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here is a PDE I am having problems with, on the domain $[0,a]\times[0,b]$:

\begin{equation} u_{xx}+u_{yy}=0 \\ u(x,0)=0 , u(x,b)=\sin \frac{n\pi}{a}x \\ u(0,y)=u(a,y)=0 \end{equation}

From this we see that there are homogeneous conditions on the x-variable, so its form must be trigonometric, while the x-variable is subjected to inhomogeneous boundary conditions, so it must be hyperbolic.

1. We solve first the easy one, the trigonometric form by separation of variables and obtain:

$X_n(x)=\sin(\frac{n\pi}{a}x)$

2. The y-variable however has the following solution:

$Y(y)=C\cosh\lambda y+ D\sinh\lambda y$. With initial conditions I get the following:

$Y(0)=0\longrightarrow Y(y)=D\sinh\lambda y$.

Then with $Y(b)=\sin x$ we should get:

$\sin y=D\sinh\lambda y\rightarrow D=\frac{\sin y}{\sinh\lambda y}$. If I am not mistaking, this leaves:

$Y(y)=\sin y$

and

\begin{equation} u(x,y)=\sum_{n=1}^\infty \sin y \sin\frac{n\pi}{a}x \end{equation}

But this is not an eigenfunction of the Laplace problem above, as it does not satisfy it.

Clearly, the mistake is in step 2. But how do I find that $D$?

Thanks

Luthier415Hz
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1 Answers1

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I think there was a typo in your original problem, which probably got you mixed up. Especially for the $Y$. In your case, the proper mode to use is for $n\in\mathbb N^*$ as you noted: $$ X_n = \sin(\pi nx/a) $$ this gives: $$ Y_n(y) = \frac{\sinh\left(\frac{n\pi}{a}y\right)}{\sinh\left(\frac{n\pi}{a}b\right)} $$ normalising the other extremity $Y(b)=1$

Your modes are therefore: $$ u_n(x,y) = \sin(\pi nx/a)\frac{\sinh\left(\frac{n\pi}{a}y\right)}{\sinh\left(\frac{n\pi}{a}b\right)} $$

I think you meant the boundary condition: $u(x,b) = \sin(\pi x/a)$. In this case the solution is directly the mode $u_1$.

If you really want the arbitrary boundary condition $u(x,b) = \sin x$ as you wrote, you'll need to expand out $\sin$ in terms of the $X_n$. Using: $$ C_n := \int_0^a\sin x \sin\left(\pi n_x \frac{x}{a}\right) dx \\ = \int_0^a\frac{1}{2}\left(\cos\left(\left(1- \frac{\pi n_x}{a}\right)x\right)-\cos\left(\left(1+ \frac{\pi n_x}{a}\right)x\right)\right)dx \\ = \frac{1}{2(1-\pi n_x/a)}\left(\sin\left(\left(1- \frac{\pi n_x}{a}\right)a\right)-\frac{1}{2(1+\pi n_x/a)}\sin\left(\left(1+ \frac{\pi n_x}{a}\right)a\right)\right)\\ $$

You have: $$ \sin(x) = 2\sum_{n=1}^\infty C_n X_n(x) $$

so

$$ u = 2\sum_{n=1}^\infty C_n \sin(\pi nx/a)\frac{\sinh\left(\frac{n\pi}{a}y\right)}{\sinh\left(\frac{n\pi}{a}b\right)} $$

Hope this helps.

LPZ
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    $u(x,b)$ is a function of $x$, not $y$. If I understood correctly, $b$ is the extremity of the domain right? – LPZ Jul 06 '22 at 12:44
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    also, are you sure it's $\sin$ ? shouldn't you normalise it because the $2\pi$ periodicity of $\sin$ in completely uncorrelated to your domain dimensions $a,b$. Unless this is a purely mathematical problem. – LPZ Jul 06 '22 at 12:47
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    You got it too. The normalisation is unimportant, however this is not the step where you use BC. You use it in the mode expansion – LPZ Jul 06 '22 at 12:49
  • thanks, that was fast ;) – LPZ Jul 06 '22 at 12:56
  • Regarding the Y-part : $Y(y)=D\sinh\frac{n\pi}{a}y$. I order to find D: $\sin\frac{n\pi}{a}x=D\sinh\frac{n\pi}{a}b\rightarrow D=\frac{\sin\frac{n\pi}{a}x}{\sinh\frac{n\pi}{a}b}$. That gives $Y(y)=\frac{\sin\frac{n\pi}{a}x}{\sinh\frac{n\pi}{a}b}\sinh\frac{n\pi}{a}y$. It seems you forgot the $\sin\frac{n\pi}{a}x$ term in your solution – Luthier415Hz Jul 06 '22 at 13:10
  • Once again, when finding the $Y$, ie the modes, there is no preferred normalisation, your $D$ is irrelevant. Furthermore, $Y$ is a function of $y$ only (separation of variables), so $D$ cannot depend on $x$. I chose the normalisation such as $Y(b)=1$ which is convenient for future calculations. The coefficients in front of the modes $C_n$ are then determined from the BC. – LPZ Jul 06 '22 at 15:22
  • btw, check the final formula numerically (truncating the series of course), you'll find the correct BC. – LPZ Jul 06 '22 at 15:24