D'Alembert solution with f(x)=0, g(x)=x

[yungman]

Homework Statement

This is an example I copy from the book. The book showed the steps of solving and provide the answer. I don't understand the book at all. Below I show the question and the solution from the book. Then I am going to ask my question at the bottom.


Question
Use D'Alembert method to solve the wave equation with boundary and initial value:

$$\frac{\partial^2 u}{\partial t^2} \;=\; c^2 \frac{\partial^2 u}{\partial x^2} \;\;,\;\; u(0,t)=u(L,t)=0 \;\;,\;\; u(x,0)=f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0)=g(x)$$

Given: f(x)=0 & g(x)=x for 0<x<1. c=1, L=1.

Homework Equations

D'Alembert:
$$u(x,t)=\frac{1}{2}[f^*(x-ct)+f^*(x+ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds$$

The Attempt at a Solution

f(x) = 0 $$.\;\;\Rightarrow\; u(x,t)\; =\; \frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds \;=\; \frac{1}{2c}[G(x+t) \;-\; G(x-t)]$$(1)

Where $$g^*$$ is the odd 2-period extension of g and G is the antiderivative of $$g^*.$$

Let .$$G(x)=\int_{-1}^x g^*(z)dz$$ (2)

To complete the solution, we must determine G. We know G on any interval of length 2. Since $$g^*(x)=x$$ on the interval (-1,1), we obtain

.$$G(x)=\int_{-1}^x g^*(z)dz \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2}$$. for x in (-1,1) (3). Hence

$$G(x) \;=\; \frac{1}{2}x^2 \;-\; \frac{1}{2} \;\;\;if\;-1<x<1$$

$$G(x) \;=\; G(x+2) \;\;otherwise.$$



My question:

1) How does it jump from (1) to (3)??

2) Is (2) just the first Fundamental Theorem of Calculus where -1 is the lower limit of x on [-1,1]?

3) Where is (3) come from? How are (x+t) and (x-t) change to x and -1 respectively?

I don't even understand the answer of the book! Can anyone explain to me?

Thanks

 

[yungman]

Anybody please?

 

[yungman]

Anyone?

 

[snipez90]

I'm not sure about the context but I'll try to answer your questions.

First off, in (1), you seem to have copied down the final expression incorrectly. Your limits of integration at first are x + ct and x - ct, but then the c's have disappeared in the final expression.

(2) is not a theorem, but it's certainly related to the fundamental theorem of calculus. As mentioned in the line above (2), G is simply the antiderivative of g*. In particular, note that

$$\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds = \frac{1}{2c}\left(\int_{x-ct}^{-1} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right) = \frac{1}{2c}\left(-\int_{-1}^{x-ct} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right),$$

which follows from basic properties of the integral.
Can you relate (2) and (1) now?

(3) just looks like a definite integral computation of g*(z) = z, nothing too complicated.

 

[yungman]

I'm not sure about the context but I'll try to answer your questions.

First off, in (1), you seem to have copied down the final expression incorrectly. Your limits of integration at first are x + ct and x - ct, but then the c's have disappeared in the final expression.

(2) is not a theorem, but it's certainly related to the fundamental theorem of calculus. As mentioned in the line above (2), G is simply the antiderivative of g*. In particular, note that

$$\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds = \frac{1}{2c}\left(\int_{x-ct}^{-1} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right) = \frac{1}{2c}\left(-\int_{-1}^{x-ct} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right),$$

which follows from basic properties of the integral.
Can you relate (2) and (1) now?

(3) just looks like a definite integral computation of g*(z) = z, nothing too complicated.

Thanks for your reply.

1) In the question from the book, c=1. Therefore I remove the c and just (x+t) and (x-t).

2) I was playing around on the integration you wrote:

$$\frac{1}{2c}\int_{x-ct}^{x+ct} g^*(s)ds = \frac{1}{2c}\left(\int_{x-ct}^{-1} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right) = \frac{1}{2c}\left(-\int_{-1}^{x-ct} g^*(s)ds + \int_{-1}^{x+ct} g^*(s)ds\right),$$

Look like it really does not matter where -1 in reference (x+ct) and (x-ct). ei. Whether -1 is on the left of or in between or on the right side
(x+ct) and (x-ct).


3) In fact, looks like I can use any number other than -1. Any number a is as good as other because it get cancel out when adding the two integrations.

4) I continue to work out the problem and this is the answer. Tell me I am right or not.

$$\Rightarrow\; u(x,t) \;=\;-\; [\frac{1}{4c} \; s^2|_{-1}^{x-ct} \;\;] \;+\; [\frac{1}{4c} \; s^2|_{-1}^{x+ct} \;\;] \;=\; [\frac{1}{4c} \; [(x+ct)^2 \;-\; (x-ct)^2]$$

$$\Rightarrow\; u(x,t)=\; [\frac{1}{4c} \; (x+ct)^2] \;-\; [\frac{1}{4c} \; [(x-ct)^2]$$. Where u(x,t) is two wave travel left and right respectively.

But as you can see from the original post, the book integrate from -1 to x alone. this is shown on (3). I even check another example and the book did the same. I just don't understand this.
The book gave the equation (3). I have double check with the book and I type everything correctly.


Thanks so much of your time.

 

[yungman]

I read more and looked at other examples, seems like the book keep putting -1 as the dummy variable for -1<x<1 and it appear on both side of the traveling wave. Why?

 

[yungman]

Anyone please?

 

[yungman]

No body?