If i have a region $x>0$ that is initially liquid at constant temp $T_0$ above freezing temp $T_f$ (so $T_0 > T_f$). The surface at $x=0$ stays at $T_1$ below freezing temp (so $T_1<T_f<T_0$) And $T_s(x,t)$ and $ T_L(x,t)$ satisfy:
$\frac{\partial {T_s}}{\partial{t}} = D_s\frac{\partial^2{T_s}}{\partial x^2}$ and $\frac{\partial {T_L}}{\partial{t}} = D_L\frac{\partial^2{T_L}}{\partial x^2}$
For constants $D_s$ and $D_L$, with $T_s(x,t)=T_1$ and $T_L(x,0)=T_0$. The boundary between liquid and solid is $x=s(t)$ with $s(0)=0$
So the boundary conditions i found were:
At $x=0$ $T=T_1$ and is maintained at $T_1$ until it starts to change to solid at $x=s(t)$ where $T=T_f$
Using boundary conditions That $T_s$ and $T_L$ must satisfy show the solution can be written as:
$T_s(x,t) = T_1+\frac{T_f-T_1}{erf(m)}erf(\frac{x}{\sqrt{4D_Lt}})$
$T_L(x,t) = T_0-\frac{T_0-T_f}{1-erf(m\sqrt{\frac{D_s}{D_L}})}(1-erf(\frac{x}{\sqrt{4D_Lt}}))$
That took flipping ages to write... anyway i understand how to find the boundary conditions, that's the easy part however i have zero clue how to go about showing these solutions are true, any help would be absolutely amazing! Thanks in advance!
This is how I managed to work it out, if anyone notices an inconsistency please point it out, I'm definitely no expert.
Boundary Conditions $$T_S(0,t)=T_1$$ $$T_S(s(t),t)=T_L(s(t),t)=T_f$$ $$T_L(x,0)=T_0$$
Solution will be in the form of $$T(x,t)=c_1\mbox{erf}(\frac{x}{\sqrt{4Dt}})+c_2$$
Therefore
$$T_S(x,t)=c_1\mbox{erf}(\frac{x}{\sqrt{4D_St}})+c_2$$ $$T_L(x,t)=c_3\mbox{erf}(\frac{x}{\sqrt{4D_Lt}})+c_4$$
Using the first boundary condition
$$T_S(0,t)=c_1\mbox{erf}(\frac{0}{\sqrt{4D_St}})+c_2 =c_1\mbox{erf}(0)+c_2 =c_2=T_1$$
Using the second boundary condition
$$T_S(s(t),t)=c_1\mbox{erf}(\frac{s(t)}{\sqrt{4D_St}})+T_1 =T_f$$
In this case, $c_1$ can only be a constant if $$s(t)=m\sqrt{4D_St}$$
Therefore
$$T_S(s(t),t)=c_1\mbox{erf}(m)=T_f-T_1$$ $$c_1=\frac{T_f-T_1}{\mbox{erf}(m)}$$
Therefore
$$\mbox{erf}(\frac{x}{\sqrt{4D_St}})$$
$$T_S(x,t)=(\frac{T_f-T_1}{\mbox{erf}(m)})\mbox{erf}(\frac{x}{\sqrt{4D_St}})+T_1$$
Which is the first part of the answer rearranged but I'll leave it as is for continuity's sake
Using the third boundary condition
$$T_L(x,0)=c_3\mbox{erf}(\infty)+c_4=T_0$$
$$c_3+c_4=T_0$$
For ease of use later, let's rearrange it to: $$c_4=T_0-c_3$$
Using the second boundary condition again $$T_L(s(t),t)=c_3\mbox{erf}(\frac{s(t)}{\sqrt{4D_Lt}})+T_4=T_f$$
$$T_f=c_3\mbox{erf}(\frac{m\sqrt{4D_St}}{\sqrt{4D_Lt}})+(T_0-C_3)$$
$$T_f-T_0=c_3\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})-c_3$$
$$T_f-T_0=-c_3(1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}}))$$
$$c_3=\frac{T_0-T_f}{1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})}$$ Therefore
$$c_4=T_0-\frac{T_0-T_f}{1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})}$$
Put both results for the liquid state together
$$T_L(x,t)=\frac{T_0-T_f}{1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})}\mbox{erf}(\frac{x}{\sqrt{4D_Lt}})+T_0-\frac{T_0-T_f}{1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})}$$
Which simplifies
$$T_L(x,t)=T_0-(\frac{T_0-T_f}{1-\mbox{erf}(m\sqrt{\frac{D_S}{D_L}})})(1-\mbox{erf}(\frac{x}{\sqrt{4D_Lt}})$$
At this point, we should have both equations, again please comment any problems you can find.