Steady-State Solution to PDE

[Metronome]

I am trying to solve the following problem...

I know how to use separation of variables to solve the PDE without the forcing function [imath]-g[/imath], so I assume this problem should be solved by adding a particular solution to that homogeneous (no forcing function) solution, but it is not clear that the problem is leading me in that direction. The explanation of steady-state solutions in this chapter is as follows...

However, this doesn't seem to help with this problem, at least if the boundary conditions referenced in part (a) are the ones later stated in part (b), because these boundary conditions are already homogeneous, so the steady-state solution is [imath]u(x,\ t) = 0[/imath].

I also do not know whether [imath]g[/imath] is a constant (the Newtonian constant of gravitation perhaps?) or something else, so I wouldn't know what to guess in order to use Undetermined Coefficients.

Can I get some clarification on how to start this problem?

 

[HallsofIvy]

A "steady state" solution is, as stated here, one that does not change over time. Here the two independent variables are "x", position, and "t", time. The equation is \(\displaystyle u_{tt}= c^2u_{xx}- g\) and, because we want the steady state solution, \(\displaystyle u_{tt}= 0\) so the equation is simply \(\displaystyle c^2u_{xx}- g= 0\) or \(\displaystyle u_{xx}= \frac{g}{c^2}\).
Just integrate, with respect to x, twice.

Yes, of course, c and g are constants.

 

[Metronome]

Okay, so this means the steady state solution is [imath]v = \frac{gx^2 - Lgx}{2c^2}[/imath], so the PDE for the transient part of the solution should be [imath]w_{tt} = c^2 w_{xx} - g, w(0,\ t) = 0, w(L,\ t) = 0, w(x,\ 0) = \frac{Lgx - gx^2}{2c^2}, w_t(x,\ 0) = 0[/imath].

This does not get rid of the forcing function though. Do I need to apply some technique from solving inhomogeneous ODEs (undetermined coefficients, annihilator method, reduction of order, variation of parameters, etc.)?

 

[nasi112]

Okay, so this means the steady state solution is [imath]v = \frac{gx^2 - Lgx}{2c^2}[/imath], so the PDE for the transient part of the solution should be [imath]w_{tt} = c^2 w_{xx} - g, w(0,\ t) = 0, w(L,\ t) = 0, w(x,\ 0) = \frac{Lgx - gx^2}{2c^2}, w_t(x,\ 0) = 0[/imath].

This does not get rid of the forcing function though. Do I need to apply some technique from solving inhomogeneous ODEs (undetermined coefficients, annihilator method, reduction of order, variation of parameters, etc.)?

I will show you how to get rid of the forcing function \(\displaystyle g\). I don't know why you call it a function. Usually, in problems like this if \(\displaystyle g\) is a function, it will be written as \(\displaystyle g(x)\) or \(\displaystyle g(x,t)\). And because this was not the case, we would treat it as a constant.

The original problem is:

\(\displaystyle u_{tt} = c^2u_{xx} - g, \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq L, \ \ \ t > 0,\)

\(\displaystyle u(0,t) = 0,\)
\(\displaystyle u(L,t) = 0,\)
\(\displaystyle u(x,0) = 0,\)
\(\displaystyle u_t(x,0) = 0,\)

We can get rid of \(\displaystyle g\) if we split the solution into two parts. One will be the transient state while the other will be the steady state. We will choose \(\displaystyle w\) and \(\displaystyle v\) for the states respectively. The steady state function, \(\displaystyle v\), will take care of the inhomogeneous part which in our case is \(\displaystyle g\). The solution will look like this:

\(\displaystyle u(x,t) = w(x,t) + v(x) \ \ \ \ \ \ \ \ \ \ (1)\)

Plugging this into the original differential equation, we get:

\(\displaystyle w_{tt} = c^2(w_{xx} + v_{xx}) - g\)

With a little simplification, we get:

\(\displaystyle w_{tt} = c^2w_{xx} + c^2v_{xx} - g\)

The idea at this point is to force \(\displaystyle c^2v_{xx} - g = 0\) to get a homogenous transient equation \(\displaystyle w_{tt} = c^2w_{xx}\).

When we apply the boundary and initial conditions to \(\displaystyle (1)\), we get:

\(\displaystyle u(0,t) = w(0,t) + v(0) = 0\)
\(\displaystyle u(L,t) = w(L,t) + v(L) = 0\)
\(\displaystyle u(x,0) = w(x,0) + v(x) = 0\)
\(\displaystyle u_t(x,0) = w_t(x,0) = 0\)

Finally, we can set the two equations to solve based on the conditions above.


\(\displaystyle w_{tt} = c^2w_{xx} \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq L, \ \ \ t > 0, \ \ \ \ \ \ \ \ \ (2)\)

\(\displaystyle w(0,t) = 0,\)
\(\displaystyle w(L,t) = 0,\)
\(\displaystyle w(x,0) = -v(x),\)
\(\displaystyle w_t(x,0) = 0,\)




\(\displaystyle c^2v_{xx} - g = 0 \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq L \ \ \ \ \ \ \ \ \ \ (3)\)

\(\displaystyle v(0) = 0\)
\(\displaystyle v(L) = 0\)


As you can see, equation \(\displaystyle (2)\) is homogeneous with homogeneous boundary conditions (representing the transient state) while equation \(\displaystyle (3)\) is ordinary (representing the steady state) which can be solved straightforward.