Does v1(x,t) converge for x=0?

[Jamin2112]

Homework Statement

Currently working on a heat equation problem for an applied math class. I got a solution

v(x,t) = Ʃ_n≥1(2/n∏)sin(n∏x)cos(n∏ct), x ε [0,1], t≥0,​

which seems wrong since, for any fixed x and t, v(x,t)=∞.

Homework Equations

Governing equation is u_tt=c²u_xx, where c is the speed of sound and u(x,t) is the longitudinal pressure distribution along a tube that is open at x=0 and has a rigid surface at x=1. The boundary condition at x=0 is u(0,t)=g(t)=cos(ωt) where ω is the frequency of the emitted sound. On the other hand, the rigid surface at x=1 implies a vanishing pressure at that point, and thus the boundary condition u(1,t)=0. Assume the initial conditions of u_tt=c²u_xx are both zero.

The Attempt at a Solution

I'm supposed to "Use the transformation u(x,t) = v(x,t) + ψ(x)g(t) to homogenize the boundary conditions by enforcing linearity on ψ." Then I'm supposed to "Solve for the homogeneous solution v₁(x,t) of the transformed equation."

I've checked over my work several times and I keep getting

v₁(x,t) = Ʃ_n≥1(2/n∏)sin(n∏x)cos(n∏ct),
​

what has to be wrong.

Thoughts? Concerns?

 

[clamtrox]

I don't think that's a problem at all. You see, the solution you have there is a superposition of all possible solutions, and since you don't have a momentum cutoff in your system, there are infinitely many modes that can live in your system in the ultraviolet (short wavelength) limit. This is quite typical for physical problems, especially in quantum field theory, where most of your time gets spent on trying to talk these infinities down.

 

[Jamin2112]

Cool!

 

[Jamin2112]

I have a question on my next homework problem. The setup is:

u_tt = u_xx
u(0,t)=0
u_t(x,0)=0
u(x,0)=xe^-x2.

I used separation of variables u(x,t)=X(x)T(t), so that we have XT'' = X''T = k. In the case of k > 0 and k = 0, when I attempted to solve for u I reached an inconsistency with u(x,0)=xe^-x2. In the case where k = -λ² < 0, I have found it impossible to determine the eigenvalue λ. We have X(x) = Asin(λx) + Bcos(λx) and T(t) = Csin(λt) + Dcos(λt). Since u_t(x,0) = 0, we need T'(0) = -λC = 0 if we want a nontrivial solution; hence u(x,t) = [Asin(λx) + Bcos(λx)]cos(λt). Then u(0,t) = Bcos(λx)cos(λt) = 0 implies B = 0; hence u(x,t) = Asin(λx)cos(λt). The last thing we could use would be u(0,t) = 0, but this gives us no information.

What can I do?

 

[clamtrox]

You need to use the full solution - you certainly can't have u = x*exp(-x^2) with just one sine and cosine term!
X(x) = \sum_\lambda A_{\lambda} \sin(\lambda x)
T(t) = \sum_\lambda D_{\lambda} \cos(\lambda t)

A Fourier series of x*exp(-x^2) might be useful next!

 

[HallsofIvy]

Homework Statement

Currently working on a heat equation problem for an applied math class. I got a solution

v(x,t) = Ʃ_n≥1(2/n∏)sin(n∏x)cos(n∏ct), x ε [0,1], t≥0,​

which seems wrong since, for any fixed x and t, v(x,t)=∞.

It is quite obvious that this does converge for x= 0.

Homework Equations

Governing equation is u_tt=c²u_xx, where c is the speed of sound and u(x,t) is the longitudinal pressure distribution along a tube that is open at x=0 and has a rigid surface at x=1. The boundary condition at x=0 is u(0,t)=g(t)=cos(ωt) where ω is the frequency of the emitted sound. On the other hand, the rigid surface at x=1 implies a vanishing pressure at that point, and thus the boundary condition u(1,t)=0. Assume the initial conditions of u_tt=c²u_xx are both zero.

The Attempt at a Solution

I'm supposed to "Use the transformation u(x,t) = v(x,t) + ψ(x)g(t) to homogenize the boundary conditions by enforcing linearity on ψ." Then I'm supposed to "Solve for the homogeneous solution v₁(x,t) of the transformed equation."

I've checked over my work several times and I keep getting

v₁(x,t) = Ʃ_n≥1(2/n∏)sin(n∏x)cos(n∏ct),
​

what has to be wrong.

Thoughts? Concerns?