# D'Alembert's Wave. Seems Easy Or Maybe I'm Missing Something???

• Jul 13th 2008, 03:41 PM
mathfied
D'Alembert's Wave. Seems Easy Or Maybe I'm Missing Something???
Hi I have this D'Alembert's question. What I've done seems easy so I wanted to clarify just in case I'm missing something, cus it seems so easy. Second part I don't understand and would appreciate a guidance.

(PART A)

D'Alembert's Solution is:
$\displaystyle z(x,t) = \frac{1} {2}\left[ {F(x + ct) + F(x - ct)} \right] + \frac{1} {{2c}}\int\limits_{x - ct}^{x + ct} {G(s)ds}$
of the one dimensional wave equation:
$\displaystyle \frac{{\partial ^2 z}} {{\partial x^2 }} = \frac{1} {{c^2 }}\frac{{\partial ^2 z}} {{\partial t^2 }},{\text{ }} - \infty < x < \infty ,{\text{ t}} \geqslant {\text{0,}}$

with "c" a constant, when the initial conditions are:
$\displaystyle z(x,0) = F(x),{\text{ }}z_t (x,0) = G(x),{\text{ }} - \infty < x < \infty$

Evaluate the solution if F(x)=0 for $\displaystyle -\infty < x < \infty$ and
$\displaystyle G(x) = \left\{ {\begin{array}{*{20}c} {\frac{1} {{1 + x}}} & {\left\{ {{\text{x}} \geqslant {\text{0}}} \right\}} \\ {\text{0}} & {\left\{ {{\text{x}} < {\text{0}}} \right\}} \\ \end{array} } \right.$

(PART B)

Show the values of z(x,t) for t>0 in different regions of the (x,t)- plane. Show that the value of z(x,t) is continuous for all values of x and t, gives the correct initial value of $\displaystyle z_t$ and $\displaystyle z_t \to 0$ as |x|$\displaystyle \to 0$

------------------------------------------------
SOLUTION TO PART A:
------------------------------------------------

GRAPH:
http://img82.imageshack.us/img82/172...olutionoy2.jpg

$\displaystyle z(x,t) = \left\{ {\begin{array}{*{20}c} {\int\limits_0^{x + ct} {\frac{1} {{1 + x}}} } & {\left\{ {x + ct > 0,x - ct < 0} \right\}} & {{\text{Region I}}} \\ 0 & {\{ x + ct < 0,x - ct < 0\} } & {{\text{Region II}}} \\ {\int\limits_{x - ct}^{x + ct} {\frac{1} {{1 + x}}} } & {\{ x + ct > 0,x - ct > 0\} } & {{\text{Region III}}} \\ \end{array} } \right.$

Working out region 1:
-----------------------
$\displaystyle \begin{gathered} {\text{Region I}}:{\text{ }}\int\limits_0^{x + ct} {\frac{1} {{1 + x}}} = \left[ {\ln (1 + x)} \right]_0^{x + ct} \hfill \\ = \ln (1 + x + ct) - \ln 1 \hfill \\ = \ln \left( {\frac{{1 + x + ct}} {1}} \right) = \ln (1 + x + ct) \hfill \\ \end{gathered}$

Working out region 2:
-----------------------
Region 2 = 0.

Working out region 3:
-----------------------
$\displaystyle \begin{gathered} {\text{Region III: }}\int\limits_{x - ct}^{x + ct} {\frac{1} {{1 + x}}} = \left[ {\ln (1 + x)} \right]_{x - ct}^{x + ct} \hfill \\ = \ln (1 + x + ct) - \ln (1 + x - ct) \hfill \\ = \ln \left( {\frac{{1 + x + ct}} {{1 + x - ct}}} \right) \hfill \\ \end{gathered}$

FINAL SOLUTION:
-----------------------------

$\displaystyle z(x,t) = \left\{ {\begin{array}{*{20}c} {\ln (1 + x + ct)} & {\left\{ {x + ct > 0,x - ct < 0} \right\}} & {{\text{Region I}}} \\ 0 & {\{ x + ct < 0,x - ct < 0\} } & {{\text{Region II}}} \\ {\ln \left( {\frac{{1 + x + ct}} {{1 + x - ct}}} \right)} & {\{ x + ct > 0,x - ct > 0\} } & {{\text{Region III}}} \\ \end{array} } \right.$

Thats part a done. As far as I know that's all there is to evaluating the question. Is there anythin else I need to add???

PART B: any ideas on this please?