If you solve Laplace's equation
$$
\nabla^{2}u=0,\\
u(0,y)=0,\;\;\; u_{x}(1,y)=0,\\
u(x,0)=0,\;\;\; u(x,1) = 100,
$$
then $v=T-u$ will satisfy
$$
v_{t}=\alpha \nabla^{2}v,\\
v(t,0,y)=0,\;\; v_{x}(t,1,y)=0,\\
v(t,x,0)=0,\;\; v(t,x,1)=0,\\
v(0,x,y)=-u(x,y).
$$
The Laplace equation solution can be written as
$$
u(x,y)=\sum_{n=0}^{\infty}A_n\sin(\lambda_n x)\sinh(\lambda_n y)
$$
where $\lambda_n$ is chosen so that $u_{x}(1,y)=0$, i.e. $\cos(\lambda_n)=0$, i.e.
$$
\lambda_n = \frac{\pi}{2}+n\pi = \frac{(2n+1)\pi}{2}.
$$
The constants $A_n$ are chosen so that
$$
u(x,1)=\sum_{n=0}^{\infty}A_n\sin((n+1/2)\pi x)\sinh((n+1/2)\pi) = 100.
$$
Using the orthogonality of the functions $\sin((n+1/2)\pi x)$ on $[0,1]$ gives
$$
A_n =\frac{100}{\sinh((n+1/2)\pi)}\frac{\int_{0}^{1}\sin((n+1/2)\pi x)dx}{\int_{0}^{1}\sin^{2}((n+1/2)x)dx}
$$
You should be able to solve for $v$ because that's a solution of the standard heat equation with homogeneous boundary conditions, and then let $T=v+u$. There are obvious convergence issues of $u$ at the corners of the region, but nowhere else.

When solving the homogeneous problem on a finite domain with time dependent boundary conditions you want to get the equilibrium temperature. So if we have

$$\begin{cases} u_{t} = k u_{xx} & 0 \leq x \leq L , t>0 \\ u(x,0) = f(x) , & 0 \leq x \leq L \\ u(0,t) = A(t) , u(L,t) = B(t) & t > 0 \end{cases}$$

In this case we need a candidate solution for the equilibrium temperature so we create a function called the **reference temperature**, $r(x,t)$ and we want the function to satisfy.

$$ r(0,t) = A(t) \\ r(L,t) = B(t) $$

we can create one like so

$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$

Now, instead we have

$$ v(x,t) = u(x,t) - r(x,t) $$

and this solution $v(x,t)$ will solve the homogeneous problem.

Ok, now that it has been set up we'll let

$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$

and now we replace $A(t) = \frac{1}{t+a^{-1}} + a$ and $B(t) = \frac{-1}{t+a^{-1}} + a $

$$ r(x,t) = \frac{1}{t+a^{-1}} + a + \frac{x}{L} \big [\frac{-2}{t+a^{-1}} \big]$$

Now, instead we have this homogeneous problem you should be able to solve

$$\begin{cases} v_{t} = k v_{xx} & 0 \leq x \leq L , t>0 \\ v(x,0) = f(x) - r(x) , & 0 \leq x \leq L \\ v(0,t) = 0 , v(L,t) = 0 & t > 0 \end{cases}$$

The solution to this is

$$ v(x,t) = \sum_{i=1}^{\infty} a_{n} \sin(\frac{n \pi x}{L}) e^{-k t(\frac{n \pi}{L})^{2} }$$

you solve for the coefficients $a_{n}$ and we get

$$ a_{n} = \frac{2}{L} \int_{0}^{L} \big[ f(x) - r(x) \big] \sin(\frac{n \pi x}{L}) dx $$

where $f(x)$ is your function

$$ f(x) = \begin{cases} u_{1} & x < 1 \\ 0 & x \geq 1 \end{cases}$$

which looks like the complement of a Heaviside function multiplied by the constant $u_1$. Try that and see what happens.

This about the reference temperature function $r(x,t)$. We need a function such that it obeys the boundary conditions.

If $u(x,t) = v(x,t) + r(x,t)$, we have already said that $v(x,t)$ is the homogeneous solution. So $u(0,t) = v(0,t) + r(0,t)$ and $u(L,t) = v(L,t) + r(L,t)$. But $v(0,t) = 0$ and $v(L,t) = 0$.

$ x = 0$ we get

$$ r(0,t) = A(t) + \frac{0}{L}[B(t) - A(t)] = A(t)$$

and at $x=L$ we have

$$ r(L,t) = A(t) + \frac{L}{L}[B(t) -A(t)] = B(t) + A(t) - A(t) = B(t) $$

which is what we wanted.

## Best Answer

Hint:

We can guess that $u(x,y) = X(x)Y(y)$ Then we get

$$u_{xx} = 4u_y = X^{''}(x)*Y(y) = 4Y^{'}(y)X(x)$$

From there, we isolate the functions to get

$$\frac{X^{''}(x)}{X(x)}=4\frac{Y^{'}(y)}{Y(y)}$$

However, a function of x can only equal a function of y if both functions are constant. Let's call this function $-\lambda$

Then

$$\frac{X^{''}(x)}{X(x)}=-\lambda$$

$$4\frac{Y^{'}(y)}{Y(y)}=-\lambda$$

See if you can solve from here!