is, since it is constant, it just "goes away," and the derivative of kx+C is k, which is what we wanted.

So, what I'm saying is, the arbitrary constants are just that. You can't solve for them, because they're true for all real numbers ie, they're arbitrary.

 

[jaketodd]

I see what Tyler means. So those equations in earlier posts - are they enough to solve for y? Could someone put it all together for me? $50 on the line.

 

[jaketodd]

A lot of thought went into this. I don't know if it's right, but it's the best I can do right now...

Do I put these:

$$X(x)=C_1e^2+C_2e^-x$$
$$T(t)=C_3e^{vt}+C_4e^{-vt}$$

BTW: shouldn't the first line be: $$X(x)=C_1e^x+C_2e^{-x}$$

Into these:

$$X''(x)-kX(x)=0,\quad T''(t)-\nu^{2}kT(t)=0$$

Then, solve for X(x) and T(t), and then multiply them together?

How do the C's go away? And how do I solve for k?

Thanks,

Jake

 

[TylerH]

Yes. It should be e^-x

No. The equations you quoted from me are the solutions to those you quoted from hunt_mat. To get y, you multiply X(x) and T(t), like I did in the third equation in my post you quoted.

The C's don't go away. They're part of the answer. You CAN'T get the answer in terms of ONLY the original variables. They HAVE to be there.

 

[jaketodd]

Yes. It should be e^-x

No. The equations you quoted from me are the solutions to those you quoted from hunt_mat. To get y, you multiply X(x) and T(t), like I did in the third equation in my post you quoted.

The C's don't go away. They're part of the answer. You CAN'T get the answer in terms of ONLY the original variables. They HAVE to be there.

Mkay...

Is there a way of solving for the C's? I can define all the variables when y = 0. However, would that make all the C's zero if there is a time, t, when y = 0?

Thanks,

Jake

 

[hunt_mat]

The two solutions of the differential equations should be:
$$X(x)=Ae^{x\sqrt{k}}+Be^{-x\sqrt{k}},\quad T(t)=Ce^{\nu t\sqrt{k}}+De^{-\nu t\sqrt{k}}$$
So the solution for y becomes:
$$y=(Ae^{x\sqrt{k}}+Be^{-x\sqrt{k}})(Ce^{\nu t\sqrt{k}}+De^{-\nu t\sqrt{k}})$$
But there are two things you need to consider, the sign of k, as this will decide what sort of solutions you will get (do you want oscillatory solutions ($k=-\lambda^{2}$), then there is the fact that the equation is linear and you can just add solutions together to find a new solution, so if y_1 and y_2 solve your equation then so does y_3=y_1+y_2 (exercise). So the next thing you require is initial and boundary conditions.

 

[jaketodd]

So for oscillatory solutions with k=-lambda^2, wouldn't that produce imaginary numbers?

Thanks,

Jake

 

[hunt_mat]

Go back to the original equations:
$$X''(x)+\lambda^{2}X(x)=0,\quad T''(t)+\lambda^{2}T(t)=0$$
What are the solutions to these equations?

 

[HallsofIvy]

An n^th order ordinary differential equation always involves n unknown constants. An n^th order partial differential equation always involves n unknown functions.

I pretty much told you before that the general solution to that differential equation is
y= F(x- vt)+ G(x+ vt) where F and G are arbitrary, twice differentiable, functions.

The simple fact is that you are trying to do a problem that is far beyond your knowledge. If you do not know how to solve basic ordinary differential equations, no one is going to be able to tell you how to solve partial differential equations in a few sentences.

 

[timthereaper]

I agree with Ivy. Everyone here is trying to help you (the OP) solve the problem, but it's your inexperience that's stopping you from understanding the solution process. PDEs are fairly difficult to solve a lot of the time even for people with a fair amount of math under their belt. I do appreciate your bravado in tackling the wave equation and applaud your desire to extend your mathematical prowess, however I think you should spend a bit more time learning the fundamentals.

It's for a physics paper I'm working on; I don't need to understand all the mathematical concepts, I just need the equation solved for y please.

About this, you can't just expect to post a problem and get a detailed solution back. It doesn't work that way. IMHO, the forums are really an aid to your own problem solving. If you're looking for the other kind of problem solving, try http://www.wolframalpha.com/" .