# linear hyperbolic PDE with some BCs at infinity

The question is as below

$U_{tt}+aU_t=b^2U_{xx}$

$U(t,0)=2\cos wt$

$U(t,x)\to0~\text{as}~x\to\infty$

$x$ is equal or larger than $0$ and less than infinity

$w$, $a$ positive and real

1) find $u(t,x)=e^{iwt+kx}+e^{iwt+kw}~;$

2) find $k$ in terms of $w$ and $a$ $\left(\text{assume}\dfrac{a}{w}<<1\right)$

I first tried to convert $U(t,x)=V(t,x)+2\cos wt$ but cannot work it out since the last BC involves infinity limit. Also, it didn't give an initial condition, or that doesn't matter

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Hello there, try putting your equations in latex, as explained here: meta.math.stackexchange.com/questions/107/… – akkkk Dec 16 '12 at 21:10
thanks a lot for converting the question into Latex, Martin : ) I'm still learning Latex and will try in next question – user53344 Dec 16 '12 at 21:35

Let $U(t,x)=T(t)X(x)$ ,

Then $T''(t)X(x)+aT'(t)X(x)=b^2T(t)X''(x)$

$(T''(t)+aT'(t))X(x)=b^2T(t)X''(x)$

$\dfrac{T''(t)+aT'(t)}{T(t)}=\dfrac{b^2X''(x)}{X(x)}=\dfrac{4s^2-a^2}{4}$

$\begin{cases}T''(t)+aT'(t)+\dfrac{a^2-4s^2}{4}T(t)=0\\X''(x)+\dfrac{a^2-4s^2}{4b^2}X(x)=0\end{cases}$

$\begin{cases}T(t)=\begin{cases}c_1(s)e^{-\frac{at}{2}}\sinh ts+c_2(s)e^{-\frac{at}{2}}\cosh ts&\text{when}~s\neq0\\c_1te^{-\frac{at}{2}}+c_2e^{-\frac{at}{2}}&\text{when}~s=0\end{cases}\\X(x)=\begin{cases}c_3(s)e^{\frac{x\sqrt{4s^2-a^2}}{2|b|}}+c_4(s)e^{-\frac{x\sqrt{4s^2-a^2}}{2|b|}}&\text{when}~s\neq\pm\dfrac{a}{2}\\c_3x+c_4&\text{when}~s=\pm\dfrac{a}{2}\end{cases}\end{cases}$

$\therefore U(t,x)=\int_\frac{a}{2}^\infty C_1(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_2(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds+\int_\frac{a}{2}^\infty C_3(s)e^{-\frac{a|b|t-x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_4(s)e^{-\frac{a|b|t-x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds$

$U(t,x)\to 0$ as $x\to\infty$ :

$C_3(s)=0$ , $C_4(s)=0$

$\therefore U(t,x)=\int_\frac{a}{2}^\infty C_1(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\sinh ts~ds+\int_\frac{a}{2}^\infty C_2(s)e^{-\frac{a|b|t+x\sqrt{4s^2-a^2}}{2|b|}}\cosh ts~ds$

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Why is it that it is not necessary to consider the condition U(t, 0) in constructing the solution? – Bitrex Dec 24 '12 at 1:10
@Bitrex, I still have no idea about when subsitute the condition of $U(t,0)$ , not it is not necessary to consider the condition of $U(t,0)$ . – doraemonpaul Dec 24 '12 at 11:01