Solving Heat Equation with Non-Homogeneous BCs 1: Time-Indepdendent BCs

Consider the following 1-dimensional heat IBVP:
\begin{align*}
&u_t=\alpha^2 u_{xx},\ 0<x<L,\ t>0\\
&{\rm BCs:}\ \left\{\begin{aligned}
u(0,t)&=c_1,\ t>0\\
u(L,t)&=c_2,
\end{aligned}
\right.\\
&{\rm IC:}\ u(x,0)=\phi(x),\ 0<x<L.
\end{align*}
Since the BCs are not homogeneous, the method of separation of variables cannot be used. Let us assume that $u(x,t)$ can be decomposed to
$$u(x,t)=S(x,t)+U(x,t)$$
where $S(x,t)$ satisfies the non-homogeneous BCs i.e.
\begin{align*}
S(0,t)&=c_1\ \ \ \ \ (1)\\
S(L,t)&=c_2
\end{align*}
This implies that $U(x,t)$ satisfies the homogeneous BCs
$$U(0,t)=U(L,t)=0$$
We don’t know the function $S(x,t)$ but we can make a guess. We can first try a possibly simplest one, a linear function in $x$
$$S(x,t)=A(t)x+B(t)$$
If this works, it is our luck day. If not, try something else that is next simplest one, say a quadratic function in $x$. It turns out that our guess works! This means that we can determine the coefficients $A(t)$ and $B(t)$ from the BCs (1) and we find that
$$A(t)=\frac{c_2-c_1}{L},\ B(t)=c_1$$
Thus
$$S(x,t)=\frac{c_2-c_1}{L}x+c_1$$
$U(x,t)$ satisfies the homogeneous IBVP:
\begin{align*}
&U_t=\alpha^2 U_{xx},\ 0<x<L,\ t>0\\
&{\rm BCs:}\ \left\{\begin{aligned}
U(0,t)&=0,\ t>0\\
U(L,t)&=0,
\end{aligned}
\right.\\
&{\rm IC:}\ U(x,0)=\phi(x)-S(x,0)=\phi(x)-\frac{c_2-c_1}{L}x-c_1,\ 0<x<L.
\end{align*}
We already know how to solve the homogeneous IBVP (see here) and
$$U(x,t)=\sum_{n=1}^\infty A_ne^{-\alpha^2\frac{n^2\pi^2}{L^2}t}\sin\frac{n\pi}{L}x$$
where
$$A_n=\frac{2}{L}\int_0^L[\phi(x)-\frac{c_2-c_1}{L}x-c_1]\sin\frac{n\pi}{L}xdx,\ n=1,2,\cdots$$

Note that $\lim_{t\to\infty}U(x,t)=0$ and so $\lim_{t\to\infty}u(x,t)=S(x)$. $U(x,t)$ is called the transient and $S(x)$ is called the steady state. The steady state is eventual state for large time.

Example. Solve the IBVP:
\begin{align*}
&u_t=\alpha^2 u_{xx},\ 0<x<1,\ t>0\\
&{\rm BCs:}\ \left\{\begin{aligned}
u(0,t)&=c_1,\ t>0\\
u(1,t)&=c_2,
\end{aligned}
\right.\\
&{\rm IC:}\ u(x,0)=0,\ 0<x<1.
\end{align*}

Solution: First we solve
\begin{align*}
&U_t=\alpha^2 U_{xx},\ 0<x<1,\ t>0\\
&{\rm BCs:}\ \left\{\begin{aligned}
U(0,t)&=0,\ t>0\\
U(1,t)&=0,
\end{aligned}
\right.\\
&{\rm IC:}\ U(x,0)=-c_1-(c_2-c_1)x,\ 0<x<1.
\end{align*}
The transient $U(x,t)$ is given by
$$U(x,t)=\sum_{n=1}^\infty A_ne^{-\alpha^2n^2\pi^2 t}\sin n\pi x$$
where
\begin{align*}
A_n&=-2\int_0^1[c_1+(c_2-c_1)]\sin n\pi xdx\\
&=\frac{2}{n\pi}[c_2(-1)^n-c_1].
\end{align*}
Hence the solution is
$$u(x,t)=(c_2-c_1)x+c_1+2\sum_{n=1}^\infty\frac{[c_2(-1)^n-c_1]}{n\pi}e^{-\alpha^2n^2\pi^2 t}\sin n\pi x$$

References:

David Betounes, Partial Differential Equations for Computational Science with Maple and Vector Analysis, TELOS, Springer-Verlag

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