Recipe

Shift a nonhomogeneous boundary value problem to a homogeneous one

Use when: the heat equation's boundary values are fixed constants that aren't both zero, such as $u(0,t)=T_1$ , $u(L,t)=T_2$

Goal: a complete $u(x,t)$ , built from a linear reference profile plus the usual zero-boundary sine series

Taught in Nonhomogeneous boundary data: shifting to homogeneous

Quick recipe

Check whether the boundary data is actually homogeneous
Build the reference solution $\psi(x)$
Substitute $u=v+\psi$ and derive $v$ 's homogeneous problem
Solve $v$ 's problem exactly as a Dirichlet problem
Recover $u$ by adding $\psi$ back

1
Look at the right side of each boundary condition. Only a right side of exactly $0$ lets separation apply directly.
- both boundary values are $0$ — separate directly, no shift needed.
- at least one boundary value is a nonzero constant — a reference solution has to be subtracted off first.
Why: a nonzero constant on the boundary can never be written as $X(0)T(t)$ for every $t$ . Separation fails structurally here, not just computationally.
2
Solve $\psi''=0$ subject to the two given boundary values; $\psi(x)=A+Bx$ , with $A,B$ fixed by matching each endpoint.

Example

For $\psi(0)=20$ , $\psi(4)=8$ : $A=20$ and $B=(8-20)/4=-3$ , so $\psi(x)=20-3x$ .

Why: $\psi$ is the one piece of the problem, depending on $x$ alone, that can carry the mismatched boundary values. It isn't required to vanish there.
3
Write $v_t=kv_{xx}$ , $v(0,t)=0$ , $v(L,t)=0$ , $v(x,0)=f(x)-\psi(x)$ . Here $\psi_{xx}=0$ cancels $\psi$ from the PDE, and matching endpoints cancels it from the boundary data.

Example

With $f(x)=20$ and $\psi(x)=20-3x$ : $v(x,0)=20-(20-3x)=3x$ , already zero at both ends.

Why: subtracting a function that already carries the nonzero data leaves exactly the zero-boundary problem the earlier sine-series method already solves.
4
Apply the earlier Dirichlet heat equation recipe to $v$ . Use $f-\psi$ as the initial data, to get its coefficients $b_n$ and $v(x,t)$ .

Example

$b_n=\frac24\int_0^4 3x\sin(n\pi x/4)\,dx=\frac{24(-1)^{n+1}}{n\pi}$ , so $b_1\approx7.639$ , $b_2\approx-3.820$ , $b_3\approx2.546$ .

Why: $v$ 's problem is literally the earlier Dirichlet case, with a different initial function. So the same machinery applies unmodified.
5
Set $u(x,t)=\psi(x)+v(x,t)$ . Then check that as $t\to\infty$ the series decays and $u$ approaches $\psi(x)$ , not zero.

Example

$u(x,t)=(20-3x)+\sum_{n=1}^\infty\frac{24(-1)^{n+1}}{n\pi}\sin\!\left(\frac{n\pi x}4\right)e^{-0.5(n\pi/4)^2t}$ ; as $t\to\infty$ , $u(x,t)\to20-3x$ .

Why: $\psi$ was only subtracted temporarily, to make the problem separable. The physical solution needs it added back, to recover the actual steady state.

Common mistakes

Trying to separate $u=X(x)T(t)$ directly when a boundary value is a nonzero constant, instead of shifting first.
Computing the sine coefficients from the original $f(x)$ instead of from $f(x)-\psi(x)$ .
Assuming the rod settles to zero as $t\to\infty$ , instead of to $\psi(x)$ .
Swapping which endpoint's value goes into $A$ versus $B$ when building $\psi(x)=A+Bx$ .