Solving PDE & Plotting -20 < x < 20, t = 0,1,2,...10

[jianxu]

Homework Statement

Hi, so the initial problem was:
given [tex]\left.\frac{d^{2}u}{dt^{2}} = \frac{d^{2}u}{dx^{2}}}[/tex]

[tex]\left.-\infty \leq x \leq \infty[/tex]

[tex]\left.u(x,0)=\frac{x}{1+x^{3}} , \frac{du}{dt}(x,0) = 0[/tex]

Solve the PDE(did this part already) and plot the solution for -20 < x <20 and t = 0,1,2,...10

Homework Equations

The Attempt at a Solution

So I've already solved for the PDE which came out to be:
[tex]\left.u(x,t) = \frac{x}{2(1+(x-t)^{3})}+ \frac{x}{2(1+(x+t)^{3})}[/tex]

I am having trouble understanding the implications of the plots though. I've attached images of the several plots(included t = 0, t = 2, t = 4 since they're all the same except the x position of the line decreases with an increase in t)

Does this mean for each t there is only one location it can be in? It seems strange because we've been doing wave motion and this doesn't look like a wave so...Thanks!

 

[Mark44]

Your attached images are pending approval, so I can't tell if they're useful just yet. Keep in mind that a graph of u(x, t) will require three dimensions. Some things that might help you visualize the graph would be the trace of the graph in a few planes, like t = 0 (the x-z plane), x = 0 (the t-z plane) and z = 0 (the x-t plane).

Also, a few contours might be helpful, such as z = k, for several values of k, both positive and negative. IOW, you'll need to plot graphs of u(x, t) = k.

Hope these ideas are helpful.

 

[jianxu]

Are you familiar with maple at all because that's what I used to graph these and they were in 2 dimensions...I simply plugged in t and used the given equation with only the x, as my function for graphing. Would you happen to know how to graph in 3D. I've never really used maple prior to this so I'm completely lost on applying 3D

 

[Mark44]

I haven't used Maple for a number of years, so I can't offer much help there. The best advice I can give is for you to look at the documentation to learn how to graph a function of two variables.

Your 2D graphs (which are still pending approval) are like slices of the surface in space that your function represents. If you have enough of these slices, you can get a fair idea of what the surface itself looks like.

 

[jianxu]

hm about the 3D graph though, this PDE including the way I solved for it(D'alembert's wave principle) only applies to 1 dimensional waves though?(we've only begun talking about 1 dimensional waves in class, motion of a "string") so it seems weird that this would require a 3D graphical representation?

 

[Mark44]

The solution you found is not a pictorial representation of the string, which is a one-D object. The graph of your solution function u(x, t) is two-dimensional surface in three-dimensional space. Two of the dimensions are the position x along the string and time t. The third is the the vertical distance of a point x on the string at time t.

I don't think I understand your term "one-dimensional wave." The string can be thought of as one-D, but its oscillatory motion is in two dimensions. To make this more concrete, think of a guitar string at rest - the string is in a straight line.

Now, pluck the string. While the string is vibrating (and sound is being emitted by the guitar), take a picture of the string's motion with a very high-speed camera, which will "freeze" the motion of the string. At that instant in time, the string will assume some sort of sinusoidal shape, which takes two dimensions to represent. If you take a bunch of pictures at different times, you will get a bunch of different wave forms.

The graph of u(x, t) is in essence giving you all of those different wave forms at different times. Hope that's clear.

 

[jianxu]

But time isn't really considered a dimension is it? Or do we have to make it a dimension in this case since, the function is based on both x, and the t?

Also, I guess I was hoping to see a more sin like wave instead of two straight lines which looks very weird

 

[Mark44]

Yes, it certainly is considered a dimension.

 

[jianxu]

Hi,

this is sort of related to my previous question. I have a 1-D wave given by
[tex]\left.u(x,t)= sin(\pi x)cos( \pi t) + \frac{1}{2}sin(3 \pi x)cos(3 \pi t) + 3sin(7 \pi x)cos(7 \pi t)[/tex]
, and through graphing I know there are certain places(6 points excluding the ends of the wave going from 0 < x <1) where the x values are fixed when the wave moves. now say at u(x,0) where all the cosine terms goto one, when I take the derivative of that function, that would give me the velocity function that represents the movement of the wave. If I set it to zero, that would mean I'm solving for the x values that are not moving.

so the derivative when t = 0 would be:
[tex]\left.u(x,0)= \pi cos (\pi x) + \frac{3}{2}\pi cos (3 \pi x) + 21 \pi cos (7 \pi x) = 0[/tex]

Now, I'm not very good at solving trig stuff so, would I have to use some kind of trig identity to simply this first before I solve? Or what suggestions would you have for solving something like this? Thanks!

 

[Mark44]

I don't if trig identities would be much help in solving that equation. I think the best you're going to be able to do is to approximate the x-intercepts using graphing or math software. BTW, I see 7 intercepts between 0 and 1, not 6.

 

[jianxu]

I talked to my professor and he says I should be solving for the x values, he says since it's a motion of waves and we only graph from t=0..10, the graph can only help support your intuition of where the non moving x values are but ultimately we're suppose to solve it, and I am pretty lost on how I should approach solving for the x values