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Eigenvalues and eigenfunctions

  • Thread starter ted12
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Homework Statement



Hello!

I don't know how to solve this problem: find eigenvalues and eigenfunctions of quadratic membrane which is fixed in three edges. Fourth edge is flexible bended in the middle (at this edge membrane is in the shape of triangular). Surface tension of membrane is γ, mass of edge is m.

Homework Equations



So I have u(x,y,t). Boundary conditions are: u(x,0,t)=u(x,a,t)=u(0,y,t)=0 and u(a,y,t)=k(t)*y for y < a/2 and u(a,y,t) = k(t)(a-y) for a/2 < y < a, where k(t)<<1 (because of that I don't have to account change of length of edge).

The Attempt at a Solution



Till now I put first three boundary conditons in solved wave equation and got u(x,y,t) = sin(k1*x)*sin(k2*y)*exp(-i*omega*t), where k2 = m*pi/a.
I don't know how to write 4. boundary condition (triangular) to get equation for k1.

If anybody can help me?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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6

Homework Statement



Hello!

I don't know how to solve this problem: find eigenvalues and eigenfunctions of quadratic membrane which is fixed in three edges. Fourth edge is flexible bended in the middle (at this edge membrane is in the shape of triangular). Surface tension of membrane is γ, mass of edge is m.

Homework Equations



So I have u(x,y,t). Boundary conditions are: u(x,0,t)=u(x,a,t)=u(0,y,t)=0 and u(a,y,t)=k(t)*y for y < a/2 and u(a,y,t) = k(t)(a-y) for a/2 < y < a, where k(t)<<1 (because of that I don't have to account change of length of edge).

The Attempt at a Solution



Till now I put first three boundary conditons in solved wave equation and got u(x,y,t) = sin(k1*x)*sin(k2*y)*exp(-i*omega*t), where k2 = m*pi/a.
I don't know how to write 4. boundary condition (triangular) to get equation for k1.

If anybody can help me?
Hi Ted12, welcome to PF!:smile:

Shouldn't you have a sum over all suitable values of [itex]m[/itex], and some factor representing the amplitude of each term? Something like:

[tex]u(x,y,t)=\sum_{m=1}^{\infty} A_m \sin(k_1x)\sin \left( \frac{m\pi y}{a} \right) e^{-i\omega t}[/tex]

Your 4th boundary condition then tells you that

[tex]\sum_{m=1}^{\infty} A_m \sin(k_1 a)\sin \left( \frac{m\pi y}{a} \right) e^{-i\omega t} = \left\{ \begin{array}{lr} k(t)y & \quad , 0<x<\frac{a}{2} \\ k(t)(a-y) & \quad , \frac{a}{2} < x < a \end{array} \right.[/tex]

What do you get if you multiply both sides of the equation by [itex]\sin \left( \frac{n\pi y}{a} \right) [/itex] and integrate from [itex]y=0[/itex] to [itex]y=a[/itex]?
 

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