**Problem**

Consider the Laplace equation in the half-strip:

$$ \Delta u = u_{xx} + u_{yy} = 0 : \{x > 0, -1 < y < 1\} $$

with the boundary conditions:

$$ u(x,-1) = u(x,1) = 0 $$

$$ u(0,y) = 1 - |y|, \phantom{\ } \max_{\{x,y\}} |u| < \infty $$

**Part a.** Write the associated eigenvalue problem

**Answer **

First we separate variables in $\{x,y\}$. Let: $(x,y) = X(x)Y(y)$.

$$ \implies u_{xx} = X''(x)Y(y), \phantom{\ } u_{yy} = X(x)Y''(y) $$

$$ u_{xx} + u_{yy} = 0 \rightarrow X''(x)Y(y) + X(x)Y''(y) = 0 $$

$$ \implies \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} = \lambda $$

For some constant $\lambda$, as each side depends only on one variable and so is constant with repsect to the other. Then we have two ODEs with BCs, and our eigenvalue problem is:

$$ X''(x) - \lambda X(x) = 0, \phantom{\ } Y''(y) + \lambda Y(y) = 0 $$

$$ Y(-1) = Y(1) = 0, \phantom{\ } \max_{\{x,y\}} |X(x)Y(y)| < \infty, \phantom{\ } u(0,y) = 1 - |y| \phantom{\ } \square $$

**Part b.** Find all eigenvalues and corresponding eigenfunctions

**Answer **As previously derived, we make the assumption that all eigenvalues $\lambda_n = \beta_n^2 > 0$. Solving first for $Y(y)$ in:

$$ \{Y(y) \phantom{\ } | Y''(y) + \lambda Y(y) = 0, \phantom{\ } Y(-1) = Y(1) = 0\} $$

Let: $\lambda = \beta^2$. Then the ODE $Y''(y) + \lambda Y(y) = Y''(y) + \beta^2 Y(y) = 0$ has solution $Y(y) = A \cos(\beta y) + B \sin(\beta y)$ for some constants $ \{A,B\} \in \mathbb{R} $. Plugging in our BCs $\{Y(-1) = Y(1) = 0\}$ yields:

$$ Y(-1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=-1} = A \cos(-\beta) + B \sin(-\beta) = 0 $$

$$ Y(1) = (A \cos(\beta y) + B \sin(\beta y))\bigr|_{y=1} = A \cos(\beta) + B \sin(\beta) = 0 $$

Using that $\sin$ is odd, and $\cos$ is even, we have the equations:

$$ A \cos(\beta) - B \sin(\beta) = A \cos(\beta) + B \sin(\beta) = 0 $$

Adding and subtracting these equations, then dividing by $2$, gives us:

$$ A \cos(\beta) = B \sin(\beta) = 0 $$

Now, as $\nexists \beta \in \mathbb{R}$ where $\cos(\beta) = \sin(\beta) = 0 $, and we discard the trivial solution $X(x) \equiv 0$ where $A = B = 0$, we must have either $ A = 0, B \ne 0, \sin(\beta) = 0 \implies \beta = n \pi $, or $ A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$, for $n \in \mathbb{N}$. Because our BC $u(0,y) = 1 - |y|$ is clearly an even function in $y$, we choose the case where $ A \ne 0, B = 0, \cos(\beta) = 0 \implies \beta = (n \pi - \frac{\pi}{2})$. Then our eigenfunction for $Y_n(y)$, and eigenvalues $\lambda_n = \beta_n^2 > 0$, are given by:

$$ n \in \mathbb{N} : \lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2, \phantom{\ } Y_n(y) = \cos(\beta_n y) = \cos((n \pi - \frac{\pi}{2}) y)$$

Next, using that $\lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2$, we solve for $X(x)$ in:

$$ \{X(x) \phantom{\ } | X''(x) - (n \pi - \frac{\pi}{2})^2 X(x) = 0, \phantom{\ } \max_{\{x\}} |X(x)| < \infty \} $$

This ODE has solutions in the form $X(x) = C e^{(n \pi - \frac{\pi}{2}) x} + D e^{-(n \pi - \frac{\pi}{2}) x} $. But, because $ x > 0$, and $ (n \pi - \frac{\pi}{2}) > 0$, we must have that $C = 0$ since we require $ |X(x)| $ to be bounded as $x \rightarrow \infty$. Otherwise: $ C \ne 0 \implies \lim_{x \to \infty} |X(x)| \ge |C e^{(n \pi - \frac{\pi}{2}) x}| \rightarrow \infty $. So each $X_n(x)$ is given by:

$$ n \in \mathbb{N} : X_n(x) = e^{-(n \pi - \frac{\pi}{2}) x} $$

Thus, our eigenvalues are given by:

$$ n \in \mathbb{N} : \lambda_n = (n \pi - \frac{\pi}{2})^2 $$

With corresponding eigenfunctions:

$$ u_n(x,y) = A_n X_n(x) Y_n(y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y), \phantom{\ } A_n \in \mathbb{R} \phantom{\ } \square$$

**Part c.** Write the solution in the form of a series expansion

**Answer**

We have for $n \in \mathbb{N}, A_n \in \mathbb{R}$ the eigenvalues $ \lambda_n = \beta_n^2 = (n \pi - \frac{\pi}{2})^2 $ and eigenfunctions $u_n(x,y) = A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) $. Then the series expansion of our solution $u(x,y)$ is given by:

$$ u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) $$

Applying our final BC: $u(0,y) = 1 - |y|$:

$$ u(0,y) = (\sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y))\bigr|_{x=0} $$

$$ = \sum_{n=1}^{\infty} A_n \cos(\pi(n - \frac{1}{2}) y) = 1 - |y| : \{ -1 < y < 1 \} $$

Proceeding; recall that the set of half-integer cosines:

$$ \{ X_n(x) = \cos(\pi(n - \frac{1}{2}) x) : n \in \mathbb{N} \} $$

$$\text{With inner product: } \langle f,g \rangle = \int_{-1}^{1} f(x)\overline{g(x)} dx $$

forms an orthonormal basis for the space of even functions which converge on $[-1,1]$ in the $L^2$ sense: $ \| f \| = \langle f,f \rangle ^{\frac{1}{2}} < \infty $. Then the projection of $f(y) = 1 - |y| $ onto the function space is given by:

$$ f(y) = 1 - |y| = \sum_{n=1}^{\infty} A_n X_n(y) = \sum_{n=1}^{\infty} \langle f,X_n \rangle \frac{X_n(y)}{\| X_n \| ^2} $$

$$ \implies A_n = \langle f,X_n \rangle \cdot \frac{1}{\| X_n \|^2} = \int_{-1}^{1} f(x)\overline{X_n(x)}dx \cdot 1 $$

$$ = \int_{-1}^{1} (1-|x|)\cos(\pi(n - \frac{1}{2}) x)dx = 2 \int_{0}^{1} (1-x)\cos(\pi(n - \frac{1}{2}) x)dx $$

$$ = \frac{8 - \sin(n \pi)}{(2 n \pi - \pi)^2} = \frac{8}{\pi^2(2 n - 1)^2} = A_n$$

Finally then, our solution $u(x,y)$ is given by the series:

$$ u(x,y) = \sum_{n=1}^{\infty} A_n e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) = \sum_{n=1}^{\infty} \frac{8}{\pi^2(2 n - 1)^2} e^{-(n \pi - \frac{\pi}{2}) x} \cos(\pi(n - \frac{1}{2}) y) \phantom{\ } \blacksquare $$ \\ \\