PDE: Sep of Vars Non homogenous BCs

[Saladsamurai]

Homework Statement

I have Laplace's equation that I need to solve. I was told that it can be solved by separtion of variables and that it should yield sinh and cosh solutions. As it stands, my current set of BCs are not homogeneous. So I need to find the proper way to assume my solution.

$$\frac{\partial^2{\phi}}{\partial{x}^2} + \frac{\partial^2{\phi}}{\partial{y}^2} = 0 \qquad(1)$$

Subject to the BCs:

$$\frac{\partial{\phi}}{\partial{y}}(x, y =-h) = 0\qquad(2a)$$

$$\frac{\partial{\phi}}{\partial{y}}(x, y =0) = -\cos(x - T)\qquad(2a)$$


I am not really sure how to start this. I tried assuming that

$$\phi (x,y) = F(x,y) + G(y)$$

and then working with the G(y) alone I thought I could apply the BCs on G(y) such that the superposition of F and G would give homogenous BCs on F(x,y) which could then be solved by separation.

The problem with that approach is that G(y) is linear i.e: G(y) = C₁y +C₂ and since my given BCs are both derivative BCs, I don't think I can ever recover C₂ (since it vanishes) to get the complete solution.

Any other ideas?

 

[Saladsamurai]

Hmmm... come to think of it, do I need more boundary conditions than 2? Do I actually need 4 BCs to solve this problem?

 

[fzero]

You should use regular separation of variables with $$\phi(x,y) = X(x)Y(y)$$. Your boundary conditions are consistent with a solution of this form. You don't need any additional boundary conditions.

 

[Saladsamurai]

Hi fzero!

Hmmm...that seems to be the same response I just got from my professor in an email. Unfortunately, I have never taken a clas in PDEs and I have only solved a handful of them by brute force. I could swear that in order to use separation, the BCs needed to be homogeneous. But I am also thinking that for the problems I have solved, I have always had a nonzero IC, so perhaps my nonzero BC can act like an IC? But I am still unclear as to why I don't need any more BCs?

 

[fzero]

Hi fzero!

Hmmm...that seems to be the same response I just got from my professor in an email. Unfortunately, I have never taken a clas in PDEs and I have only solved a handful of them by brute force. I could swear that in order to use separation, the BCs needed to be homogeneous. But I am also thinking that for the problems I have solved, I have always had a nonzero IC, so perhaps my nonzero BC can act like an IC? But I am still unclear as to why I don't need any more BCs?

The boundary conditions need to be compatible with the solutions of the equation. For example,

$$\frac{\partial{\phi}}{\partial{y}}(x, y =0) =x^4$$

would not be compatible, but the $$\cos$$ is, because there is a solution $$X(x)=\cos(x+x_0)$$.

Now, we obtain 2 integration constants from each 2nd order ODE, so, naively, we need initial conditions at a minimum of 4 points, which cannot all be at the same values of $$x$$ or $$y$$. The boundary conditions we're given actually specify the solution at an infinite number of points at two distinct values of $$y$$, so we appear to have enough information to fix all constants.

The above counting is a bit naive, since it usually turns out that the constant of separation becomes a discrete set of "eigenvalues." There are therefore infinite families of solutions and the general solution is an infinite series expansion in them, leading to an infinite number of coefficients that must be solved for. Nevertheless, it's typically the case that 4 points are enough to fix the solution through a Fourier transform.

The above discussion might be a little clearer after you try working through the problem.

 

[Saladsamurai]

The boundary conditions need to be compatible with the solutions of the equation. For example,

$$\frac{\partial{\phi}}{\partial{y}}(x, y =0) =x^4$$

would not be compatible, but the $$\cos$$ is, because there is a solution $$X(x)=\cos(x+x_0)$$.

Now, we obtain 2 integration constants from each 2nd order ODE, so, naively, we need initial conditions at a minimum of 4 points, which cannot all be at the same values of $$x$$ or $$y$$. The boundary conditions we're given actually specify the solution at an infinite number of points at two distinct values of $$y$$, so we appear to have enough information to fix all constants.

The above counting is a bit naive, since it usually turns out that the constant of separation becomes a discrete set of "eigenvalues." There are therefore infinite families of solutions and the general solution is an infinite series expansion in them, leading to an infinite number of coefficients that must be solved for. Nevertheless, it's typically the case that 4 points are enough to fix the solution through a Fourier transform.

The above discussion might be a little clearer after you try working through the problem.

Hi fzero! Yes. I am working through it now and I hope that the above does become clearer. For now, I have a quick question:

By assuming that φ(x,y) = F(x)G(y) I end up with two ODEs

(1/F) * F'' = - (1/G) * G'' = something constant

Now, I am working with G first:

- (1/G) * G'' = C

or

- G'' - C*G = 0

Now, I have alway been told what my separation constant is; I have never had to choose one and so I do not have any insight as to how to choose one? If I let C = -A I get one solution and if I let C = +A I will get another. Surely this cannot be. What am I missing conceptually? How can I choose my separation constant wisley?

 

[fzero]

Yes, you get different solutions if C is positive or negative. It turns out to be simpler to parameterize these cases as $$C= k^2$$ or $$C=-k^2$$. What you'll find is that only one class of solutions is compatible with the boundary conditions that you're given. The other class would be compatible with different boundary conditions.

 

[Saladsamurai]

Yes, you get different solutions if C is positive or negative. It turns out to be simpler to parameterize these cases as $$C= k^2$$ or $$C=-k^2$$. What you'll find is that only one class of solutions is compatible with the boundary conditions that you're given. The other class would be compatible with different boundary conditions.

Interesting. So with that in mind, I am thinking that since I have a BC of the form G(y=-h) = 0 that I should go with the case where C = + k^2 since this will reveal an oscillatory solution that has zeros, whereas the -k^2 case reveals exponentials which are never zero (on a finite domain).

Sound good?

 

[fzero]

Interesting. So with that in mind, I am thinking that since I have a BC of the form G(y=-h) = 0 that I should go with the case where C = + k^2 since this will reveal an oscillatory solution that has zeros, whereas the -k^2 case reveals exponentials which are never zero (on a finite domain).

Sound good?

If G is a trig function, then F is a hyperbolic trig function and vice versa. You should write down solutions for both cases and check both boundary conditions to see which case is needed. Also note that

$$\sinh(k(y+h)) = \frac{1}{2} \left( e^{k(y+h)} - e^{-k(y+h)}\right)$$

vanishes at $$y=-h$$.

 

[Saladsamurai]

Okay! I am ready to attack this one:

Let me start off by adding some new information that might help. This equation comes from a Fluid Mechanics problem involving Gravity Waves in Liquids where there is a free surface (i.e. air above the liquid).

The PDE I am trying to solve in (1) with BCs (2a,b) are Laplace's Equation for potential flow.

I know that the solution is:

$$\phi = -\frac{\cosh(y + h)}{\sinh h}\cos(x-T)\qquad (4)$$

from the text.

**********************************************************************************************
**********************************************************************************************

To recap: After assuming the product solution (x,y) = F(x)G(y) I end up with two ODEs given by:

$$-\frac{1}{F} * F'' = \frac{1}{G} * G'' = k^2 \qquad(5)$$

where I have moved the (-) sign to F and assumed a positive separation constant under the advisement of my professor (though he said he wasn't 100% on that).

************************************************** ********************************************
************************************************** ********************************************

From the LHS of (5): $-\frac{1}{F} * F'' = k^2$ we obtain from the characteristic equation r² + k² = 0 → r = {k*i , -k*i} so that

$$F(x) = C_1\cos(kx) + C_2\sin(kx) \qquad(6)$$From the RHS of (5): $\frac{1}{G} * G'' = k^2$ we obtain from the characteristic equation r² - k² = 0 → r = {k , -k} so that

$$G(y) = C_3e^{-ky} + C_4e^{ky} \qquad(7)$$

************************************************** ********************************************
************************************************** ********************************************

From the 2nd BC:

$$\frac{\partial{\phi}}{\partial{y}}(x, y =0) =F(x)*G'(y=0) = -\cos(x - T) \qquad(8)$$

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3 + C_4)] = -\cos(x - T) \qquad(9)$$

From the 1st BC:

$$\frac{\partial{\phi}}{\partial{y}}(x, y =-h) = F(x)*G'(y=-h)= 0 \qquad(10)$$

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3e^{kh} + C_4e^{-kh})] = 0 \qquad(11)$$************************************************** ********************************************
************************************************** ********************************************

This is as far as I have gotten. I was told to let k = 1, so I will do so for now and then hopefully I can come back later and show that there is a k = 1. So (9) and (11) become:$$\left [ C_1\cos(x) + C_2\sin(x) \right ]* (-C_3 + C_4) = -\cos(x - T) \qquad(9')$$

$$\left [ C_1\cos(x) + C_2\sin(x) \right ]* (-C_3e^{h} + C_4e^{-h}) = 0 \qquad(11')$$

And it was also suggested that the trig identity: cos(a - b) = cos(a)cos(b) + sin(a)sin(b) might help me to do something with (9') so that I could write

$$\left [ C_1\cos(x) + C_2\sin(x) \right ]* (-C_3 + C_4) = -\cos(x) \cos(T) - \sin(x)\sin(T) \qquad(12)$$

But I have not figured out the utility in that one just yet.

Any thoughts as to the next move?

 

[fzero]

From the 2nd BC:

$$\frac{\partial{\phi}}{\partial{y}}(x, y =0) =F(x)*G'(y=0) = -\cos(x - T) \qquad(8)$$

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3 + C_4)] = -\cos(x - T) \qquad(9)$$

From the 1st BC:

$$\frac{\partial{\phi}}{\partial{y}}(x, y =-h) = F(x)*G'(y=-h)= 0 \qquad(10)$$

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3e^{-kh} + C_4e^{kh})] = -\cos(x - T) \qquad(11)$$

The last equation should be

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3e^{kh} + C_4e^{-kh})] = 0 \qquad(11)$$

Note the signs in the exponents and the zero on the RHS. This equation can be used to solve for $$C_3,C_4$$. It's more convenient to solve for them in terms of a new coefficient $$C$$ instead of for one in terms of another, but you can do it either way.

This is as far as I have gotten. I was told to let k = 1, so I will do so for now and then hopefully I can come back later and show that there is a k = 1.

Yeah, it's unecessary to set k=1 because that follows immediately from (9). You can see this by multiplying both sides by $$\cos kx$$ and integrating over $$x$$ (say over half a period).

And it was also suggested that the trig identity: cos(a - b) = cos(a)cos(b) + sin(a)sin(b) might help me to do something with (9') so that I could write

$$\left [ C_1\cos(x) + C_2\sin(x) \right ]* (-C_3 + C_4) = -\cos(x) \cos(T) - \sin(x)\sin(T) \qquad(12)$$

But I have not figured out the utility in that one just yet.

Any thoughts as to the next move?

$$\sin$$ and $$\cos$$ are linearly independent (check by multiplying by one or the other and integrating), so we can use this equation to solve for the remaining unknown coefficients.

 

[Saladsamurai]

The last equation should be

$$\Rightarrow \left [ C_1\cos(kx) + C_2\sin(kx) \right ]* [k(-C_3e^{kh} + C_4e^{-kh})] = 0 \qquad(11)$$

Note the signs in the exponents and the zero on the RHS. This equation can be used to solve for $$C_3,C_4$$. It's more convenient to solve for them in terms of a new coefficient $$C$$ instead of for one in terms of another, but you can do it either way.

Hi fezero. I am not sure what I am noting about the signs of the exponents here? Nor am I seeing how to rewrite C1 and C2 as a new coefficient? Sorry, I don't see what you are alluding to?

 

[fzero]

Hi fezero. I am not sure what I am noting about the signs here? Nor am I seeing how to rewrite C1 and C2 as a new coefficient? Sorry, I don't see what you are alluding to?

From your (7)

$$\left. G'(y) \right|_{y=-h} = k\left( -C_3e^{kh} + C_4e^{-kh} \right),$$

but you had the opposite signs in the exponents in (11).

What I meant is that (11) is solved by

$$-C_3e^{kh} + C_4e^{-kh} = 0 .$$

We can just solve this for $$C_4$$ in terms of $$C_3$$, but the algebra later on turns out to be a bit simpler if we solve it via

$$C_3 = C e^{-kh}, ~ C_4 = C e^{kh}.$$

 

[Saladsamurai]

From your (7)

$$\left. G'(y) \right|_{y=-h} = k\left( -C_3e^{kh} + C_4e^{-kh} \right),$$

but you had the opposite signs in the exponents in (11).

Oops! Edited! Nice catch

What I meant is that (11) is solved by

$$-C_3e^{kh} + C_4e^{-kh} = 0 .$$

We can just solve this for $$C_4$$ in terms of $$C_3$$, but the algebra later on turns out to be a bit simpler if we solve it via

$$C_3 = C e^{-kh}, ~ C_4 = C e^{kh}.$$

I am not really following what you are doing there? Why can we use the same C for both equalities?

EDIT: Looks like you said

$C_3e^{kh} = C_4e^{-kh}\rightarrow C_3 = \frac{C_4}{e^{kh}}e^{-kh} = C e^{-kh}$ where we let $C=\frac{C_4}{e^{kh}}$ so that $C_3e^{kh} = Ce^{-kh}e^{kh} = C_4e^{kh}\rightarrow C_4 = Ce^{kh}$

So now we have that:

$$G(y) = C[e^{k(-y-h)} + e^{k(y+h)}] \qquad(13)$$

Is that as far as we take it with (13) ? I am going to solve for the C1 and C2 now, but do we ever solve for C explicitly?

 

[Saladsamurai]

Combining $C_3 = C e^{-kh}, ~ C_4 = C e^{kh}$ with
(12) creates a mess of products of C, C1 and C2. So I am thining I do need to solve for C explicitly somewhere?

 

[fzero]

:
So now we have that:

$$G(y) = C[e^{k(-y-h)} + e^{k(y+h)}] \qquad(13)$$

Is that as far as we take it with (13) ? I am going to solve for the C1 and C2 now, but do we ever solve for C explicitly?

You'll be able to solve for the products $$C C_1$$ and $$C C_2$$. The final answer will only depend on $$h$$ and $$T$$.

 

[Saladsamurai]

You'll be able to solve for the products $$C C_1$$ and $$C C_2$$. The final answer will only depend on $$h$$ and $$T$$.

OK then! Onward! Starting with (12) and $C_3 = C e^{-h}, ~ C_4 = C e^{h}$ (I am officially dropping the k's for consistency since my BC does not contain one)

$$\left [ C_1\cos(x) + C_2\sin(x) \right ]* (-C_3 + C_4) = -\cos(x) \cos(T) - \sin(x)\sin(T) \qquad(12)$$

$$\Rightarrow \left [ C_1\cos(x) + C_2\sin(x) \right ]* (C_3 - C_4) = \cos(x) \cos(T) + \sin(x)\sin(T)$$

$$\Rightarrow \left [ C_1\cos(x) + C_2\sin(x) \right ]* C(e^{-h} - e^{h}) = \cos(x) \cos(T) + \sin(x)\sin(T)$$

$$\Rightarrow \left C C_1\cos(x) + C C_2\sin(x) = \frac{\cos(T) }{e^{-h} - e^{h}}\cos(x) +\frac{\sin(T)}{e^{-h} - e^{h}}\sin(x)$$

$$\Rightarrow C C_1 = \frac{\cos(T) }{e^{-h} - e^{h}} \quad\text{ and }\quad C C_2 = \frac{\sin(T)}{e^{-h} - e^{h}} \qquad(14)$$

So now we have that F(x):

$$F(x) = \frac{1}{C} \left [ \frac{\cos(T)}{e^{-h} - e^{h}} \cos x + \frac{\sin(T)}{e^{-h} - e^{h}}\sin(x) \right ]\qquad(15)$$

How we looking? Looks like those C's will cancel when i form the product of F*G So

$$F*G = \frac{e^{-(y+h)} + e^{y+h}}{e^{-h} - e^h}\cos(x-T) = -\frac{\cosh(y + h)}{\sinh h}\cos(x-T)\qquad(16)$$

Woot! Now I just have to take the derivative a couple of times to get some velocity components and we're done! Thanks again fzero! I'll be posting back to discuss some of the details that came up and this regarding sign choices and what not. Maybe if you have time, you could chime in when I do.