This problem is related to this question. If you can answer this, then you might be able to also answer the other question, so please have a look at it.
I am trying to solve the following problem:
$$\dfrac{\partial{\phi}}{\partial{t}} = \kappa \dfrac{\partial^2{\phi}}{\partial{x}^2}, \ x > 0, t > 0$$
$$\phi(x, 0) = 0, \ x > 0$$
$$\phi(0, t) = T_0e^{-bt}, \ t > 0$$
Taking the Laplace transform (in $t$, of course) gives
$$\mathcal{L} \left\{ \dfrac{\partial{\phi}}{\partial{t}} \right\} = \kappa \mathcal{L} \left\{ \dfrac{\partial^2{\phi}}{\partial{x}^2} \right\}$$
$$\therefore s \mathcal{L}\{\phi\} - \phi(0) = \kappa \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2}$$
And since $\phi(x, 0) = 0$, we have
$$s \mathcal{L} \{ \phi \} = \kappa \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2}$$
This is an ODE with constant coefficients.
From now on, let $\mathcal{L}\{ \phi \} = \bar{\phi}$ (for the sake of conciseness).
So our constant-coefficient ODE is
$$\dfrac{d^2 \bar{\phi}}{dx^2} - \dfrac{s}{\kappa} \bar{\phi} = 0$$
This is a homogeneous ODE.
The characteristic polynomial is
$$m^2 - \dfrac{s}{\kappa} = 0$$
$$\therefore m = \pm \sqrt{\dfrac{s}{\kappa}}$$
Therefore, the solution to the constant-coefficient homogeneous ODE is
$$\bar{\phi}(x) = Ae^{x\sqrt{\dfrac{s}{\kappa}}} + Be^{-x\sqrt{\dfrac{s}{\kappa}}}$$
Here is where I am stumped.
One way that I have seen other problems proceed is by assuming another boundary condition:
$$\lim\limits_{x \to \infty} \phi(x, t) = 0$$
This makes sense from a physical standpoint, but it was not specified in the problem statement, so I am unsure if it is valid for me to use it?
Also, I don't have full solutions for these problems, so I have no way of checking the intermediate steps.
I would greatly appreciate it if people could please help me solve this problem.
If we proceed with assuming the extra boundary condition (that is, if we proceed with assuming that $\lim\limits_{x \to \infty} \phi(x, t) = 0$), then we continue from above as follows:
Now, since $\lim\limits_{x \to \infty} \phi(x, t) = 0$ and
$$\bar{\phi}(x, s) = \int_0^\infty \phi(x, t) e^{-st} \ dt,$$
we have that $\bar{\phi}(x, s) \to 0$ as $x \to \infty$, and hence $A = 0$, since $Be^{-x\sqrt{\dfrac{s}{\kappa}}} \to 0$ as $x \to \infty$.
Therefore, we now have
$$\bar{\phi}(x, s) = Be^{-x\sqrt{\dfrac{s}{\kappa}}}.$$
We now use the given boundary conditions.
Letting $x = 0$ in the Laplace transform of $\phi$ gives
$$\bar{\phi}(0, s) = \int_0^\infty \phi(0, t)e^{-st} \ dt = \int_0^\infty T_0 e^{-bt}e^{-st} \ dt = \int_0^\infty T_0 e^{-t(b + s)} \ dt$$
$$= \dfrac{-T_0}{b + s} \int_0^{-\infty} e^u \ du = \dfrac{T_0}{b + s},$$
since $\phi(0, t) = T_0 e^{-bt}$ for all $t > 0$.
Inserting this boundary condition for $\bar{\phi}$ gives
$$B = \dfrac{T_0}{b + s},$$
Is this correct?
Therefore, our solution for $\bar{\phi}$ is
$$\bar{\phi}(x, s) = \dfrac{T_0}{b + s} e^{-x \sqrt{\dfrac{s}{\kappa}}},$$
where $b > 0$
Again, is this correct?