Solve u_t=a^2u_(xx) with u(0,t)=0 , u_x(l,t)=-Hu(l,t)…
Solve $u_t=a^2u_{xx}$ with $u(0,t)=0$, $u_x(l,t)=-Hu(l,t)$, $u(x,0)=\varphi(x)$, and interpret physically.
Explanation of Method
Separation of variables leads to a transcendental eigenvalue equation (no closed form for $\mu_n$); the eigenfunctions are still mutually orthogonal, with a norm computable directly from the eigenvalue equation.
Step 1 · Eigenvalue problem
$X(0)=0$ gives $X=A\sin\mu x$; the Robin condition $X'(l)=-HX(l)$ gives the transcendental equation \[ \tan\mu_nl = -\frac{\mu_n}{H}, \] with infinitely many positive roots $\mu_n$ (found graphically/numerically) and eigenfunctions $X_n=\sin\mu_nx$.
Step 2 · Norm and coefficients
Using $\tan\mu_nl=-\mu_n/H$ to evaluate $\sin2\mu_nl$: $\|X_n\|^2=\frac{l}{2}+\frac{H}{2(\mu_n^2+H^2)}$, so \[ u(x,t)=\sum_{n=1}^\infty C_ne^{-a^2\mu_n^2t}\sin\mu_nx,\qquad C_n = \frac{2(\mu_n^2+H^2)}{l(\mu_n^2+H^2)+H}\int_0^l\varphi(x)\sin\mu_nx\,dx. \]
Verification
$u(0,t)=0$ trivially; at $x=l$, $u_x=\sum C_ne^{-\cdots}\mu_n\cos\mu_nl$ and $Hu=\sum C_ne^{-\cdots}H\sin\mu_nl$, and the eigenvalue equation $\mu_n\cos\mu_nl=-H\sin\mu_nl$ gives $u_x(l,t)=-Hu(l,t)$. \checkmark
Physical interpretation
Left end clamped at zero temperature; right end loses heat by Newton convective cooling into a medium at temperature $0$, with $H=\alpha/k$ (heat-transfer coefficient over conductivity). As $H\to\infty$, $\mu_n\to n\pi/l$ (fixed-fixed rod); as $H\to0$, $\mu_n\to(2n-1)\pi/(2l)$ (fixed-insulated rod).
Result
The solution is: