PDE - Two Dimensional Wave Equation

[erok81]

Homework Statement

Solve the boundary value problem (1)-(3) with a=b=1, c=1/Π

$$f(x)=sin(3 \pi x) sin(\pi y),g(x)=0$$

$$(1)\frac{\partial^{2}u}{\partial t^{2}}=c^{2}\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right)$$ 0 < x < a, 0< y <b, t > 0

(2) u(0,y,t)=0 and u(a,y,t)=0 for 0 ≤ y ≤ b, t ≥ 0
(2) u(x,0,t)=0 and u(x,b,t)=0 for 0 ≤ y ≤ b, t ≥ 0

(3) u(x,y,0)=f(x,y) and u_t(x,y,0)=g(x,y)

Homework Equations

$$B_{mn}=\frac{4}{ab}\int^{1}_{0} \int^{1}_{0}f(x,y)sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy$$

$$B^{*}_{mn}=\frac{4}{ab \lambda_{mn}}\int^{1}_{0} \int^{1}_{0} g(x,y)sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy$$

$$\lambda_{mn}=c \pi \sqrt{\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}$$

There is one more to combine it all, but I'm not there yet.

The Attempt at a Solution

So we had a huge storm here and long story short, I missed class when we went over this. Which sucks because I have no idea how to do these. I am fine deriving most of up until actually finding values. Here is where I am...

I started finding the B_mn's first.

$$B^{*}_{mn}=\frac{4}{ab \lambda_{mn}}\int^{1}_{0} \int^{1}_{0} 0*sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy$$

Since g(x,y)=0, this should also equal zero.

$$B_{mn}=\frac{4}{ab}\int^{1}_{0} \int^{1}_{0}sin(3 \pi x) sin(\pi y) sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy$$

This gave me a solution of...

$$B_{mn}=\frac{12sin(\pi n)sin(\pi m)}{\pi^{2} (-9n^{2}+9+m^{2}n^{2}-m^{2}}$$

This will give me zero as well, except when m=n=0 (which I just realized typing this out.

Next up I have λ_mn=√(m²+n²)

So first, how do I represent that B value when it's only valid when n=m=0?

The last part of the problem is combining everything. The model is:

$$\sum^{\infty}_{n=1} \sum^{\infty}_{m=1}\left(B_{mn} cos\lambda_{mn}t+B^{*}_{mn}sin\lambda_{mn}t)sin(m \pi x) sin(n \pi y)$$

So...I'm not how to represent my answer as I mentioned above.

The book shows an answer of u(x,y,t)=sin3πx sin πy cos√10t

I definitely don't see how to get here. Mainly because the sum terms go away. Could that be because it's only valid when n=m=0? Also, the model contains m and n's but the final answer doesn't. The only way I can see they got values for those is from the given f(x). In that case m=3 and n=1 - which matches the book. Is that correct way for the last part?

 

[LCKurtz]

I haven't checked all the missing steps, but the short answer to your question is this.

If you put t = 0 in your solution you get

$$u(x,y,0)=\sum^{\infty}_{n=1} \sum^{\infty}_{m=1}\left(B_{mn}^* )sin(m \pi x) sin(n \pi y) = f(x,y) = \sin(3 \pi x) \sin(\pi y)$$

Notice that the term on the right is one of the terms on the left, namely when m = 3 and n = 1. That tells you B*₃₁ = 1 and the others are all zero. Then do a similar thing with the initial condition on u_t.

 

[erok81]

All of the missing steps are correct. I tried deriving it all to make sure I could. But everything - with the exception of my B_mn solution since I found that.

Now I see what you are doing with t = 0. Is that how solves for the mn's? If not, all I can see that giving me is just to affirm my givens.

On another note: How would I express my B_mn how it is only valid when m=n=0

 

[LCKurtz]

All of the missing steps are correct. I tried deriving it all to make sure I could. But everything - with the exception of my B_mn solution since I found that.

Now I see what you are doing with t = 0. Is that how solves for the mn's? If not, all I can see that giving me is just to affirm my givens.

On another note: How would I express my B_mn how it is only valid when m=n=0

It is simple to get the Fourier coefficients on this problem because your f(x,y) is already a finite Fourier series. So most of the B's are zero and you just get one term.

Generally you have to solve for the coefficients using the double Fourier series formulas for the coefficients. Your text probably talks about this or you can look here for example:

http://planetmath.org/encyclopedia/FourierSineAndCosineSeries.html

 

[erok81]

Thanks, that link helped.

I guess the only part I am hung on up on how this one ends with an non-summation answer. All of the 1D wave equation problems I did all ended with some sort of summation series keeping n/m in tact. This one is u(x,y,t)=sin3πx sin πy cos√10t.
EDIT
Disregard what I said above. The answer in the book isn't the full answer. That's why the summation is missing.

There are really only two things I don't get. Solving for the m and n. Do I just use the IC like you did above? But then does that only apply to some of the m/n's? If they were all computed, that would make the summation pointless.

And then the legit values for when m=n=0. My B_mn is zero at all values except when they equal zero. I don't see anywhere in my text or that link on how to resolve this.

 

[erok81]

Oh wow. This is what happens when you study all day.

My B_mn is always zero. Not just when m=n=0.

So now my only problem is my solution is zero since both B's are zero.

 

[LCKurtz]

Oh wow. This is what happens when you study all day.

My B_mn is always zero. Not just when m=n=0.

So now my only problem is my solution is zero since both B's are zero.

They aren't all zero. Read reply #2 again.

 

[erok81]

For my B term I have these as solutions. Are these incorrect?

$$B_{mn}=\frac{12sin(\pi n)sin(\pi m)}{\pi^{2} (-9n^{2}+9+m^{2}n^{2}-m^{2}}$$

$$B^{*}_{mn}=\frac{4}{ab \lambda_{mn}}\int^{1}_{0} \int^{1}_{0} 0*sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy = 0$$

I see the reply stating I should end up with one term. I don't see how though. Whatever value I use I get zero.

 

[LCKurtz]

I haven't checked all the missing steps, but the short answer to your question is this.

If you put t = 0 in your solution you get

$$u(x,y,0)=\sum^{\infty}_{n=1} \sum^{\infty}_{m=1}\left(B_{mn}^* )sin(m \pi x) sin(n \pi y) = f(x,y) = \sin(3 \pi x) \sin(\pi y)$$

Notice that the term on the right is one of the terms on the left, namely when m = 3 and n = 1. That tells you B*₃₁ = 1 and the others are all zero. Then do a similar thing with the initial condition on u_t.

[Edit] I see a mistake in the above quote of post #2. It should read if you put t = 0 you get

$$u(x,y,0)=\sum^{\infty}_{n=1} \sum^{\infty}_{m=1}\left(B_{mn} )sin(m \pi x) sin(n \pi y) = f(x,y) = \sin(3 \pi x) \sin(\pi y)$$

with no * on the B. And the conclusion is that B₃₁ = 1, not B*₃₁ = 1.

For my B term I have these as solutions. Are these incorrect?

$$B_{mn}=\frac{12sin(\pi n)sin(\pi m)}{\pi^{2} (-9n^{2}+9+m^{2}n^{2}-m^{2}}$$

$$B^{*}_{mn}=\frac{4}{ab \lambda_{mn}}\int^{1}_{0} \int^{1}_{0} 0*sin\frac{m \pi x}{a}sin\frac{n \pi x}{b}dxdy = 0$$

I see the reply stating I should end up with one term. I don't see how though. Whatever value I use I get zero.

Your formula for B_mn doesn't work when m = 3 and n = 1. But you are missing the point about the corrected series for u(x,y,0) above. You don't have to use the formulas to figure out the B_mn. Just look at it and you will see that if B₃₁ = 1 and the other B's are zero you have equality. And since you have shown that the B^*s are all zero, you do have just that one term in your solution.