Wave Equation with Continuous Piecewise Initial Velocity

[Claud123]

Homework Statement

Hello I am asked to find the solution to the following equation no infinite series solutions allowed. We are given that there is a string of length 4 with the following...

y_tt=y_xx

With y(0,t) = 0 y(4,t) = 0 y(x,0) = 0 y_t(x,0) = x from [0,2] and (4-x) from [2,4].

Homework Equations

given none

The Attempt at a Solution

Since I can't use an infinite series to solve this separation of variables and Fourier series is out. So instead I think we are expected to use d'Almbert's solution.

y = $$\frac{1}{2}$$[g(x-ct) + g(x+ct)] + $$\frac{1}{2c}$$$$\int h(s) ds$$ from x-ct to x+ct.

Well we are given that the initial position function is just 0 so the g terms all drop out. However, I am stuck at the integration part. How exactly am I supposed to integrate d'Almbert's solution over this initial velocity?

***Note: I just noticed I used "Discontinuous in my title to describe the initial velocity which it is not. I understand why, I just used wrong wording at the time I wrote this. I guess this a continuous piecewise initial velocity would be a better description.

 

[HallsofIvy]

What is the difficulty? To integrate a discontinuous function, integrate whatever formula you are given up to the discontinuity, then continue with the next formula past the discontinuity.

If h(x)= x from 0 to 2, then
[tex]\int_{x-t}^{x+ t} h(s)ds= \int_{x- t}^{x+ t} s ds= \frac{1}{2}[(x+t)^2- (x- t)^2= 4xt[/math] for $x+ t\le 2$
Note that if x+ t= 2, then t= 2- x so $4xt= 4x(2- x)= 8x- 4x^2$.

and then, for x+ t> 2,
$$8x- 4x^2+ \int_{2x- 2}^{x+ t} (4- s)ds$$

 

[Claud123]

What is the difficulty? To integrate a discontinuous function, integrate whatever formula you are given up to the discontinuity, then continue with the next formula past the discontinuity.

If h(x)= x from 0 to 2, then
[tex]\int_{x-t}^{x+ t} h(s)ds= \int_{x- t}^{x+ t} s ds= \frac{1}{2}[(x+t)^2- (x- t)^2= 4xt[/math] for $x+ t\le 2$
Note that if x+ t= 2, then t= 2- x so $4xt= 4x(2- x)= 8x- 4x^2$.

and then, for x+ t> 2,
$$8x- 4x^2+ \int_{2x- 2}^{x+ t} (4- s)ds$$

Thank you for your response. However, I am running into a boundary condition issue when I run your integral for the regions [2,4], mainly that I need Y(x,t=0) = 0 .

I was thinking something along these lines would work.

$$\int_{x-t}^{x+t} f(s) ds =$$
on region [0,2]
$$\int_{x-t}^{2} sds$$
and on region [2,4]
$$\int_{x-t}^{2} sds + \int_{2}^{x+t}(4-s)ds$$

but again I running into BC issues.