Parabolic equation with variable coefficients I want to solve this equation
$$\frac{{\partial u}}{{\partial t}} = 2ct\frac{{\partial u}}{{\partial x}} + \frac{1}{2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}$$
with initial data $u(x,0) = \varphi (x)$ and $u(0,t) = 0$ where $x \in [0, + \infty )$
At first, I wanted to use Fourier Transformation and I got that in the case $x \in ( - \infty , + \infty )$ the solution is
$$u(x,t) = \frac{1}{{\sqrt {2\pi t} }}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x + c{t^2} - \xi )}^2}}}{{2t}}}}\varphi (\xi )d\xi } $$
but in this case we can't use the method of images to get a solution in a semi-interval from all-interval solution...
Another way - using a substitution $y = x + c{t^2}$. We get the equation ${u_t} = \frac{1}{2}{u_{xx}}$ but with moving boundary condition $$u(c{t^2},t) = 0$$ and I think it's not easier to solve than the original equation.
Any ideas? Thanks for any help.
P.S.Sorry for my terrible English
 A: Hint:
Let $\begin{cases}p=x+ct^2\\q=t\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=2ct\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\therefore2ct\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}=2ct\dfrac{\partial u}{\partial p}+\dfrac{1}{2}\dfrac{\partial^2u}{\partial p^2}$
$\dfrac{\partial u}{\partial q}=\dfrac{1}{2}\dfrac{\partial^2u}{\partial p^2}$
Let $u(p,q)=P(p)Q(q)$ ,
Then $P(p)Q'(q)=\dfrac{1}{2}P''(p)Q(q)$
$\dfrac{2Q'(q)}{Q(q)}=\dfrac{P''(p)}{Q'(q)}=-s^2$
$\begin{cases}\dfrac{Q'(q)}{Q(q)}=-\dfrac{s^2}{2}\\P''(p)+s^2P(p)=0\end{cases}$
$\begin{cases}Q(q)=c_3(s)e^{-\frac{qs^2}{2}}\\P(p)=\begin{cases}c_1(s)\sin ps+c_2(s)\cos ps&\text{when}~s\neq0\\c_1p+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{qs^2}{2}}\sin ps~ds+\int_0^\infty C_2(s)e^{-\frac{qs^2}{2}}\cos ps~ds=\int_0^\infty C_1(s)e^{-\frac{ts^2}{2}}\sin((x+ct^2)s)~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{2}}\cos((x+ct^2)s)~ds$
