# Boundary value problem for Laplace equation

• MHB
• evinda
In summary, the conversation discusses solving a boundary value problem for the Laplace equation in a rectangular domain with Dirichlet boundary conditions. The solution is shown to be the sum of four respective problems with homogeneous Dirichlet boundary conditions on three sides of the rectangle. These four problems are defined by simplifying the original problem and using a different boundary condition on each one. The solution is then shown to satisfy the original problem's boundary conditions. The fact that the given function, $h$, is equal to $0$ at the vertices of the rectangle is not used in the final solution.
evinda
Gold Member
MHB
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
The 4th side will be the function $h$, won't it?

It would indeed be $h$ plus possibly a constant.

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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 like Serena said:
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)

evinda said:
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?

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)

evinda said:
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)

I like Serena said:
Huh?

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?

I like Serena said:
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?

evinda said:
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)

evinda said:
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)

I like Serena said:
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)

## 1) What is the Laplace equation and why is it important in boundary value problems?

The Laplace equation is a partial differential equation that describes the behavior of a scalar field in space. It is important in boundary value problems because it allows us to find the solution to a physical system by specifying the boundary conditions, which are the conditions that the solution must satisfy at the boundaries of the system.

## 2) What is a boundary value problem for the Laplace equation?

A boundary value problem for the Laplace equation is a mathematical problem that involves finding the solution to the Laplace equation subject to specific boundary conditions. These boundary conditions can be specified as either Dirichlet boundary conditions (the value of the solution is known at the boundary) or Neumann boundary conditions (the derivative of the solution is known at the boundary).

## 3) Can you give an example of a physical system that can be modeled using the Laplace equation?

One example of a physical system that can be modeled using the Laplace equation is the flow of heat in a solid object. The temperature distribution in the object can be described by the Laplace equation, with the boundary conditions being the temperature at the surface of the object.

## 4) What are some methods for solving boundary value problems for the Laplace equation?

Some common methods for solving boundary value problems for the Laplace equation include separation of variables, Fourier series, and numerical methods such as finite differences or finite elements. These methods involve breaking down the problem into simpler parts and using mathematical techniques to find the solution.

## 5) How are boundary value problems for the Laplace equation related to other differential equations?

Boundary value problems for the Laplace equation are a special case of more general boundary value problems involving other types of differential equations, such as the Poisson equation or the Helmholtz equation. The Laplace equation is also closely related to the diffusion equation, which describes the behavior of a scalar quantity over time.

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