Solving $u_t = \alpha^2 u_{xx}$
with boundary and initial conditions
$u(0,t)=0$,
$u_x(1,t)+h u(1,t)=0$,
$u(x,0)=x$.
(Following the book by Farlow, "PDEs for scientists and engineers", page 54)
Before applying the initial condition the solution is
$u(x,t) = \Sigma_{n=1}^{\infty} A_n exp(-(\lambda_n \alpha)^2 t) \sin(\lambda_n x)$
where $\lambda_n$ are solutions to $tan(\lambda) = -\lambda/h.$
To find $A_n$ we multiply above by $\sin(\lambda_m x)$ and integrate $x$ between 0 and 1.
Now, in the book right hand side becomes:
$\Sigma_{n=1}^{\infty} A_n \int_0^1 \sin(\lambda_n x) \sin(\lambda_m x) dx = A_m \int_0^1 \sin^2(\lambda_m x) dx$
My question is: Why? And how can I prove it? I'm familiar with the concept of orthogonality of trig functions, but I didn't know the same rules follow for real coefficients of of $x$ ($\lambda_n$ and $\lambda_m$ rather than integers $m$ and $n$). Any hints or pointers towards the general theory will be much appreciated.
The eigenfunctions are orthogonal whenever the boundary conditions are symmetric, meaning that $$(\phi'\psi)|_0^1 = (\phi\psi')|_0^1 \tag{1}$$ holds for all pairs $\phi,\psi$ satisfying these conditions. This follows from double integration by parts shown here. To verify that (1) holds in your case, observe that the terms at $x=0$ are zero, and the condition at $x=1$ turns (1) into the identity $$-h\phi(1)\psi(1) = -h\phi(1)\psi(1)$$