On $\{0<x<1,t>0\}$ I want to solve this mixed problem:
$\left\{ \begin{gathered} {u_{tt}} - {u_{xx}} = 0 \\ u(0,t) = 0,{u_x}(1,t) = \frac{{\sin t}}{{2017}} \\ u(x,0) = 0,{u_t}(x,0) = 0 \\ \end{gathered} \right.$
It's obvious that I should use separating variables, but I find it hard to determine the eigenvalues... Could anyone show me a solution as an example for problems of this type?
Thanks in advance.
Try a substitution of the form $$ u(x,t)=v(x,t)+a(x)\frac{\sin t}{2017}. $$ The condition $u(0,t)=0$ becomes $v(0,t)=0$ if $a(0)=0$. The condition $u_{x}(1,t)=\sin t/2017$ becomes $v_{x}(1,t)=0$ if $a'(1)=1$. And the equation $u_{tt}-u_{xx}=0$ becomes $v_{tt}-v_{xx}=0$ if $$ a(x)+a''(x)=0. $$ So you want a solution $a(x)$ of $$ a''(x)+a(x) = 0,\;\; a(0)=0,\; a'(1)=1. $$ That leads to $a(x)=\sin(x)/\cos(1)$. Everything works out to give a new equation for $v$ of the form: $$ v_{tt}-v_{xx} = 0 \\ v(0,t)=0, \;\; v_{x}(1,t)=0 \\ v(x,0)=0,\;\; v_{t}(x,0)=-\frac{\sin(x)}{2017\cos(1)}. $$ This looks like a standard problem now. After solving for $v$, the original solution is $$ u(x,t) = v(x,t) + \frac{\sin x}{\cos 1}\frac{\sin t}{2017}. $$