Heat Equation, Part 4

The One-Dimensional Heat Equation

Part 4: Unequal Boundary Conditions

Now we consider the case where the boundary conditions may assume values other than 0. In our example, we'll assume that the left end of the rod is kept at 1 and the right at -2. Here is a statement of the current requirements.

The initial condition h is graphed below

Let w be given by

w(x) = 1 - (3/2) x
Show that w is the unique steady-state solution of the heat equation with the boundary conditions w(0) = 1 and w(2) = -2, i.e., the unique function satisfying the heat equation, the boundary conditions, and the condition that the partial derivative of w with respect to t is 0.

Now suppose that u(x,t) is a solution of the initial/boundary value problem

. Explain why

v(x,t) = u(x,t) - w(x)

is a solution of

Calculate the Fourier sine expansion of h(x) - w(x) and compare the graphs of the Fourier approximations with the graph of h(x) - w(x).

Now examine the time evolution snap shots for the approximate solutions.

Explain in terms of the symbolic form of the approximate solution why the graph of the solution approaches the graph of the linear function w.
Explain in terms of the symbolic form of the approximate solution why this long-term behavior does not depend on the initial condition.

Next we consider a problem where the boundary conditions are not equal and the initial condition is not compatible with the boundary conditions. Graph the new initial condition function h2 and compare that graph to the graph of the steady-state solution w(x).

Compute the corresponding approximate solution of this initial/boundary value problem. For small values of t (between 0.001 and 0.1) compare time snapshots of the approximate solution to the initial condition. Then consider time snapshots for larger values of t (between 0.1 and 3).

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