1-D Wave equation with mixed boundary conditions

In summary, the homework statement is to solve the following boundary and initial conditions: u_{x}(0,t) = 0, u(L,t) = 0. The attempted solution is to find u(x,t) by applying the Fourier cosine series to the initial condition u(x,0) = f(x)g(t).
  • #1
KEØM
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Homework Statement


Solve, [tex]u_{t} = u_{xx}c^{2}[/tex]

given the following boundary and initial conditions

[tex]u_{x}(0,t) = 0, u(L,t) = 0[/tex]

[tex]u(x,0) = f(x) , u_{t}(x,0) = g(x)[/tex]

Homework Equations



[tex]u(x,t) = F(x)G(t)[/tex]

The Attempt at a Solution


I solved it, I am just not sure if it is right.

[tex]u(x,t) = \sum_{n=1}^\infty(a_{n}cos(\lambda_{n}t) + b_{n}sin(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x)

, \lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c [/tex]

[tex]a_{n} = \frac{2}{L}\int_0^L f(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx,

b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx
[/tex]

Can someone please verify this for me?

Thanks in advance,
KEØM
 
Last edited:
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  • #2
Why does your formula for bn[/b] involve cosine rather than sine?
 
  • #3
Thanks for replying HallsofIvy.

Well to get [tex]b_{n}[/tex] I applied the initial condition

[tex]u_{t}(x,0) = g(x).[/tex]

So I first find [tex]u_{t}[/tex] which comes to

[tex]u_{t} = \sum_{n=1}^\infty(-a_{n}\lambda_{n}sin(\lambda_{n}t) + b_{n}\lambda_{n}cos(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x) [/tex]

now evaluating [tex]u_{t}[/tex] at t = 0 and setting it equal to g(x) we get,

[tex]u_{t}(x,0) = \sum_{n=1}^\infty b_{n}\lambda_{n}cos((n-\frac{1}{2})\frac{\pi}{L}x) = g(x) [/tex]

But that is just the Fourier cosine series of g(x) just with the extra [tex]\lambda_{n}[/tex] and with the zero term equal to zero

so [tex]b_{n} = \frac{2}{\lambda_{n}}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

where [tex]\lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c
[/tex]

[tex] \Rightarrow b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

Please let me know if I did something wrong.

Thanks again,
KEØM
 

FAQ: 1-D Wave equation with mixed boundary conditions

1. What is the 1-D Wave equation with mixed boundary conditions?

The 1-D Wave equation with mixed boundary conditions is a mathematical model that describes the behavior of waves in a one-dimensional space. It takes into account both the initial displacement and the boundary conditions of the wave.

2. What are mixed boundary conditions in the 1-D Wave equation?

Mixed boundary conditions in the 1-D Wave equation refer to a combination of fixed and free boundary conditions. This means that some points on the boundary are constrained while others are allowed to move freely.

3. How is the 1-D Wave equation with mixed boundary conditions solved?

The 1-D Wave equation with mixed boundary conditions can be solved using a variety of techniques, such as the method of separation of variables, the Laplace transform method, or numerical methods such as finite difference or finite element methods.

4. What are some real-world applications of the 1-D Wave equation with mixed boundary conditions?

The 1-D Wave equation with mixed boundary conditions has many applications in physics and engineering, such as modeling the behavior of sound waves in a tube with one end closed and one end open, studying the vibrations of a guitar string, or predicting the movement of seismic waves in the Earth's crust.

5. Are there any limitations to the 1-D Wave equation with mixed boundary conditions?

Like any mathematical model, the 1-D Wave equation with mixed boundary conditions has its limitations. It assumes that the medium through which the wave travels is continuous and homogeneous, and that the wave is propagating in a straight line. It also does not take into account any external forces acting on the system.

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