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