Solve u_t=a^2u_(xx) with u_x(0,t)=0 , u(l,t)=u_0…
Solve $u_t=a^2u_{xx}$ with $u_x(0,t)=0$, $u(l,t)=u_0$, $u(x,0)=\varphi(x)$, and interpret physically.
Explanation of Method
Subtract off the constant steady state $u_0$ to homogenise the boundary condition at $x=l$; the resulting insulated–fixed eigenvalue problem has cosine eigenfunctions of odd quarter-wavelength.
Step 1 · Homogenise
With $u=u_0+v$: $v_t=a^2v_{xx}$, $v_x(0,t)=0$, $v(l,t)=0$, $v(x,0)=\varphi(x)-u_0$.
Step 2 · Eigenvalue problem
$X'(0)=0$, $X(l)=0$ gives $X=\cos\mu x$ with $\cos\mu l=0$: \[ \mu_n=\frac{(2n-1)\pi}{2l},\qquad X_n=\cos\frac{(2n-1)\pi x}{2l}. \]
Step 3 · Coefficients
\[ v(x,t)=\sum_{n=1}^\infty C_ne^{-a^2(2n-1)^2\pi^2t/(4l^2)}\cos\frac{(2n-1)\pi x}{2l}, \] \[ C_n = \frac2l\int_0^l[\varphi(x)-u_0]\cos\frac{(2n-1)\pi x}{2l}\,dx. \]
Verification
$v_x(0,t)=0$ since $\sin0=0$; $v(l,t)=0$ since $\cos\frac{(2n-1)\pi}{2}=0$. \checkmark As $t\to\infty$, $u\to u_0$ (uniform steady state). \checkmark
Physical interpretation
A rod with an insulated (no heat flux) left end and the right end held at fixed temperature $u_0$ by a reservoir. Heat redistributes from the arbitrary initial profile $\varphi(x)$ toward the uniform equilibrium $u_0$; higher modes decay fastest, so the fundamental mode $n=1$ dominates the long-time approach.
Result
The solution is: