I'm asked to find the solution to the following problem. I was going to use Method of Seperation of Variables -and Fourier series- to deal with it. i.e I thought of assuming this equation is seperable, so $U(x,t)=X(x)T(t)$. Then I just calculate $U_{xx} = X''(x)T(t)$ and $U_t = X(x)T'(t)$ and plug these ito the following differential equation. However, I am not sure about the given conditions. It resembles Neumann BC but we don't have $u_x(0,t)$. My question is that is there a problem - instead of $u_x(0,t)$, $u(0,t)$ is given-? Or can I solve this? \begin{array} { c } u _ { t } - u _ { x x } = 0 , \quad 0 < x < 1 , t > 0 \\ u ( x , 0 ) = 1 \\ u ( 0 , t ) = 0 \quad u _ { x } ( 1 , t ) = 0 \quad t \geq 0 \end{array}
EDIT: Using Method of Separation of Variables would work. Assume $U(x,t)=X(x)T(t)$. Then plug this into the original pde, get $XT'-X''T=0$. Then apply boundary conditions get, $X(0)=X'(1)=0$. The rest will done. However, I have trouble now when it comes to finding coefficients. I got $U(x,t)= u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \cdot \sin \left( \frac { \pi n } { 2 } x \right) \cdot e ^ { - \left( \frac { n \pi } { 2 } \right) ^ { 2 } t } $.
However, now when I try to find $b _ { n } = \int _ { - 1 } ^ { 1 } f ( x ) \cdot \sin \left( \frac { \pi n } { 2 } x \right) \cdot d x = \int _ { - L } ^ { 1 } \sin \left( \frac { \pi n } { 2 } x \right) \cdot d x = 0$ But, $f(x)=1=u(x,0)$. How am I supposed to find nonzero $b_n$s?
EDIT2: Read comment section.