Look at a solution $v(x,y)$ of the Neumann problem
\begin{align*}
v_{xx}+v_{yy}& =0\text{ in the unit square} \\
v_{x}(0,y)=v_{x}(1,y)& =0 \\
v_{y}(x,1)& =0 \\
v_{y}(x,0)& =-\pi \cos \pi x.
\end{align*}
Now obtain a solution $u(x,y)$ of your Dirichlet problem as the curve
integral
\begin{equation*}
u(x,y)=\int_{\varphi }\left( -v_{y}\mathrm{\,d}x+v_{x}\,\mathrm{d}y\right) ,
\end{equation*}
where $\varphi $ is any curve in the unit square connecting $(0,0)$ to $(x,y)
$. By the Gauss-Green theorem and the heat equation satisfied by $v$, the
value of this integral is independent of the path. For example,
\begin{equation*}
u(x,y)=-\int_{0}^{x}v_{y}(s,0)\mathrm{\,d}s+\int_{0}^{y}v_{x}(x,t)\,\mathrm{d
}t.
\end{equation*}
From this follows that
\begin{align*}
u_{x}& =-v_{y} \\
u_{y}& =v_{x},
\end{align*}
and hence
\begin{equation*}
u_{xx}+u_{yy}=-v_{yx}+v_{xy}=0.
\end{equation*}
Also the Dirichlet boundary values are easily checked.
This is very involved problem. To solve, you'll need to break the solution into successively smaller pieces. First, separate the steady-state and transient solutions, then split up the boundary conditions in order to use separation of variables.
To start off, I'm going to label
$$ T(x,y,t) = u(x,y,t) + v(x,y) $$
where $v(x,y)$ is the time-independent, steady-state solution, and $u(x,y,t)$ is the decaying, transient solution.
The steady-state solution should satisfy $\nabla^2v = 0$ and all the boundary conditions as listed. Since all boundaries are inhomogeneous, we need so split it up further
$$ v(x,y) = v_1(x,y) + v_2(x,y) $$
such that
\begin{matrix}
\begin{cases}
\nabla^2 v_1 = 0 \\ \\
v_1(0,y) = T_1,\ v_1(L,y) = T_0 \\ \\
\dfrac{\partial}{\partial y}v_1(x,0) = \dfrac{\partial}{\partial y}v_1(x,L) = 0
\end{cases}
&&&
\begin{cases}
\nabla^2v_2 = 0 \\ \\
v_2(0,y) = v_2(L,y) = 0 \\ \\
\dfrac{\partial}{\partial y}v_2(x,0) = \dfrac{\partial}{\partial y}v_2(x,L) = a
\end{cases}
\end{matrix}
The homogeneous boundaries allow us to use separation of variables to solve each individual problem
- The first one is easier since you can intuitively guess that it's constant in $y$ and linear in $x$. This turns out to be
$$ v_1(x,y) = T_0\frac{x}{L} + T_1\frac{L-x}{L} = \frac{(T_0-T_1)x}{L} + T_1 $$
- For the second problem, the homogeneous boundary condition on $x$ returns a series solution of the form
$$ v_2(x,y) = \sum_{n=1}^\infty \sin\left(\frac{n\pi x}{L}\right)\left[A_n\cosh\left(\frac{n\pi y}{L}\right) + B_n \cosh\left(\frac{n\pi(L-y)}{L}\right)\right] $$
Then, applying the remaining boundary conditions on $y$ will give
$$ A_n = -B_n = \begin{cases} \dfrac{4aL}{n^2\pi^2\sinh(n\pi)}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases} $$
For a more thorough explanation of why I'm using hyperbolic functions instead of exponentials, check out my answer for this similar problem. Short answer: It makes the math easier
The remaining transient solution is homogeneous on all boundaries and has initial conditions that cancel out the steady-state:
\begin{cases}
\dfrac{\partial u}{\partial t} = \nabla^2 u \\ \\
u(0,y,t) = u(L,y,t) = \dfrac{\partial}{\partial y}u(x,0,t) = \dfrac{\partial}{\partial y}u(x,L,t) = 0 \\ \\
u(x,y,0) = T_0 - v(x,y)
\end{cases}
Applying separation of variables once more and matching the homogeneous boundaries, we obtain
$$ u(x,y,t) = \sum_{n,m} c_{n,m} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right)\exp \left[-(n^2+m^2)\frac{\pi^2}{L^2}t\right] $$
The initial condition for this last piece will have you solve a double Fourier series in $x$, $y$. You can use linearity to simplify calculations a bit, i.e.
\begin{align}
u_1(x,y,0) &= T_0 - T_1 - \frac{T_0-T_1}{L}x = \sum_{n,m} c_{1(n,m)} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right) \\
u_2(x,y,0) &= -\sum_{n=2k+1} \frac{4aL}{n^2\pi^2\cosh(n\pi)} \sin\left(\frac{n\pi x}{L}\right)\left[\cosh\left(\frac{n\pi y}{L}\right) - \cosh\left(\frac{n\pi(L-y)}{L}\right)\right] \\ &\qquad = \sum_{n,m} c_{2(n,m)} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right)
\end{align}
The first boundary function is constant in $y$ and the second is already a partial Fourier series, so they simplify to
\begin{align}
T_0 - T_1 - \frac{T_0-T_1}{L}x &= \sum_{n=1}^\infty c_{1(n,0)} \sin\left(\frac{n\pi x}{L}\right) \\
-\frac{4aL}{n^2\pi^2\sinh(n\pi)} \left[\cosh\left(\frac{n\pi y}{L}\right) - \cosh\left(\frac{n\pi(L-y)}{L}\right)\right] &= c_{2(n,0)} + \sum_{m=1}^\infty c_{2(n,m)}\cos\left(\frac{m\pi y}{L}\right)
\end{align}
The last round of integration finishes the job
Best Answer
This is because you are considering the equation on the half-line. Your domain is essentially $\Omega=[0,\infty)\times [0,y]$
If you look at the heat equation and wave equation on a bounded domain, you will see the solutions are typically given as Fourier Series, but on unbounded domains, the solutions look a lot different. You cannot use boundary methods and separation of variables to solve the PDE here.
Generally speaking, we want solutions $u(x,y)$ which "vanish at infinity" (tend towards $0$ as $||u(x,y)||_{L^2}\rightarrow \infty)$. Functions which are smooth and have compact support (denoted by $C_{c}^{\infty}(\Omega)$) include those that vanish at infinity $C_{0}(\Omega)$. These conditions imposed on the solutions are desirable because they make physical sense. The temperature/potential in a rod is not going to shoot off to infinity, it's going to decrease with time until it reaches a steady-state.
Now to actually solve the PDE, I don't think there is a slick and easy way to do it. Solutions to Laplace's equation in domains which aren't on a disk $D^n$ tend to be ad-hoc and very specific to the domain. Polynomials wouldn't work because they do not vanish at infinity. I would try to look for solutions which use sines and cosines. Alternatively, you could look for a harmonic function on the domain $[0,\infty)\times [0,y]$ (you don't necessarily need to work in the complex plane for this). Start by looking for qualitative solutions and then try to get a concrete, quantitative one.