Laplace's equation in an infinite strip - bounded vs. unbounded solution

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Thanks to @Ian! I figured this out.

Note that $v(x,y) = \sin( \pi y ) \big( \alpha e^{\pi x} + \beta e^{-\pi x} \big)$ is a solution to Laplace's equation for all $\alpha, \beta \in \mathbb{R}$.

In addition to this, note that $v(x,0)=0$ and $v(x,1)=0$.

Then define the function:

$u(x,y) = (B-A)y + A + v(x,y) $

We see that $\Delta u=0$ and $u(x,1)=A$ and $u(x,1)=B$. This function is also unbounded.

For any $\alpha, \beta \in \mathbb{R}$, we find that $u(x,y)$ is a solution to the BVP in the case that $u$ need not be bounded.

We have found infinitely many solutions to the BVP, and so uniqueness is lost.

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QuantumEyedea
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Updated on August 01, 2022

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  • QuantumEyedea
    QuantumEyedea over 1 year

    Let $D = \{\ ( x,y )\in \mathbb{R}^{2} \ | \ 0<y<1\ \}$. Let $A,B\in\mathbb{R}$. Consider the boundary value problem:

    $\Delta u = 0$

    $u(x,0)=A\ $ and $\ u(x,1)=B\ $ for all $x \in \mathbb{R}$

    Restriction: $u$ is bounded

    $\ $

    I already know that in this case (that $u$ is bounded), then the solution is $u(x,y)=(B-A)y + A$. Furthermore, this solution is UNIQUE.

    $\ $

    The next thing I need to show is that in the case that we DROP the restriction, meaning in the case that we do not force $u$ to be bounded - then there is no longer a single unique solution to the BVP. This is where I get stuck.

    $\mathbf{MY\ ATTEMPT:}$

    By separation of variables, $u(x,y)=X(x)Y(y)$, and out of $\Delta u = 0$ we get the two ODEs:

    $\frac{d^{2}X}{dx^{2}} = \lambda^{2} X(x)$

    $\frac{d^{2}Y}{dy^{2}} = -\lambda^{2} Y(x)$

    In the above, $\lambda^{2}$ is my non-negative seperation constant.

    In the case that $\lambda=0$, I get $X(x)=C_{1}x+C_{2}$ and $Y(y)=C_{3}y+C_{4}$.

    In the case that $\lambda\neq 0$, I get $X(x)=C_{1}\sinh(\lambda x)+C_{2}\cosh(\lambda x)$ and $Y(y)=C_{3}\sin(\lambda y)+C_{4}\cos(\lambda y)$.

    $\ $

    It was easy force boundedness on $u$, because it just killed the $\lambda \neq 0$ solution. It also forced $C_{1}=0$ in the $\lambda=0$ solution.

    I am at a loss for how to solve for the constants with the boundary conditions in this case that $u$ is not bounded. Am I missing a detail? How can I simplify the problem? What can I do?

    Maybe I need to do a Fourier series of some sort? But I can't figure out how to do this since $\lambda$ doesn't necessarily have to be equal to $n \pi$.

    • Ian
      Ian about 8 years
      Write the solution as your linear solution plus something. Check that the something is another solution but with homogeneous boundary conditions at $y=0,y=1$.
    • QuantumEyedea
      QuantumEyedea about 8 years
      Thanks for your response. So I write a new solution $\widetilde{u}(x,y) = (B-A)y + A + v$. The "something" v, is it right that v=v(x,y) - or just v=v(x) or v=v(y)? I'm a little confused.