FEM Method for the Wave Equation

In summary, the author is trying to understand how to apply the finite element method for a simple 1D wave equation with four elements. The author assumes that the user knows what the function ##f(x)## is, but if ##f(x)## is ##T(x)## then the equation becomes \begin{multline*}\frac{d^2T}{dx^2} + kT(x) = -f(x).\end{multline*} which can be solved by using the method shown in Numerical Methods for Engineers by Chapra and Canale.f
  • #1

bob012345

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How to set up the Finite Element Method of a 1D Wave Equation.
I am trying to understand how to apply the finite element method for a simple 1D wave equation with four elements for learning purposes.

$$\frac{d^2 T(x)}{dx^2} = -f(x)$$

I am stuck because the structure of the equations set up in Numerical Methods for Engineers by Chapra and Canale as shown here seems to assume one knows what the function ##f(x)## is. What if ##f(x)## is ##T(x)## as in the wave equation? Is that case doable? Thanks.

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  • #2
I assume [itex]N_i[/itex] are test functions with [itex]T[/itex] approximated by [itex]T_1N_1(x) + T_2N_2(x)[/itex]. If so, then [tex]\frac{d^2T}{dx^2} + kT(x) = -f(x)[/tex] where [itex]f[/itex] is a known source term gives [tex]
\begin{multline*}
\int_{x_1}^{x_2} N_i(x)\frac{d^2T}{dx^2} + kN_i(x)T(x) + N_i(x)f(x)\,dx \\=
\left[ N_i(x)\frac{dT}{dx}\right]_{x_1}^{x_2}
+\int_{x_1}^{x_2} -\frac{dN_i}{dx}\frac{dT}{dx} + kN_i(x)T(x) + N_i(x)f(x)\,dx
\end{multline*}[/tex] which results in [tex]
\begin{multline*}
\begin{pmatrix}
\int_{x_1}^{x_2} \left(\frac{dN_1}{dx}\right)^2\,dx & \int_{x_1}^{x_2} \frac{dN_1}{dx}\frac{dN_2}{dx} \,dx \\
\int_{x_1}^{x_2} \frac{dN_1}{dx}\frac{dN_2}{dx} \,dx & \int_{x_1}^{x_2} \left(\frac{dN_2}{dx}\right)^2\,dx
\end{pmatrix}
\begin{pmatrix} T_1 \\ T_2 \end{pmatrix} \\
- k\begin{pmatrix} \int_{x_1}^{x_2} N_1^2(x)\,dx & \int_{x_1}^{x_2} N_1(x)N_2(x)\,dx \\
\int_{x_1}^{x_2} N_1(x)N_2(x)\,dx & \int_{x_1}^{x_2} N_2^2(x)\,dx \end{pmatrix}
\begin{pmatrix} T_1 \\ T_2 \end{pmatrix} \\
=
\begin{pmatrix}
N_1(x_2) \frac{dT}{dx}(x_2) - N_1(x_1) \frac{dT}{dx}(x_1) \\
N_2(x_2) \frac{dT}{dx}(x_2) - N_2(x_1) \frac{dT}{dx}(x_1) \end{pmatrix} +
\begin{pmatrix}
\int_{x_1}^{x_2} N_1(x)f(x)\,dx \\
\int_{x_1}^{x_2} N_2(x)f(x)\,dx
\end{pmatrix}.\end{multline*}[/tex] I'm sure the authors will give an example of this at some point.

The wave equation is actually a PDE, so [itex]T_1[/itex] and [itex]T_2[/itex] are not constants but functions of time. You therefore end up with the system of ODEs [tex]
A\begin{pmatrix} \frac{d^2T_1}{dt^2} \\ \frac{d^2T_2}{dt^2} \end{pmatrix} = -c^2 B \begin{pmatrix} T_1 \\ T_2 \end{pmatrix} + (\mbox{boundary terms})[/tex] where [itex]B[/itex] is the stiffness matrix and [tex]
A = \begin{pmatrix} \int_{x_1}^{x_2} N_1^2(x)\,dx & \int_{x_1}^{x_2} N_1(x)N_2(x)\,dx \\
\int_{x_1}^{x_2} N_1(x)N_2(x)\,dx & \int_{x_1}^{x_2} N_2^2(x)\,dx \end{pmatrix}.[/tex]
 
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  • #3
I worked it out and it works great! Thanks. Here is a plot of the points calculated by the Finite Element Method as compared to the exact analytical solution ##\psi(x)= \sqrt{\frac{2}{10}}sin(\frac{\pi x}{10})## of an infinite potential well of width 10 units.;

desmos-graph (7).png
 
Last edited:
  • #4
What does the input function ##f(x)## represent in the case of a wave equation? I know from experience it is a constant and it is necessary to solve the linear equations. I also noticed it can be scaled to get the amplitude I want but I am having trouble justifying it and what value to give it.
 

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