Let $u(t,x)=T(t)X(x)$ ,
Then $T'(t)X(x)=x^2T(t)X''(x)$
$\dfrac{T'(t)}{T(t)}=\dfrac{x^2X''(x)}{X(x)}=-\dfrac{4s^2+1}{4}$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{4s^2+1}{4}\\x^2X''(x)+\dfrac{4s^2+1}{4}X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-\frac{t(4s^2+1)}{4}}\\X(x)=\begin{cases}c_1(s)\sqrt{x}\sin(s\ln x)+c_2(s)\sqrt{x}\cos(s\ln x)&\text{when}~s\neq0\\c_1\sqrt{x}\ln x+c_2\sqrt{x}&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$
$u(0,x)=u_0(x)$ :
$C_1\sqrt{x}\ln x+C_2\sqrt{x}+\int_0^\infty C_3(s)\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)\sqrt{x}\cos(s\ln x)~ds=u_0(x)$
$C_1\ln x+C_2+\int_0^\infty C_3(s)\sin(s\ln x)~ds+\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}$
$\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}-C_1\ln x-C_2-\int_0^\infty C_3(s)\sin(s\ln x)~ds$
$\int_0^\infty C_4(s)\cos xs~ds=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\int_0^\infty C_3(s)\sin xs~ds$
$\mathcal{F}_{c,s\to x}\{C_4(s)\}=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\mathcal{F}_{s,s\to x}\{C_3(s)\}$
$C_4(s)=\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}+C_1\delta'(s)-C_2\delta(s)-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}$
$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_1\delta'(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty C_2\delta(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-C_1e^{-\frac{t}{4}}\sqrt{x}\ln x-C_2e^{-\frac{t}{4}}\sqrt{x}-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$
The function not only exist for at least $x\in(0,\infty)$ , but also has one arbitrary function left.
To determine whether the function exist at $x=0$ or not, we just need to consider $\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)$ .
Since both $\sin(s\ln x)$ and $\cos(s\ln x)$ are bounded,
$\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)=0$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)=0$
So the function not only also exist at $x=0$ , but also equals to $0$ when $x=0$ .