# Trying to get this PDE in terms of 'y'

1. Jun 23, 2011

### 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

2. Jun 23, 2011

### 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.

3. Jun 23, 2011

### jaketodd

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

4. Jun 23, 2011

### Hootenanny

Staff Emeritus
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,

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

5. Jun 23, 2011

### hunt_mat

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

Last edited: Jun 23, 2011
6. Jun 23, 2011

### HallsofIvy

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

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

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

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.

7. Jun 23, 2011

### 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.

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.

Last edited: Jun 23, 2011
8. Jun 23, 2011

### 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?

9. Jun 23, 2011

### jaketodd

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.

10. Jun 23, 2011

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

11. Jun 23, 2011

### 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.

12. Jun 23, 2011

### 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.

13. Jun 23, 2011

### TylerH

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.

Last edited: Jun 23, 2011
14. Jun 23, 2011

### 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.

15. Jun 23, 2011

### 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.

16. Jun 23, 2011

### 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

17. Jun 24, 2011

### 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?

18. Jun 24, 2011

### jaketodd

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.

19. Jun 24, 2011

### 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.

20. Jun 24, 2011

### jaketodd

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.