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[SOLVED] Laplace equation

dwsmith

Well-known member
Feb 1, 2012
1,673
Consider Laplace's equation $\nabla^2u = 0$ on the rectangle with the following boundary conditions:
$$
u_y(x,0) = f(x)\quad u(L,y) = 0\quad u(x,H) = 0\quad u(0,y) = g(y).
$$

How does one of the boundary conditions being defined by a derivative alter the solving of this problem? I have never done a Laplace equation with derivative BC.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Consider Laplace's equation $\nabla^2u = 0$ on the rectangle with the following boundary conditions:
$$
u_y(x,0) = f(x)\quad u(L,y) = 0\quad u(x,H) = 0\quad u(0,y) = g(y).
$$

How does one of the boundary conditions being defined by a derivative alter the solving of this problem? I have never done a Laplace equation with derivative BC.
Hi dwsmith, :)

Let me suggest a way to solve this problem. The first boundary condition is redundant and you can replace it with \(u(x,0)=h(x)\). Then you will have to solve the Laplace's equation separately for boundary conditions,

\[u_{1}(x,0) = f(x)\quad u_{1}(L,y) = 0\quad u_{1}(x,H) = 0\quad u_{1}(0,y) = 0\]

and

\[u_{2}(x,0) = 0\quad u_{2}(L,y) = 0\quad u_{2}(x,H) = 0\quad u_{2}(0,y) = g(y)\]

The solution to the original problem will then be,

\[u(x,y)=u_{1}(x,y)+u_{2}(x,y)\]

Further details of this method can be found >>here<<.

Kind Regards,
Sudharaka.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith, :)

Let me suggest a way to solve this problem. The first boundary condition is redundant and you can replace it with \(u(x,0)=h(x)\). Then you will have to solve the Laplace's equation separately for boundary conditions,

\[u_{1}(x,0) = f(x)\quad u_{1}(L,y) = 0\quad u_{1}(x,H) = 0\quad u_{1}(0,y) = 0\]

and

\[u_{2}(x,0) = 0\quad u_{2}(L,y) = 0\quad u_{2}(x,H) = 0\quad u_{2}(0,y) = g(y)\]

The solution to the original problem will then be,

\[u(x,y)=u_{1}(x,y)+u_{2}(x,y)\]

Further details of this method can be found >>here<<.

Kind Regards,
Sudharaka.
So we can disregard the fact that it is a partial derivative?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
So we can disregard the fact that it is a partial derivative?
I think you are referring to the first boundary condition. When you differentiate \(u(x,y)\) with respect to \(y\) and then substitute \(y=0\) it is obvious that you get a function of \(x\) since we have substituted for \(y\). Hence I don't see any usefulness in that boundary condition. In other words, even if it's not given we know that \(u_{y}(x,0)=\mbox{a function of }x\). So we don't know anything about \(u(x,0)\) except the fact that it's a function of \(x\).
 

chisigma

Well-known member
Feb 13, 2012
1,704
Consider Laplace's equation $\nabla^2u = 0$ on the rectangle with the following boundary conditions:
$$
u_y(x,0) = f(x)\quad u(L,y) = 0\quad u(x,H) = 0\quad u(0,y) = g(y).
$$

How does one of the boundary conditions being defined by a derivative alter the solving of this problem? I have never done a Laplace equation with derivative BC.
The 'classical' approach permits to find a solution in the form...

$\displaystyle u(x,y)= v(x)\ w(y)$ (1)

... and consists in several steps. The first step is to compute from (1)...

$\displaystyle \frac{\partial^{2} u}{\partial^{2} x}= v^{'\ '} (x)\ w(y)\ ,\ \frac{\partial^{2} u}{\partial^{2} y}= v(x)\ w^{'\ '} (y)\ $ (2)

... that permits to write the original PDE as a pair of ODE...

$\displaystyle \frac{v^{'\ '}(x)}{v(x)}= - \frac{w^{'\ '}(y)}{w(y)}= \lambda$ (3)

... where $\lambda$ is a constant. Then, as suggested by Sudharaka,You set $u(x,y)= u_{1}(x,y)+u_{2}(x,y)$ and solve separately the two PDE...

$\displaystyle \nabla^2 u_{1} = 0\ ,\ u_{1}(x,0)= f(x)\ ,\ u_{1}(L,y)= 0\ ,\ u_{1}(x,H)= 0\ ,\ u_{1}(0,y)= 0 $

$\displaystyle \nabla^2 u_{2} = 0\ ,\ u_{2}(x,0)= 0\ ,\ u_{2}(L,y)= 0\ ,\ u_{2}(0,y)= 0\ ,\ u_{2}(0,y)= g(y) $ (4)

Other details will be given in a successive post...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I solved for $u_1$. To solve for $u_2$, I just need to swap x and y and make the substitution $x^* = L - x$.
Is $u_1$ correct and the approach for $u_2$ correct as well?

If we integrate the first boundary condition, we will have $u(x,y) = yf(x) + h(x)$.
Adding in the fact that $y = 0$, tells us that $u(x,0) = h(x)$.
Therefore, we can solve the equation with the following boundary conditions:
$$
u(x,0) = h(x)\quad u(L,y) = 0\quad u(x,H) = 0\quad u(0,y) = g(y).
$$
The general solution for the generic boundary conditions of $u(x,0) = 0, u(L,y) = 0, u(x,H) = f(x),$ and $u(0,y) = 0$ is
$$
\sum_{n = 1}^{\infty}A_n\sin\lambda_n x\sinh\lambda_n y,
$$
where $\lambda_n = \frac{n\pi}{L}$ and
$$
A_n = \frac{2}{L\sinh\lambda_n H}\int_0^Lf(x)\sin\lambda_n xdx.
$$
For the boundary conditions $u(x,0) = h(x), u(L,y) = 0, u(x,H) = 0$, and $u(0,y) = 0$, we need to make the change of coordinates $x = x^*$ and $y^* = H - y$.
Therefore,
$$
u_1(x,y) = \sum_{n = 1}^{\infty}A_n\sin\lambda_n x\sinh[\lambda_n (H - y)]
$$
where $\lambda_n = \frac{n\pi}{L}$ and
$$
A_n = \frac{2}{L\sinh\lambda_n H}\int_0^Lh(x)\sin\lambda_n xdx.
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Here is the full solution I obtained (Is it correct?):
$$
u(x,y) = \sum_{n=1}^{\infty}\left[A_n\sin\frac{n\pi x}{L}\sinh\left(\frac{n\pi}{L}(H-y)\right) + B_n\sin\frac{n\pi y}{H}\sinh\left(\frac{n\pi}{H}(L-x)\right)\right]
$$
where
$$
A_n = \frac{2}{L\sinh\frac{n\pi H}{L}}\int_0^Lh(x)\sin\frac{n\pi x}{L}dx
$$
and
$$
B_n = \frac{2}{H\sinh\frac{n\pi L}{H}}\int_0^Hg(y)\sin\frac{n\pi y}{H}dy
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Here is the full solution I obtained (Is it correct?):
$$
u(x,y) = \sum_{n=1}^{\infty}\left[A_n\sin\frac{n\pi x}{L}\sinh\left(\frac{n\pi}{L}(H-y)\right) + B_n\sin\frac{n\pi y}{H}\sinh\left(\frac{n\pi}{H}(L-x)\right)\right]
$$
where
$$
A_n = \frac{2}{L\sinh\frac{n\pi H}{L}}\int_0^Lh(x)\sin\frac{n\pi x}{L}dx
$$
and
$$
B_n = \frac{2}{H\sinh\frac{n\pi L}{H}}\int_0^Hg(y)\sin\frac{n\pi y}{H}dy
$$
My solution is in terms of $h(x)$ not $f(x)$ though. How can I get back $f(x)$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
If we integrate the first boundary condition, we will have $u(x,y) = yf(x) + h(x)$.
This is incorrect. You have,

\[u_{y}(x,0)=\left.\frac{\partial}{\partial y}u(x,y)\right|_{y=0}=f(x)\]

You cannot integrate this to obtain, \(u(x,y)\) because there is a substitution \(y=0\). Of course you can write,

\[\int u_{y}(x,0)\,\partial y=\int\left.\frac{\partial}{\partial y}u(x,y)\right|_{y=0}\,\partial y=yf(x)+h(x)\]

but I don't think this will give you any useful information. :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
This is incorrect. You have,

\[u_{y}(x,0)=\left.\frac{\partial}{\partial y}u(x,y)\right|_{y=0}=f(x)\]

You cannot integrate this to obtain, \(u(x,y)\) because there is a substitution \(y=0\). Of course you can write,

\[\int u_{y}(x,0)\,\partial y=\int\left.\frac{\partial}{\partial y}u(x,y)\right|_{y=0}\,\partial y=yf(x)+h(x)\]

but I don't think this will give you any useful information. :)
Ok so what do I do then?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Ok so what do I do then?
I think you should read >>this<< to understand how to solve a Laplace's equation with three homogeneous boundary conditions. Examples 1 and 2 provides you a complete walk through, so I don't want to write them again here. :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I think you should read >>this<< to understand how to solve a Laplace's equation with three homogeneous boundary conditions. Examples 1 and 2 provides you a complete walk through, so I don't want to write them again here. :)
The problem isn't solving the equation. The problem is the $u_y$ piece.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The problem isn't solving the equation. The problem is the $u_y$ piece.
Please read posts #2 and #5 carefully. There is no need to do anything with the first boundary condition and the problem is equivalent to solving the two equations,

\[\displaystyle \nabla^2 u_{1} = 0\ ,\ u_{1}(x,0)= f(x)\ ,\ u_{1}(L,y)= 0\ ,\ u_{1}(x,H)= 0\ ,\ u_{1}(0,y)= 0\]

and

\[\displaystyle \nabla^2 u_{2} = 0\ ,\ u_{2}(x,0)= 0\ ,\ u_{2}(L,y)= 0\ ,\ u_{2}(0,y)= 0\ ,\ u_{2}(0,y)= g(y)\]

The final solution will be,

\[u(x,y)=u_{1}(x,y)+u_{2}(x,y)\]
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Please read posts #2 and #5 carefully. There is no need to do anything with the first boundary condition and the problem is equivalent to solving the two equations,

\[\displaystyle \nabla^2 u_{1} = 0\ ,\ u_{1}(x,0)= f(x)\ ,\ u_{1}(L,y)= 0\ ,\ u_{1}(x,H)= 0\ ,\ u_{1}(0,y)= 0\]

and

\[\displaystyle \nabla^2 u_{2} = 0\ ,\ u_{2}(x,0)= 0\ ,\ u_{2}(L,y)= 0\ ,\ u_{2}(0,y)= 0\ ,\ u_{2}(0,y)= g(y)\]

The final solution will be,

\[u(x,y)=u_{1}(x,y)+u_{2}(x,y)\]
So it is wont affect the solution to let $u(x,0) = f(x)$ and solve it as the condition $u_y(x,0)=f(x)$ isn't there?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
So it is wont affect the solution to let $u(x,0) = f(x)$ and solve it as the condition $u_y(x,0)=f(x)$ isn't there?
Correct. :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So I have as my final solution. Correct?

$$
A_n = \frac{2}{L\sinh\frac{n\pi H}{L}}\int_0^Lf(x)\sin\frac{n\pi x}{L}dx.
$$

$$
B_n = \frac{2}{H\sinh\frac{n\pi L}{H}}\int_0^Hg(y)\sin\frac{n\pi y}{H}dy.
$$

$$
u(x,y) = u_1 + u_2 = \sum_{n = 1}^{\infty}\left[A_n\sin\frac{n\pi x}{L}\sinh\left[\frac{n\pi}{L} (H - y)\right] + B_n\sin\frac{n\pi y}{H}\sinh\left[\frac{n\pi}{H}(L - x)\right]\right]
$$