I posed myself the following PDE because it would be interesting to graph: $ u_t=u_{xx},\qquad0
Now, since the $x$-dependent problem does not have two homogeneous boundary conditions, I wonder if it is possible to solve this PDE as it stands.
I posed myself the following PDE because it would be interesting to graph: $ u_t=u_{xx},\qquad0
Now, since the $x$-dependent problem does not have two homogeneous boundary conditions, I wonder if it is possible to solve this PDE as it stands.
The way I was taught to solve boundary value problems with non-homogeneous boundary value conditions is via the introduction of a second term to satisfy the boundary, i.e. set $ u(x,t) = \phi(x,t) + v(x,t)$ where $\phi(x,t)$ satisfies the boundary conditions.
So, we need $\phi_x(0,t)=0 \text{ and }\phi(L,t)=\sin^2\frac{t}{2}.$ The simplest function that satisfies these conditions is $\phi(x,t)=\sin^2\frac{t}{2},$ which means $u$ is of the form $u(x,t)=\sin^2\frac{t}{2}+v(x,t).$
The pde now required to solve, for $v$ after substituting into your equation is $v_t=v_{xx}-\cos\frac{t}{2}.$ Substituting into the boundary conditions also gives $v_x(0,t)=0\text{ and } v(L,t)=0$ and the initial condition condition $v(x,0)=0.$
This transforms the original problem with non-homogeneous boundary conditions into one with homogeneous boundary conditions, but a non-homogeneous pde, which is easier to solve.
Can you continue from this point?
Here is a different approach, called the method of eigenfunction expansions.
I correct your differentiation mistake,
$(\sin^2(\frac{t}{2}))' = 2\sin(\frac{t}{2})(\frac{1}{2})\cos(\frac{t}{2}) = \sin(\frac{t}{2})\cos(\frac{t}{2}).$