Understanding PDE Solutions: N=0 vs N≥1 Cases

[member 428835]

Hey PF!

I have a quick question. When I was solving a PDE via separation of variables, I was able to come up with a same format solution for $n \geq 1$ but when $n=0$ I had a different "type" of solution. This doesn't really bother me since I am dealing with a linear PDE. However, I matched my last boundary condition $u(a, \theta) = f(\theta)$ where $u=u(r, \theta)$. I satisfied this boundary condition only for the equation when $n \geq 1$. So why do I even bother with the $n=0$ case?

I can give more details specific to the problem if you want.

Thanks!

Josh

 

[OldEngr63]

A little more might help.

 

[member 428835]

The problem is $\nabla^2 u(r,\theta)=0$. and $u(a,\theta) = f(\theta)$ and $u(b,\theta) = g(\theta)$. Let $u(r,\theta) = R(r)T(\theta)$. This gives rise to the following: $$\frac{r^2R''}{R} + \frac{rR'}{R} = -\frac{T''}{T} = -\lambda \neq 0$$. (The case where $\lambda = 0$ is discussed below).This gives rise to the ODE $$r^2R''+rR+\lambda R = 0$$ where we guess solutions of the form $R(r) = ar^n + br^{-n} : n \geq 1$ Then solving for $T$ (different ODE) we arrive at the following solution: $$u(r, \theta)= \sum_{n=1} A_n \sin(n \theta)a^n+B_n \sin(n \theta)a^{-n}+C_n \cos(n \theta)a^{n}+D_n \cos(n \theta)a^{-n}:n\geq 1$$
I match this using orthogonality of sines and cosines to the BC listed above and solve for the constants. However, I have not addressed the $\lambda = 0$ case, which reduces the original ODE's from above. My question is, why do I need to worry about $n=0$ if I have satisfied the BC's and the PDE?

 

[OldEngr63]

I'm not sure you do need to worry about n = 0. What would n = 0 mean in terms of its contribution to the solution?

Looks like for n = 0, the R(r) solution would just be a single constant. Then the contribution to the overall solution looks like an added constant term. I don't see why this would be necessary if you have already satisfied the BCs.

 

[member 428835]

I'm not sure you do need to worry about n = 0. What would n = 0 mean in terms of its contribution to the solution?

This is totally my question too.

Looks like for n = 0, the R(r) solution would just be a single constant. Then the contribution to the overall solution looks like an added constant term. I don't see why this would be necessary if you have already satisfied the BCs.

The solution I get is not a constant for $R(r)$ but is actually $R(r) = a+b \ln(r)$. Looking above I wrote the ODE incorrectly after the PDE. It should be $r^2 R'' + r R' + \lambda R = 0$.

At any rate, if the $n=0$ case doesn't really help, then why do we ever worry about it? For example, suppose $u(x,t)$ solves the heat equation where we find $u(x,t) = \sum_{n=0} a_n \cos(n \pi x / L) e^{-k n^2 \pi^2 t / L}$ and a boundary condition is $u(x,0) = f(x)$. Applying $u$ to this boundary condition we have $A_n = 2/L \int_0^L f(x) \cos(n \pi x / L) dx : n \neq 0$. Why do we need to worry about $n=0$ here, which obviously gives another $A_0$? Why can't we start anywhere in the integers, say at $n=57$? Like what if $f(x) = e^x$ and I started at $n=4$?

Please help me with what I am not understanding. Thanks a ton

Josh

 

[member 428835]

Can anyone help me out here? Pleeeeease :)

 

[Chestermiller]

Can anyone help me out here? Pleeeeease :)

Hi Josh. How's it goin' buddy?
There are some problems with your solution.

Let me first guess: f and g are supposed to be periodic in theta, correct?

In post #3, there should be a + sign in front of the lambda.
I also note that the equation for u as a function of r and theta should have r's in it rather than a's.
Also, the n= 0 term cannot be omitted, and you can't satisfy the boundary conditions without it.

That's enough hints for now.

Chet

 

[member 428835]

Hi Josh. How's it goin' buddy?
There are some problems with your solution.

Let me first guess: f and g are supposed to be periodic in theta, correct?

In post #3, there should be a + sign in front of the lambda.
I also note that the equation for u as a function of r and theta should have r's in it rather than a's.
Also, the n= 0 term cannot be omitted, and you can't satisfy the boundary conditions without it.

That's enough hints for now.

Chet

Hi Chet!

Shoot, you're totally right, the $a$ definitely should be $r$ in the main solution. And definitely $f$ and $g$s are periodic, a period being from $-\pi,\pi$ as are their first derivatives. A positive sign in front of $\lambda$? are you meaning in the PDE when we separate? If so, I think the $-\lambda$ is okay; you don't?

How do we know the $n=0$ term can't be omitted? I'll work more on this and see what I get and then post it. Thanks!

 

[Chestermiller]

Suppose f and g are constant. Then what do you get?

Chet

 

[member 428835]

Suppose f and g are constant. Then what do you get?

Chet

I see what you mean! So, when solving this PDE we have
$$u(r, \theta)= \sum_{n=1} A_n \sin(n \theta)r^n+B_n \sin(n \theta)r^{-n}+C_n \cos(n \theta)r^{n}+D_n \cos(n \theta)r^{-n}:n\geq 1$$
Which is true for $n \geq 1$. However, $n=0 \implies \lambda = 0$ which reduces our two ODE's to $T''(\theta) = 0 \implies T = c_1 \theta + c_2$ yet $T(-\pi)=T(\pi) \implies c_1=0$ and thus $T=const.$

However, our ODE in $r$ becomes $r^2R''(r) + rR'(r)=0 \implies R(r) = c_1 \ln r + c_2$. Thus, when taking $T(\theta)R(r)$ for the $n=0$ situation we have simply $u(r,\theta) = c_1 \ln r + c_2$.

Understanding this, our overall solution for $u(r, \theta)$ for all possible $n \geq 0$ will be $$u(r, \theta)= c_1 \ln r + c_2 + \sum_{n=1} A_n \sin(n \theta)r^n+B_n \sin(n \theta)r^{-n}+C_n \cos(n \theta)r^{n}+D_n \cos(n \theta)r^{-n}$$

Do you agree with this so far Chet?

 

[Chestermiller]

I'm looking at this on my iPhone, so I can't see the full equation, but what I do see looks correct. Now, one more question for you: if u is periodic with Θ, what kind of constraint does this place on n?

Chet

 

[member 428835]

I'm looking at this on my iPhone, so I can't see the full equation, but what I do see looks correct. Now, one more question for you: if u is periodic with Θ, what kind of constraint does this place on n?

Chet

I have no idea...I'm definitely not sure. Any clues?

And on a side note, how do I solve for $C_n$ and $D_n$? I know to multiply the equation by $\cos (m \theta)$ and that the orthogonality will clear the sine terms, but what about the natural logarithm? How do I deal with that?

 

[Chestermiller]

I have no idea...I'm definitely not sure. Any clues?

Please disregard this question. It was my mistake. Sorry.

And on a side note, how do I solve for $C_n$ and $D_n$? I know to multiply the equation by $\cos (m \theta)$ and that the orthogonality will clear the sine terms, but what about the natural logarithm? How do I deal with that?

Just integrate the whole equation with respect to theta, with no sine or cosine weighting.
Chet

 

[LCKurtz]

Hey PF!

I have a quick question. When I was solving a PDE via separation of variables, I was able to come up with a same format solution for $n \geq 1$ but when $n=0$ I had a different "type" of solution. This doesn't really bother me since I am dealing with a linear PDE. However, I matched my last boundary condition $u(a, \theta) = f(\theta)$ where $u=u(r, \theta)$. I satisfied this boundary condition only for the equation when $n \geq 1$. So why do I even bother with the $n=0$ case?

I can give more details specific to the problem if you want.

Thanks!

Josh

Generally after separation of variables you apply the BC's to get the eigenfunctions which are frequently either sines or cosines. Typically if you get only cosines, the Fourier series expansion to fit the last condition might need the constant term $a_0$ which comes from $n=0$ and which itself comes from the eigenvalue $\lambda = 0$.

[Edit] Whoa! Somehow I missed all these replies; I just saw the first reply.

 

[member 428835]

Just integrate the whole equation with respect to theta, with no sine or cosine weighting.
Chet

Ohhhhhhh! OK! So, to summarize, to find $A_n$ and $B_n$ and thus satisfy the Boundary Conditions, multiply the equation by $\sin (m \theta)$ and integrate w.r.t $\theta$ over $[-\pi,\pi]$. Then we use the other BC (since we have two) and that gives us 2 equations and 2 unknowns (Cramer's Rule maybe?).

To find $C_n$ and $D_n$ multiply the equation by $\cos (m \theta)$ and integrate w.r.t $\theta$ over $[-\pi,\pi]$. Then we use both BC 2 equations and 2 unknowns. (just like for sine)

To find $c_1$ and $c_2$ not to be confused with $C_n$ (sorry) we simply let the orthogonality of sine and cosine take care of themselves, and thus use both BC's and integrate w.r.t $\theta$, again over $[-\pi,\pi]$.

Thanks you SOOOO MUCH! And please let me know if this final post is correct.

 

[Chestermiller]

Ohhhhhhh! OK! So, to summarize, to find $A_n$ and $B_n$ and thus satisfy the Boundary Conditions, multiply the equation by $\sin (m \theta)$ and integrate w.r.t $\theta$ over $[-\pi,\pi]$. Then we use the other BC (since we have two) and that gives us 2 equations and 2 unknowns (Cramer's Rule maybe?).

To find $C_n$ and $D_n$ multiply the equation by $\cos (m \theta)$ and integrate w.r.t $\theta$ over $[-\pi,\pi]$. Then we use both BC 2 equations and 2 unknowns. (just like for sine)

To find $c_1$ and $c_2$ not to be confused with $C_n$ (sorry) we simply let the orthogonality of sine and cosine take care of themselves, and thus use both BC's and integrate w.r.t $\theta$, again over $[-\pi,\pi]$.

Thanks you SOOOO MUCH! And please let me know if this final post is correct.

Yes. Have you not learned about Fourier Series yet?

Chet

 

[member 428835]

No, we have. I just spaced integrating the whole thing w.r.t $\theta$ to find $c_1$ and $c_2$. Also, all of my friends were solving this problem differently (just taking formulas and plugging and chugging) and then looking at the answer in the back of the book. This approach is much more enlightening. I feel dumb not thinking about this integration now.

Thanks Chet!