### Author Topic: Problem 2  (Read 9952 times)

#### Calvin Arnott ##### Problem 2
« on: November 12, 2012, 12:17:59 PM »
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 ##### Re: Problem 2
« Reply #1 on: November 12, 2012, 01:48:50 PM »
Nice! BTW, note where approximation is worse. Why here, what do you think?

#### Thomas Nutz

• Full Member
•   • Posts: 26
• Karma: 1 ##### Re: Problem 2
« Reply #2 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?

#### Victor Ivrii ##### Re: Problem 2
« Reply #3 on: November 13, 2012, 06:00:07 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?

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 ##### Re: Problem 2
« Reply #4 on: November 13, 2012, 08:22:01 PM »
edit: I guess I answered my question by reading the above post again.
« Last Edit: November 13, 2012, 08:51:59 PM by Zarak Mahmud »

#### Calvin Arnott ##### Re: Problem 2
« Reply #5 on: November 14, 2012, 04:01:57 PM »
Nice! BTW, note where approximation is worse. Why here, what do you think?

Hmm, well, intuitively the Fourier series representation is quite smooth. Because it's so smooth, we would expect that whenever it tries to approximate a function that is jagged or discontinuous, it should have errors proportional somehow to how rough the approximated function is.

In this case $1-|y|$ is continuous at y=0, and at the points of periodic extension, and the derivative is piecewise continuous at these points, so we have convergence of the Fourier series. But the derivative has a jump discontinuity, so I would expect some kind of analogue to the Gibbs phenomenon, except in the estimation of the derivative of the function instead of the estimation of the function itself, where the derivative "overshoots" the estimated function's derivative, and converges slower here. I'm not sure how exactly to state that more mathematically, or to make that intuition rigorous if it is correct.

#### Victor Ivrii ##### Re: Problem 2
« Reply #6 on: November 14, 2012, 04:28:00 PM »
Nice! BTW, note where approximation is worse. Why here, what do you think?

Hmm, well, intuitively the Fourier series representation is quite smooth. Because it's so smooth, we would expect that whenever it tries to approximate a function that is jagged or discontinuous, it should have errors proportional somehow to how rough the approximated function is.

In this case $1-|y|$ is continuous at y=0, and at the points of periodic extension, and the derivative is piecewise continuous at these points, so we have convergence of the Fourier series. But the derivative has a jump discontinuity, so I would expect some kind of analogue to the Gibbs phenomenon, except in the estimation of the derivative of the function instead of the estimation of the function itself, where the derivative "overshoots" the estimated function's derivative, and converges slower here. I'm not sure how exactly to state that more mathematically, or to make that intuition rigorous if it is correct.

Yes, you are absolutely correct. Consider function s.t. $\int |f^{(s)}(x)|\,dx \le M$. Then $\int f(x)e^{i\omega x}\,dx =O(\omega^{-s})$ as $\omega\to\infty$. Using this we conclude that if a periodic continuation of $f$  obeys this inequality then Fourier coefficients are $O(n^{-s})$ and the larger $s$ is the better it converges.

Further, similar statement holds for subintervals: if $\int_\alpha^\beta |f^{(s)}(x)|\,dx \le M$ then on smaller subinterval $(\alpha',\beta')$, $\alpha <\alpha'<\beta'<\beta$, an error is $O(N^{-s})$.

#### Maxwell Adams ##### Re: Problem 2
« Reply #7 on: November 14, 2012, 09:24:13 PM »
In Calvin's solution, why did we not need to look for zero eigenvalues or negative eigenvalues?

#### Victor Ivrii ##### Re: Problem 2
« Reply #8 on: November 14, 2012, 09:26:56 PM »
In Calvin's solution, why did we not need to look for zero eigenvalues or negative eigenvalues?

Because we already solved this problem ($X''+\lambda X=0$, X(0)=X(l)=0\$) long ago and know e.v. and e.f.