APM346-2012 > Home Assignment Y

Problem 2

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

Victor Ivrii:
Nice! BTW, note where approximation is worse. Why here, what do you think?

Thomas Nutz:
Ok I did
$$
u(x,y)=\int_0^{\infty}\hat{u}(\omega,y)\sin(\omega x)d\omega
$$
and took the resulting eigenvalue problem as
$$
\hat{u}_{yy}-\omega^2\hat{u}=0
$$
which yields eigenvalues $\omega=i\pi n$
and eigenfunctions
$$
\hat{u}(\omega,y)=A(\omega) e^{i \pi n y}+B(\omega)e^{-i \pi n y}
$$
Is that wrong?

Victor Ivrii:

--- Quote from: Thomas Nutz on November 13, 2012, 05:21:52 PM ---Ok I did
$$
u(x,y)=\int_0^{\infty}\hat{u}(\omega,y)\sin(\omega x)d\omega
$$
and took the resulting eigenvalue problem as
$$
\hat{u}_{yy}-\omega^2\hat{u}=0
$$
which yields eigenvalues $\omega=i\pi n$
and eigenfunctions
$$
\hat{u}(\omega,y)=A(\omega) e^{i \pi n y}+B(\omega)e^{-i \pi n y}
$$
Is that wrong?

--- End quote ---

Well, it would be a correct approach if conditions were like this
\begin{align*}
&u(0,y)=0,\\
&u(x,-1)=g(x),\\
&u(x,1)=h(x)
\end{align*}
Then you would get
\begin{align*}
&\hat{u}_{yy}-\omega^2\hat{u}=0,\\
&\hat{u}(\omega,-1)=\hat{g}(\omega),\\
&\hat{u}(\omega,1)=\hat{h}(\omega)
\end{align*}
and you would solve BVP for ODE and make inverse sin-FT.

But you have different problem
\begin{align*}
&u(0,y)=f(y),\\
&u(x,-1)=0,\\
&u(x,1)=0.
\end{align*}

Your third equation comes from nowhere and is completely wrong.

Zarak Mahmud:
edit: I guess I answered my question by reading the above post again.

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