31

**HA5 / Re: HA5-P6**

« **on:**October 17, 2015, 06:26:23 PM »

I add question c), but I'm not sure the answer, please correct me!

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

31

I add question c), but I'm not sure the answer, please correct me!

32

and for question b, it should be *x* - *ct* instead of *x* - *vt*; otherwise, we couldn't have the solution.

33

professor, I don't understand why *Ï•*(0) = 0 only when without jumps?

35

the answer for âˆ’*c*_{2}*t*<*x*<0, Yeming Wen add one constant - 1 / 2 *Ï•*(0) in the bracket [].

let*k* be the constant multiplying *Ï•*(c_{1}/c_{2}(*x*+*c*_{2}*t*)),

plug*x* = 0 in *U* | ^{t = 0} by the initial condition, then we will have *U*(0,0) = *Ï•*(0)

so this constant should be*Ï•*(0) (1 / *k* - 1) in the bracket [].

for 0<*x*<*c*_{1}*t*, same steps used in calculate the constant in the brackets [].

let

plug

so this constant should be

for 0<

36

because plug *x* = 0 in *U*|_{t = 0} = *Ï•(x)*, we will have

*U*|_{t = 0, x = 0 } = *Ï•*(0)

and plug*x* = 0, *t* = 0 in *u*(*x,t*)=4/3 *Ï•*(12*x*+*t*), we will have

*U*(0,0) = 4/3 *Ï•*(0)

conclude (1) and (2), we will have the answer by Fei Fan Wu

(1)

and plug

(2)

conclude (1) and (2), we will have the answer by Fei Fan Wu

38

39

Now I understand its the part we discussed should be c integrate Xï¼ˆxï¼‰ from t - x/c to 0

40

My solution is Ï•(x+ct)âˆ’1/câˆ«^{ctâˆ’x}_{0}X(tâˆ’xc)+Ï•(ctâˆ’x)for for {0<x<ct}, Because the boundary condition is U_{x}|x=0=Ï‡(t), so we should integral Ï‡(t). I plug in the derivative of U respect to x, and I got this answer, may be I'm wrong? I'm not sure,

42

For 5b, I know derivative of arcsin(x) is 1/(1-x^2)^(1/2), can I use this to calculate the integral ? sorry about my typo, I haven't download matlab yet, I will download it~