I'm supposed to solve this problem
$$ \begin{align}\frac{\partial u }{\partial t} &= c^2 \frac{\partial ^2u }{\partial x^2}\\\text{ so that :} \\x&>0\\ t&>0 \\ u(x,0)&=f(x)\\ u(0,t)&=0\end{align}$$
It is similar to heat equation but I couldn't figure out how to solve it without $u(l,t)=0$ boundary condition.
Could someone help me, please?
Of course use separation of variables:
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=c^2X''(x)T(t)$
$\dfrac{T'(t)}{c^2T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{c^2T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-c^2ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-c^2ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-c^2ts^2}\cos xs~ds$
$u(0,t)=0$ :
$\int_0^\infty C_2(s)e^{-c^2ts^2}~ds=0$
$C_2(s)=0$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-c^2ts^2}\sin xs~ds$
$u(x,0)=f(x)$ :
$\int_0^\infty C_1(s)\sin xs~ds=f(x)$
$\mathcal{F}_{s,s\to x}\{C_1(s)\}=f(x)$
$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{f(x)\}$
$\therefore u(x,t)=\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{f(x)\}e^{-c^2ts^2}\sin xs~ds$