Welcome to our community

Be a part of something great, join today!

[SOLVED] Boundary value problem for Laplace equation

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hello!!! (Wave)

Let $a,b>0$ and $D$ the rectangle $(0,a) \times (0,b)$. We consider the boundary value problem in $D$ for the Laplace equation, with Dirichlet boundary conditions,

$\left\{\begin{matrix}
u_{xx}+u_{yy}=0 & \text{ in } D,\\
u=h & \text{ in } \partial{D},
\end{matrix}\right.$

where $h:[0,a] \times [0,b] \to \mathbb{R}$ given function.

Supposing that $h$ is equal to $0$ at the vertices of the rectangle, I want to prove that the solution of the problem is the sum of the solutions of four respective problems, with homogeneous Dirichlet boundary conditions in three sides of the rectangle.


If we suppose that $h$ is equal to $0$ at the vertices of the rectangle, don't we have a problem with homogeneous Dirichlet boundary conditions? Or am I wrong? So do we maybe have to suppose that $h$ is nonzero? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Hello!!! (Wave)

Let $a,b>0$ and $D$ the rectangle $(0,a) \times (0,b)$. We consider the boundary value problem in $D$ for the Laplace equation, with Dirichlet boundary conditions,

$\left\{\begin{matrix}
u_{xx}+u_{yy}=0 & \text{ in } D,\\
u=h & \text{ in } \partial{D},
\end{matrix}\right.$

where $h:[0,a] \times [0,b] \to \mathbb{R}$ given function.

Supposing that $h$ is equal to $0$ at the vertices of the rectangle, I want to prove that the solution of the problem is the sum of the solutions of four respective problems, with homogeneous Dirichlet boundary conditions in three sides of the rectangle.


If we suppose that $h$ is equal to $0$ at the vertices of the rectangle, don't we have a problem with homogeneous Dirichlet boundary conditions? Or am I wrong? So do we maybe have to suppose that $h$ is nonzero?
Hey evinda !! (Wave)

Doesn't a homogeneous Dirichlet boundary condition mean that the boundary is constant everywhere?
But we only have the constant 0 at the 4 corners... (Thinking)
And the hint suggests homogeneous Dirichlet boundary conditions at only 3 of the 4 sides, meaning that the 4th side can be anything can't it? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hey evinda !! (Wave)

Doesn't a homogeneous Dirichlet boundary condition mean that the boundary is constant everywhere?
But we only have the constant 0 at the 4 corners... (Thinking)
And the hint suggests homogeneous Dirichlet boundary conditions at only 3 of the 4 sides, meaning that the 4th side can be anything can't it? (Wondering)
The 4th side will be the function $h$, won't it? (Thinking)

But how can we find different four problems of which the sum of the solutions is the solution of our problem? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
The 4th side will be the function $h$, won't it?
It would indeed be $h$ plus possibly a constant.

But how can we find different four problems of which the sum of the solutions is the solution of our problem?
Suppose we define a new Dirichlet boundary condition $\phi$ on $\partial D$, corresponding to one of the four problems:
$$
\phi(x)=\begin{cases}h(x)&\text{if }y=0 \\0 &\text{otherwise}\end{cases}
$$
Can we solve it? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Suppose we define a new Dirichlet boundary condition $\phi$ on $\partial D$, corresponding to one of the four problems:
$$
\phi(x)=\begin{cases}h(x)&\text{if }y=0 \\0 &\text{otherwise}\end{cases}
$$
Can we solve it? (Wondering)
You mean that our initial problem is equivalent to this one?

$$u_{xx}+u_{yy}=0 \text{ in } D \\ u=\phi(x) \text{ in } \partial{D}$$

If so, why does this hold? I haven't understood it. Isn't it given that $h$ is equal to $0$ at all the vertices of the rectangle? (Worried)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
You mean that our initial problem is equivalent to this one?

$$u_{xx}+u_{yy}=0 \text{ in } D \\ u=\phi(x) \text{ in } \partial{D}$$

If so, why does this hold? I haven't understood it. Isn't it given that $h$ is equal to $0$ at all the vertices of the rectangle? (Worried)
It's not equivalent no. It's a different problem since the boundary condition is different. (Shake)
We're defining 4 problems with the same ODE but with different, and simpler, boundary conditions, like the one I've just mentioned.
If we add the 4 solutions, won't we have the solution for the original problem? (Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
It's not equivalent no. It's a different problem since the boundary condition is different. (Shake)
We're defining 4 problems with the same ODE but with different, and simpler, boundary conditions, like the one I've just mentioned.
If we add the 4 solutions, won't we have the solution for the original problem? (Thinking)
So the boundary conditions of the other three problems have to be homogeneous? But how can that be? Then we wouldn't get three other different problems... (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
So the boundary conditions of the other three problems have to be homogeneous? But how can that be? Then we wouldn't get three other different problems... (Thinking)
We're simplifying the original problem, so that it has one side equal to the original boundary condition.
And so that the other 3 sides have a homogeneous boundary condition (constant), aren't we? (Thinking)

So yes, we make 4 different simplifications of the original problem, meaning we have 4 distinct problems.


Edit: btw, for the boundary condition I suggested, I intended:
$$
\phi(x,y)=\begin{cases}h(x,y)&\text{if }y=0 \\0 &\text{otherwise}\end{cases}
$$
(Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
We're simplifying the original problem, so that it has one side equal to the original boundary condition.
And so that the other 3 sides have a homogeneous boundary condition (constant), aren't we? (Thinking)

So yes, we make 4 different simplifications of the original problem, meaning we have 4 distinct problems.


Edit: btw, for the boundary condition I suggested, I intended:
$$
\phi(x,y)=\begin{cases}h(x,y)&\text{if }y=0 \\0 &\text{otherwise}\end{cases}
$$
(Thinking)
Don't we have the following four problems?


Problem 1:
\begin{equation*}\begin{cases}u^{(1)}_{xx} + u^{(1)}_{yy} = 0 \\
u^{(1)}(x, 0) = h(x) \\ u^{(1)}(x, b) = 0 \\ u^{(1)}(0, y) = 0 \\ u^{(1)}(a, y) = 0 \end{cases}\end{equation*}

Problem 2:
\begin{equation*}\begin{cases}u^{(2)}_{xx} + u^{(2)}_{yy} = 0 \\
u^{(2)}(x, 0) = 0 \\ u^{(2)}(x, b) = h(x) \\ u^{(2)}(0, y) = 0 \\ u^{(2)}(a, y) = 0 \end{cases}\end{equation*}

Problem 3:
\begin{equation*}\begin{cases}u^{(3)}_{xx} + u^{(3)}_{yy} =0 \\
u^{(3)}(x, 0) = 0 \\ u^{(3)}(x, b) = 0 \\ u^{(3)}(0, y) = h(y) \\ u^{(3)}(a, y) = 0\end{cases}\end{equation*}

Problem 4:
\begin{equation*}\begin{cases}u^{(4)}_{xx} + u^{(4)}_{yy} = 0 \\
u^{(4)}(x, 0) = 0 \\ u^{(4)}(x, b) = 0 \\ u^{(4)}(0, y) = 0 \\ u^{(4)}(a, y) = h(y)\end{cases}\end{equation*}


And we prove that the solution to our problem is $u=u^{(1)}+u^{(2)}+u^{(3)}+u^{(4)}$, by observing that

$u_{xx}+u_{yy}=0$ and that $u(x,0)=h(x)$, $u(x,b)=h(x)$, $u(0,y)=h(y)$ and $u(a,y)=h(y)$, i.e. that $u=h$ in $\partial{D}$. Am I right?


If so, then we do not use the fact that $h$ is equal to $0$ at the vertices of the rectangle, do we? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Don't we have the following four problems?


Problem 1:
\begin{equation*}\begin{cases}u^{(1)}_{xx} + u^{(1)}_{yy} = 0 \\
u^{(1)}(x, 0) = h(x) \\ u^{(1)}(x, b) = 0 \\ u^{(1)}(0, y) = 0 \\ u^{(1)}(a, y) = 0 \end{cases}\end{equation*}

Problem 2:
\begin{equation*}\begin{cases}u^{(2)}_{xx} + u^{(2)}_{yy} = 0 \\
u^{(2)}(x, 0) = 0 \\ u^{(2)}(x, b) = h(x) \\ u^{(2)}(0, y) = 0 \\ u^{(2)}(a, y) = 0 \end{cases}\end{equation*}

Problem 3:
\begin{equation*}\begin{cases}u^{(3)}_{xx} + u^{(3)}_{yy} =0 \\
u^{(3)}(x, 0) = 0 \\ u^{(3)}(x, b) = 0 \\ u^{(3)}(0, y) = h(y) \\ u^{(3)}(a, y) = 0\end{cases}\end{equation*}

Problem 4:
\begin{equation*}\begin{cases}u^{(4)}_{xx} + u^{(4)}_{yy} = 0 \\
u^{(4)}(x, 0) = 0 \\ u^{(4)}(x, b) = 0 \\ u^{(4)}(0, y) = 0 \\ u^{(4)}(a, y) = h(y)\end{cases}\end{equation*}


And we prove that the solution to our problem is $u=u^{(1)}+u^{(2)}+u^{(3)}+u^{(4)}$, by observing that

$u_{xx}+u_{yy}=0$ and that $u(x,0)=h(x)$, $u(x,b)=h(x)$, $u(0,y)=h(y)$ and $u(a,y)=h(y)$, i.e. that $u=h$ in $\partial{D}$. Am I right?


If so, then we do not use the fact that $h$ is equal to $0$ at the vertices of the rectangle, do we? (Thinking)
Yes.
Although strictly speaking, we don't have $u^{(1)}(x,0)=h(x)$, but we have $u^{(1)}(x,0)=h(x,0)$ and so on. (Nerd)

We do need that $h$ is equal to $0$ at the vertices.
Consider what happens when we add the 4 solutions together.
If they are not zero at the vertices, we wouldn't actually get the boundary condition $h$, because at the vertices we would get isolated discontinuities. (Thinking)

That is, suppose we have $u^{(1)}(0,0)=h(0,0)\ne 0$ and $u^{(3)}(0,0)=h(0,0)\ne 0$.
If we add them we get $u(0,0)=u^{(1)}(0,0)+u^{(3)}(0,0)=2h(0,0) \ne 0$, which is not equal to $h(0,0)$ is it? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Yes.
Although strictly speaking, we don't have $u^{(1)}(x,0)=h(x)$, but we have $u^{(1)}(x,0)=h(x,0)$ and so on. (Nerd)

We do need that $h$ is equal to $0$ at the vertices.
Consider what happens when we add the 4 solutions together.
If they are not zero at the vertices, we wouldn't actually get the boundary condition $h$, because at the vertices we would get isolated discontinuities. (Thinking)

That is, suppose we have $u^{(1)}(0,0)=h(0,0)\ne 0$ and $u^{(3)}(0,0)=h(0,0)\ne 0$.
If we add them we get $u(0,0)=u^{(1)}(0,0)+u^{(3)}(0,0)=2h(0,0) \ne 0$, which is not equal to $h(0,0)$ is it? (Wondering)
I see... (Nod)

So it follows easily that $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$. But what else do we have to show in order to prove that $u=h$ in $\partial{D}$ ? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
I see... (Nod)

So it follows easily that $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$. But what else do we have to show in order to prove that $u=h$ in $\partial{D}$ ? (Thinking)
Well, if we can find the 4 solutions to those 4 problems, we should verify that:

  • The sum of the solutions satisfies the ODE, which follows from the sum rule for differentiation.
  • The sum of the 4 boundary conditions is equal to the boundary condition $h$.
    And we've picked them so that would be the case, with the condition that $h$ is zero at the vertices.
I believe that only leaves verifying that if the original problem has a solution, that the 4 sub problems also have a solution. (Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Well, if we can find the 4 solutions to those 4 problems, we should verify that:

  • The sum of the solutions satisfies the ODE, which follows from the sum rule for differentiation.
  • The sum of the 4 boundary conditions is equal to the boundary condition $h$.
    And we've picked them so that would be the case, with the condition that $h$ is zero at the vertices.
So additionally to what I wrote above, we also have to mention that $u(x,0)=h(x,0), u(x,b)=h(x,0), u(0,y)=h(0,y)$ and $u(a,y)=h(0,y)$. Right? (Thinking)


I believe that only leaves verifying that if the original problem has a solution, that the 4 sub problems also have a solution. (Thinking)
How could we show this? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
So additionally to what I wrote above, we also have to mention that $u(x,0)=h(x,0), u(x,b)=h(x,0), u(0,y)=h(0,y)$ and $u(a,y)=h(0,y)$. Right?
Huh? :confused:

Isn't that just the boundary condition of the problem in the OP?
Oh wait, shouldn't it be: $u(x,0)=h(x,0), u(x,b)=h(x,{\color{red}b}), u(0,y)=h(0,y)$ and $u(a,y)=h({\color{red}a},y)$?
Or in short: $u=h$ on $\partial D$? (Wondering)

How could we show this?
Under which conditions can we solve each of the sub problems?
Or do we otherwise know of any restrictions on a Dirichlet boundary condition? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Huh? :confused:

Isn't that just the boundary condition of the problem in the OP?
Oh wait, shouldn't it be: $u(x,0)=h(x,0), u(x,b)=h(x,{\color{red}b}), u(0,y)=h(0,y)$ and $u(a,y)=h({\color{red}a},y)$?
Or in short: $u=h$ on $\partial D$? (Wondering)
Right... So it wasn't necessary to show that $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$, when mentioning that $u(x,0)=h(x,0), u(x,b)=h(x,b), u(0,y)=h(0,y)$ and $u(a,y)=h(a,y)$, right?

Under which conditions can we solve each of the sub problems?
Or do we otherwise know of any restrictions on a Dirichlet boundary condition? (Wondering)
How can we know under which consitions we can solve each of the sub problems? Is there a relative theorem?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Right... So it wasn't necessary to show that $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$, when mentioning that $u(x,0)=h(x,0), u(x,b)=h(x,b), u(0,y)=h(0,y)$ and $u(a,y)=h(a,y)$, right?
Not quite.
The boundary condition $u=h$ on $\partial D$ means $u(x,0)=h(x,0), u(x,b)=h(x,b), u(0,y)=h(0,y)$ and $u(a,y)=h(a,y)$.
Additionally we need $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$ to find the solution through the 4 sub problems, since otherwise the boundary conditions of the sub problems do not add up to the original boundary condition. (Thinking)

How can we know under which consitions we can solve each of the sub problems? Is there a relative theorem?
I found this article that shows that we can solve $u^{(1)}$ for any integrable boundary function. We can find it by separation of variables.
By symmetry it follow that the other 3 sub problems have a solution as well.
Therefore their sum is the solution for the original problem, which then has a solution for any $h$ that is integrable on the boundary as well, provided that $h$ is zero on the vertices.

Thus, if $h$ is zero on the vertices, the original problem has a solution for any $h$ that is integrable on the boundary, and the sub problems have solutions as well, allowing us to find the solution to the original problem. (Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Not quite.
The boundary condition $u=h$ on $\partial D$ means $u(x,0)=h(x,0), u(x,b)=h(x,b), u(0,y)=h(0,y)$ and $u(a,y)=h(a,y)$.
Additionally we need $u(0,0)=u(0,b)=u(a,0)=u(a,b)=0$ to find the solution through the 4 sub problems, since otherwise the boundary conditions of the sub problems do not add up to the original boundary condition. (Thinking)



I found this article that shows that we can solve $u^{(1)}$ for any integrable boundary function. We can find it by separation of variables.
By symmetry it follow that the other 3 sub problems have a solution as well.
Therefore their sum is the solution for the original problem, which then has a solution for any $h$ that is integrable on the boundary as well, provided that $h$ is zero on the vertices.

Thus, if $h$ is zero on the vertices, the original problem has a solution for any $h$ that is integrable on the boundary, and the sub problems have solutions as well, allowing us to find the solution to the original problem. (Thinking)
I understand.... Thanks a lot!!! (Smirk)