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I posed myself the following PDE because it would be interesting to graph: $ u_t=u_{xx},\qquad00,\\ \begin{align} u(0,t)&=\sin^2\frac t2,\\ u_x(L,t)&=0,\\ u(x,0)&=0. \end{align}$ Physically, this is a rod with one end insulated and the other being periodically heated and cooled.

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.

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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?

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    Is there a general formula for finding φ(x,t) in case of more difficult problems?2016-02-05
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Here is a different approach, called the method of eigenfunction expansions.

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I correct your differentiation mistake,

$(\sin^2(\frac{t}{2}))' = 2\sin(\frac{t}{2})(\frac{1}{2})\cos(\frac{t}{2}) = \sin(\frac{t}{2})\cos(\frac{t}{2}).$

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/help/notation).2014-10-18