- #1

H Smith 94

Gold Member

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So here I have Laplace's equation with non-homogeneous, mixed boundary conditions in both [itex]x[/itex] and [itex]y[/itex]

Solve Laplace's equation \begin{equation}\label{eq:Laplace}\nabla^2\phi(x,y)=0\end{equation} for the following boundary conditions:

In two dimensions:[tex]\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}.[/tex]

[/B]

As far as I can tell, a solution exists in \begin{equation}\label{eq:phi1}\phi(x,y)=\phi_0(x) + \Delta\phi(x,y),\end{equation} where [itex]\phi_0(x)[/itex] satisfies the [itex]x[/itex]-boundaries, in that [tex]\phi_0(x)=A_0+B_0 x[/tex] and [itex]\Delta\phi(x,y)[/itex] satisfies the [itex]y[/itex]-boundaries, with a separable solution of the form [tex]\Delta\phi(x,y)=\sum_{n}\chi_n(x)\gamma_n(y).[/tex]

Applying boundary condition 1: [tex]\phi_0(0,y) = 2 \implies \boxed{A_0 = 2},[/tex] so now [itex]\phi_0= 2+B_0 x [/itex].

Applying boundary condition 2: [tex]\phi_0(1,y) = 2+ B_0 = 0 \implies \boxed{B_0 = -2}.[/tex] Hence, have that \begin{equation}\phi_0(x,y) = 2-2x.\end{equation}

Now, equation [itex]\ref{eq:phi1}[/itex] can be substituted into Laplace's equation ([itex]\ref{eq:Laplace}[/itex]) to yield [tex] \frac{\partial^2\Delta\phi(x,y)}{\partial x^2} + \frac{\partial^2\Delta\phi(x,y)}{\partial y^2} = 0.[/tex] This has the separable solution in [itex]\Delta\phi(x,y) = \chi(x)\gamma(y)[/itex]: [tex]\frac{\mathrm{d}^2\chi}{\mathrm{d}x^2} = \kappa \chi(x) \\ \frac{\mathrm{d}^2\gamma}{\mathrm{d}y^2} = -\kappa \gamma(y)[/tex] so that [tex]\begin{align}\chi(x) = E \cos{\kappa x} + F \sin{\kappa x} \\ \gamma(y) = G e^{\kappa y} + H e^{-\kappa y}, \end{align}[/tex] for which I found using the homogeneous boundary conditions produced from [itex]\phi_0(x)[/itex] that [itex]F = -E[/itex] and, in turn, [itex]E = 0[/itex] -- so [tex]\chi(x) = 0\ \forall x \implies \boxed{\Delta\phi(x,y) = 0}[/tex] and so \begin{equation}\boxed{\boxed{\phi(x,y)=2 - 2x}},\end{equation} which just cannot be correct!

I have tried solving the solution directly using separation of variables with all types of attempted forms of solutions and---after pages and pages of working---have come up with nothing. I feel like there's a very simple but essential point I'm missing here.

Thanks in advance!

**.**

1. Homework Statement1. Homework Statement

Solve Laplace's equation \begin{equation}\label{eq:Laplace}\nabla^2\phi(x,y)=0\end{equation} for the following boundary conditions:

- [itex]\phi(0, y)=2[/itex];

- [itex]\phi(1, y)=0[/itex];
- [itex]\phi(x, 0)=0[/itex];
- [itex]\frac{\partial}{\partial y}\phi(x,1)=1[/itex].

## Homework Equations

In two dimensions:[tex]\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}.[/tex]

## The Attempt at a Solution

[/B]

As far as I can tell, a solution exists in \begin{equation}\label{eq:phi1}\phi(x,y)=\phi_0(x) + \Delta\phi(x,y),\end{equation} where [itex]\phi_0(x)[/itex] satisfies the [itex]x[/itex]-boundaries, in that [tex]\phi_0(x)=A_0+B_0 x[/tex] and [itex]\Delta\phi(x,y)[/itex] satisfies the [itex]y[/itex]-boundaries, with a separable solution of the form [tex]\Delta\phi(x,y)=\sum_{n}\chi_n(x)\gamma_n(y).[/tex]

Applying boundary condition 1: [tex]\phi_0(0,y) = 2 \implies \boxed{A_0 = 2},[/tex] so now [itex]\phi_0= 2+B_0 x [/itex].

Applying boundary condition 2: [tex]\phi_0(1,y) = 2+ B_0 = 0 \implies \boxed{B_0 = -2}.[/tex] Hence, have that \begin{equation}\phi_0(x,y) = 2-2x.\end{equation}

Now, equation [itex]\ref{eq:phi1}[/itex] can be substituted into Laplace's equation ([itex]\ref{eq:Laplace}[/itex]) to yield [tex] \frac{\partial^2\Delta\phi(x,y)}{\partial x^2} + \frac{\partial^2\Delta\phi(x,y)}{\partial y^2} = 0.[/tex] This has the separable solution in [itex]\Delta\phi(x,y) = \chi(x)\gamma(y)[/itex]: [tex]\frac{\mathrm{d}^2\chi}{\mathrm{d}x^2} = \kappa \chi(x) \\ \frac{\mathrm{d}^2\gamma}{\mathrm{d}y^2} = -\kappa \gamma(y)[/tex] so that [tex]\begin{align}\chi(x) = E \cos{\kappa x} + F \sin{\kappa x} \\ \gamma(y) = G e^{\kappa y} + H e^{-\kappa y}, \end{align}[/tex] for which I found using the homogeneous boundary conditions produced from [itex]\phi_0(x)[/itex] that [itex]F = -E[/itex] and, in turn, [itex]E = 0[/itex] -- so [tex]\chi(x) = 0\ \forall x \implies \boxed{\Delta\phi(x,y) = 0}[/tex] and so \begin{equation}\boxed{\boxed{\phi(x,y)=2 - 2x}},\end{equation} which just cannot be correct!

I have tried solving the solution directly using separation of variables with all types of attempted forms of solutions and---after pages and pages of working---have come up with nothing. I feel like there's a very simple but essential point I'm missing here.

Thanks in advance!

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