A The boundary conditions in reference to Laplace's equation

[chwala]

TL;DR Summary

I am looking at Laplace equation. The beginning steps are pretty clear...kindly see attached... I need some insight on the given boundary conditions

We have inhomogenous dirichlet boundary conditions (well understood)....the laplace equation is a steady state equation and we can clearly see that in 2D..it will be defined by 4 boundary conditions and NO initial condition...having said that; kindly have a look at the continuation below...

I can follow the steps; In the attempt to solve; they made $3$ boundary conditions homogenous thereby leaving one inhomogenous boundary condition to work with to realize steps to solution...
Now my question is; Why the choice of $f_1(x)$ ? or it does not matter...could we as well have worked with $u(0,y)=g_1(y)$ and made the other $3$ conditions homogenous? The other steps are not difficult to follow....am on that now cheers!

 

[hutchphd]

TL;DR Summary: I am looking at Laplace equation. The beginning steps are pretty clear...kindly see attached... I need some insight on the given boundary conditions

Now my question is; Why the choice of f1(x) ?

The example simply chooses one of the edges to be a nonzero function. The choice is arbitrary and could be any edge (although a particular choice may make the notation simpler)

 

[chwala]

Ok I will explore this...looks like if we were to use $g_1(y)$ then I guess that we shall be dealing with solutions (both general and particular) in terms of $y$...

 

[chwala]

Hmmmmm...getting to understand this isn't a walk in the park...spent a whole 25 minutes trying to go through the problem...Will post my area of doubts/clarity later...

 

[pasmith]

Now my question is; Why the choice of $f_1(x)$ ? or it does not matter...could we as well have worked with $u(0,y)=g_1(y)$ and made the other $3$ conditions homogenous? The other steps are not difficult to follow....am on that now cheers!

The problem is linear; you can decompose it as the sun of four problems, each of which is homogenous on three sides and inhomogenous on the fourth.

 

[chwala]

I am still on this problem...i have a query on the highlighted part of the notes...can we use the alternative approach below?

We have,

...

$Y=C \cosh (λy) + D \sinh (λy)$ now on using the boundary condition $u(x,b)=0$

We can also get;

$0=C\cosh (λb)+0$

$0=C+ 0$

$⇒C=0$

Instead of the $C=-D\tanh (λb)$ given on the text. Would that also be correct?

It then follows that,

$⇒Y=D \sinh (λy)$

then the other steps may follow i.e
...
$U_n( x,y) = \sin \dfrac{nπx}{a} ⋅ \sinh \dfrac{nπy}{a}$

Just another question on the given general solution; why did they ignore/leave out the Constants B and D ?

We ought to have;

$U_n( x,y) = B\sin \dfrac{nπx}{a} ⋅ D \sinh \dfrac{nπy}{a}$

 

[pasmith]

I am still on this problem...i have a query on the highlighted part of the notes...can we use the alternative approach below?

We have,

...

$Y=C \cosh (λy) + D \sinh (λy)$ now on using the boundary condition $u(x,b)=0$

We can also get;

$0=C\cosh (λb)+0$

Why is D\sinh (\lambda b) zero?

$0=C+ 0$

$⇒C=0$

Instead of the $C=-D\tanh (λb)$ given on the text. Would that also be correct?

No. You need Y(0) \neq 0 in order to apply the boundary condition on y = 0. That means you need a non-zero coefficient of \cosh (\lambda y). What the text is saying is that <br /> \begin{split} <br /> \sinh(\lambda(b-y)) &amp;= \sinh(\lambda b)\cosh(\lambda y) - \cosh(\lambda b)\sinh(\lambda y) \\<br /> &amp;= \cosh(\lambda b)\left(\tanh(\lambda b)\cosh(\lambda y) - \sinh(\lambda y)\right) \end{split} is a solution of Y&#039;&#039; = \lambda^2 Y which satisfies Y(b) = 0 and Y(0) \neq 0 - and every non-zero multiple of that solution is also a solution.

It then follows that,

$⇒Y=D \sinh (λy)$

then the other steps may follow i.e
...
$U_n( x,y) = \sin \dfrac{nπx}{a} ⋅ \sinh \dfrac{nπy}{a}$

Just another question on the given general solution; why did they ignore/leave out the Constants B and D ?

We ought to have;

$U_n( x,y) = B\sin \dfrac{nπx}{a} ⋅ D \sinh \dfrac{nπy}{a}$

How do you determine the coefficients a_n in u(x,0) = \sum_{n} a_n U_n(x,0), and how does the value of U_n(x,0) \neq 0 affect the value of a_n?

 

[chwala]

Why is D\sinh (\lambda b) zero?
No. You need Y(0) \neq 0 in order to apply the boundary condition on y = 0. That means you need a non-zero coefficient of \cosh (\lambda y). What the text is saying is that <br /> \begin{split}<br /> \sinh(\lambda(b-y)) &amp;= \sinh(\lambda b)\cosh(\lambda y) - \cosh(\lambda b)\sinh(\lambda y) \\<br /> &amp;= \cosh(\lambda b)\left(\tanh(\lambda b)\cosh(\lambda y) - \sinh(\lambda y)\right) \end{split} is a solution of Y&#039;&#039; = \lambda^2 Y which satisfies Y(b) = 0 and Y(0) \neq 0 - and every non-zero multiple of that solution is also a solution.
How do you determine the coefficients a_n in u(x,0) = \sum_{n} a_n U_n(x,0), and how does the value of U_n(x,0) \neq 0 affect the value of a_n?

'Why is D\sinh (\lambda b) zero?'

Response;

I used the boundary condition $u(x,b)=0$ on

$Y=C \cosh (λy) + D \sinh (λy)$ to give me

$0=C \cosh (λb) + D \sinh (λb)$

$0=C\cosh (λb)+0$

$0=C+ 0$

$⇒C=0$

It then follows that,

$Y=D \sinh (λy)$

when $b=0$ then it follows that $\sinh (λb)=0$ and $\cosh (λb)=1$ .

Let me know if this reasoning is correct. I used that in arriving at my general solution as indicated above...

Kindly note that the difference in my approach and the text is on $y$ and the $(y-b)$...

$U_n( x,y) = \sin \dfrac{nπx}{a} ⋅ \sinh \dfrac{nπ(y-b)}{a}$ [text approach]

$U_n( x,y) = \sin \dfrac{nπx}{a} ⋅ \sinh \dfrac{nπy}{a}$ [My approach]

 

[chwala]

Why is D\sinh (\lambda b) zero?
No. You need Y(0) \neq 0 in order to apply the boundary condition on y = 0. That means you need a non-zero coefficient of \cosh (\lambda y). What the text is saying is that <br /> \begin{split}<br /> \sinh(\lambda(b-y)) &amp;= \sinh(\lambda b)\cosh(\lambda y) - \cosh(\lambda b)\sinh(\lambda y) \\<br /> &amp;= \cosh(\lambda b)\left(\tanh(\lambda b)\cosh(\lambda y) - \sinh(\lambda y)\right) \end{split} is a solution of Y&#039;&#039; = \lambda^2 Y which satisfies Y(b) = 0 and Y(0) \neq 0 - and every non-zero multiple of that solution is also a solution.
How do you determine the coefficients a_n in u(x,0) = \sum_{n} a_n U_n(x,0), and how does the value of U_n(x,0) \neq 0 affect the value of a_n?

If i am getting you right, we are going to use the inhomogenous boundary condition to determine the coefficient $a_n$.

My question is in reference to the step before applying the inhomogenous boundary condition...i can follow the steps quite well...i just want to know if the reason behind ignoring the coefficents $B$ and $D$ in

$U_n( x,y) = B\sin \dfrac{nπx}{a} ⋅ D \sinh \dfrac{nπ(y-b)}{a}$

is due to the fact that they are inconsequential...or they will be taken care when determining value of $a_n$. Cheers.

 

[pasmith]

'Why is D\sinh (\lambda b) zero?'

Response;

I used the boundary condition $u(x,b)=0$ on

$Y=C \cosh (λy) + D \sinh (λy)$ to give me

$0=C \cosh (λb) + D \sinh (λb)$

$0=C\cosh (λb)+0$

But you can't have D\sinh(\lambda b) = 0 withouh at least one of D, \lambda, and b being zero. We know that b \neq 0, and by assumption \lambda &gt; 0; therefore you must have D = 0. So when you subsequently conclude that C = 0 you in fact have Y \equiv 0 which is not allowed.

 

[chwala]

But you can't have D\sinh(\lambda b) = 0 withouh at least one of D, \lambda, and b being zero. We know that b \neq 0, and by assumption \lambda &gt; 0; therefore you must have D = 0. So when you subsequently conclude that C = 0 you in fact have Y \equiv 0 which is not allowed.

Thanks, let me look at this again slowly and in detail...fun going through this...though a bit confusing...

...i noted the use of the homogenous boundary conditions in this particular order $u(0,y), u(a,y),$ that were applied on $X$ and $u(x,b), u(a,y)$ that were applied on $Y$.

I will need to go all over again then finish with the last part that will make use of the inhomogenous boundary condition. Cheers mate.

 

[chwala]

But you can't have D\sinh(\lambda b) = 0 withouh at least one of D, \lambda, and b being zero. We know that b \neq 0, and by assumption \lambda &gt; 0; therefore you must have D = 0. So when you subsequently conclude that C = 0 you in fact have Y \equiv 0 which is not allowed.

...just confirm i thought we have $Y(b)=0$ thus $b$ can indeed be equal to $0$... or am i not getting it right?

 

[pasmith]

...just confirm i thought we have $Y(b)=0$ thus $b$ can indeed be equal to $0$... or am i not getting it right?

b is fixed by the problem: it's part of the definition of the domain of u. You don't get to choose it to your advantage.

 

[chwala]

b is fixed by the problem: it's part of the definition of the domain of u. You don't get to choose it to your advantage.

If you look at the text; they have indicated that

$Y=C \cosh (λy) + D \sinh (λy)$ can match $Y(b)=0$. I just followed the text and did not choose it...not unless my interpretation is wrong.

 

[pasmith]

If you look at the text; they have indicated that

$Y=C \cosh (λy) + D \sinh (λy)$ can match $Y(b)=0$. I just followed the text and did not choose it...not unless my interpretation is wrong.

I don't see that in the text. Do you have an equation number reference for it?

 

[chwala]

I don't see that in the text. Do you have an equation number reference for it?

Check on page $4$ of the attached pdf.

 

[pasmith]

I see

Y&#039;&#039; - \lambda^2 Y = 0 \Rightarrow Y(y) = C\cosh \lambda y + D\sinh \lambda y can only match Y(0) = 0 or Y(b) = 0

and

Y&#039;&#039; + \lambda^2 Y = 0 \Rightarrow Y(y) = C\cos \lambda y + D\sin \lambda y can match Y(0) = 0 = Y(b)

We are looking at the first case.

 

[chwala]

I am still on this problem...i have a query on the highlighted part of the notes...can we use the alternative approach below?

We have,

...

$Y=C \cosh (λy) + D \sinh (λy)$ now on using the boundary condition $u(x,b)=0$

We can also get;

$0=C\cosh (λb)+0$

$0=C+ 0$

$⇒C=0$

Instead of the $C=-D\tanh (λb)$ given on the text. Would that also be correct?

It then follows that,

$⇒Y=D \sinh (λy)$

then the other steps may follow i.e
...
$U_n( x,y) = \sin \dfrac{nπx}{a} ⋅ \sinh \dfrac{nπy}{a}$

Just another question on the given general solution; why did they ignore/leave out the Constants B and D ?

We ought to have;

$U_n( x,y) = B\sin \dfrac{nπx}{a} ⋅ D \sinh \dfrac{nπy}{a}$

This was wrong thinking on my part! I am on it now... actually before this step we have;

...

$u(x,b)=X(x)⋅Y(b)=0$, that is from the given boundary condition; $u(x,b)=0$, therefore, it follows that $Y(b)=0$.

Thus we shall have;

$y(b)=C \cosh(λb)+D \sinh(λb)$

$0=C \cosh(λb)+D \sinh(λb)$

$⇒C=-D\tanh (λb)$ ...

My point is that i was wrong by stating that $C=0$.

Cheers

 

[chwala]

I am conversant now with solving this problem at ease; just need some clarification on the highlighted part;

on the integral part... looks like something to do with Fourier?...or something else...cheers mate!

 

[pasmith]

Yes, it is a fourier series.

 

[chwala]

Thanks @pasmith happy festivities mate!

 

[chwala]

The example simply chooses one of the edges to be a nonzero function. The choice is arbitrary and could be any edge (although a particular choice may make the notation simpler)

Just to be absolutely clear on this; You said we can choose to make any of the boundary conditions: $u(x,0), u(x,b), u(a,y)$ and $u(0,y)$ to be inhomogenous?

 

[chwala]

I am conversant now with solving this problem at ease; just need some clarification on the highlighted part;

View attachment 319260

View attachment 319261

on the integral part... looks like something to do with Fourier?...or something else...cheers mate!

Yeah now i understand this; from fourier sine series;

$$B_n=\dfrac{1}{a}\int_{-a}^a f_1(x) \sin (\dfrac{nπx}{a})dx=\dfrac{2}{a}\int_0^a f_1(x) \sin (\dfrac{nπx}{a})dx$$ with $n=1,2.3,...$

Am looking at the Fourier series now (for both $\sin$ and $\cos$ ) -am refreshing actually...they are pretty easy! only thing challenging is the integration of parts-sometimes 2 to 3 times- phew and dealing with the constants ...here and there... otherwise, they're quite easy!