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I posed myself the following PDE because it would be interesting to graph: $$ u_t=u_{xx},\qquad0<x<L,\qquad t>0,\\ \begin{align} u(0,t)&=\sin^2\frac t2,\\ u_x(L,t)&=0,\\ u(x,0)&=0. \end{align}$$ 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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3 Answers 3

up vote 2 down vote accepted

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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Yes, I can continue from here on. Thank you very much for your help! I had not heard of that technique before. :) –  Josué Molina Oct 5 '12 at 21:51
    
In fact your approach is similar to that in maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=38. –  doraemonpaul Oct 5 '12 at 22:03

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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