$$\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0$$

$$0 < x < \infty, \ \ 0 < y < H$$

subject to these boundary conditions:

$${\frac {\partial u} {\partial y}} (x,0)=0$$

$${\frac {\partial u} {\partial y}} (x,H)=0$$

$$u(0,y)=f(y)$$ This is what I found and I understand it but is this the answer?

You will want to solve this using separation of variables.
Then ∇2(XY)=X′′(x)Y(y)+X(x)Y′′(y)=0 so that X′′X+Y′′Y=0.Hence X′′=CX, Y′′=−CY for some real constant C (known as a separation constant). The values of C we need to take depend on the boundary conditions, which are: X(0)=1, limx→∞X(x)=0 ,Y(0)=Y(h)=0 with Y(y) not identically zero (actually all that's required is X(0)≠0, but it is convenient to specify X(0)=1).

The easiest boundary condition to satisfy is that X(x)→0 as x→∞. We must have X(x)=$e^{−kx}$ for some k>0. This means that C=k$^2$ so that Y′′=−$k^2$Y subject to Y(0)=Y(h)=0 but with Y(y) not identically zero.

That can be done if we take k=(nπ)/h for some positive integer n with Y(y)=Bsin(nπyh) where the constant B cannot be determined from the boundary conditions on Y. But given the next stage of the solution we may as well take B=1. Putting this together, we have, for each positive integer n, an eigenfunction $X_n$(x)$Y_n$(y)=exp(−nπxh)sin(nπyh) and the natural thing to do is to take a linear combination of these, u(x,y)=∑$a_n$exp(−nπxh)sin(nπyh), and choose the coefficients an to satisfy the boundary condition u(0,y)=f(y). We then have f(y)=u(0,y)=∑$a_n$sin(nπyh) which is the fourier sine series for f(y) on the interval 0≤y≤h. Thus $a_n$= 2/h ∫f(y)sin(nπyh)dy.

Reference https://www.physicsforums.com/threads/laplaces-equation-inside-a-semi-infinite-strip.673334/

I understand how to compute Laplace equations but I don't understand where the semi-infinite strip comes to play. So this question is a bit difficult for me to understand. Can someone please help?

  • $\begingroup$ Show us what you have done so far so we can help. $\endgroup$ – mattos Mar 6 '15 at 10:53
  • $\begingroup$ I posted it in the question. $\endgroup$ – Jessy White Mar 6 '15 at 11:29

So the semi-infinite strip means that we are working in the region where $(x,y)\in [0,\infty)\times [0,H]$. We have Laplace's equation, which simply states:

$$\nabla^{2} \Phi = 0 \iff \frac{\partial^{2} \Phi}{\partial x^{2}}+\frac{\partial^{2} \Phi}{\partial y^{2}}=0$$

And we have $\Phi(0,y)=f(y)$ and $\lim_{x\to\infty}\Phi(x,y) = 0$ (from physical conditions) as our boundary conditions, along with:

$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,0)} = \left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,H)}=0$$

We note that using the method of characteristics we can assume that: $\Phi(x,y)=X(x)Y(y)$, and therefore we can rewrite the Laplace equation above:

$$Y(y)\frac{\partial^{2} X}{\partial x^{2}}+X(x)\frac{\partial^{2} Y}{\partial y^{2}}=0 \implies \frac{1}{X(x)}\frac{\partial^{2} X}{\partial x^{2}}=-\frac{1}{Y(y)}\frac{\partial^{2} Y}{\partial y^{2}}$$

But $x$ and $y$ are independent, and therefore we must have that:

$$\frac{1}{X(x)}\frac{\partial^{2} X}{\partial x^{2}}=k^{2} \land \frac{1}{Y(y)}\frac{\partial^{2} Y}{\partial y^{2}}=-k^{2}$$

These are ODEs with known solutions:

$$X(x)=\alpha \exp(kx)+\beta\exp(-kx) \land Y(y) = \gamma \sin(ky) + \delta \cos(ky)$$

Where $\exists \alpha,\beta,\gamma,\delta\in\mathbb{R}$. We now wish to find the value of these coefficients based on the initial conditions. We can see that our second boundary condition, $\lim_{x\to\infty}\Phi(x,y)=0$ means that $X(x)$ must have a decaying form, thus $\alpha = 0$. We can now use the differential boundary conditions:

$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,0)}=0 \implies \gamma k \cos(0) - \delta k \sin(0) = 0$$

Therefore we have $\gamma k = 0$, and therefore $\gamma = 0$ (if we assume $k \neq 0$ for non-static solution). We can now look at the second differential boundary condition:

$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,H)}=0 \implies \delta k \sin(k H)=0$$

But assuming $\delta \neq 0$ (for non-trivial solution), we must have $k = \frac{n \pi}{H}$, $\exists n \in \mathbb{N}_{0}$.

We finally turn to our last boundary condition $\Phi(0,y)=f(y)$, where $f(y)$ is an arbitrary function. We can write this as:

$$\sum_{n = 0}^{\infty} \delta_{n} \beta_{n} \sin(k_{n} y) = f(y)$$

But WLOG we can assume $\beta_{n} = 1$, $\forall n \in \mathbb{N}_{0}$ therefore we have:

$$\sum_{n=0}^{\infty}\delta_{n} \sin(k_{n} y) = f(y)$$

We can recognize this as a Fourier sine series, thus giving us coefficients:

$$\delta_{n} = \frac{2}{H}\int_{0}^{H} f(y) \sin\left(\frac{n \pi y}{H}\right)\:\mathrm{d}y\tag{1}$$

So we have in general:

$$\Phi(x,y) = \sum_{n=0}^{\infty}\delta_{n}\exp\left(-\frac{n\pi x}{H}\right)\sin\left(\frac{n \pi y}{H}\right)$$

Where $\delta_{n}$ is defined by $(1)$.

I hope this helps clear things up!

  • $\begingroup$ Thank you it makes sence ! So $e^{-kx}$ should be positive? $\endgroup$ – Jessy White Mar 6 '15 at 12:48
  • 1
    $\begingroup$ @JessyWhite Sorry, I missed out the minus sign! The minus sign is important as it means the solution decays as we approach infinity. This is the significance of the half-infinite $x$-dimension $\endgroup$ – Thomas Russell Mar 6 '15 at 12:51
  • $\begingroup$ Sorry... I feel like I am missing something here... in the equation just before (1) (near the bottom) the sum has a sine term... shouldn't this be a cosine term as Y(y) only has a cosine term $\endgroup$ – John Don Apr 22 '17 at 12:14

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