### Author Topic: TT2--P2N  (Read 1820 times)

#### Victor Ivrii

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##### TT2--P2N
« on: March 23, 2018, 06:15:25 AM »
Solve
\begin{align}
&u|_{y=0}=0,\label{2-2a} \\
&(u_y+\alpha u)|_{y=1}=g(x)=\left\{\begin{aligned} &1 &&|x|<1,\\ &0 &&|x|>1,\end{aligned}\right.
\label{2-2b}\\
&\max|u|<\infty. \label{2-3}\end{align}

Hint: Use partial Fourier transform with respect to $x$. Write solution as a Fourier integral without calculating it.

Find restriction to $\alpha$ , so that there will be no singularities.

#### Tristan Fraser

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##### Re: TT2--P2N
« Reply #1 on: March 23, 2018, 08:38:42 PM »
As usual, let's setup our partial fourier transform, shifting from x basis to k basis, i.e.

$$\hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx} f(x) dx$$

Thus equations (1,2,3) become

$$\hat{u_{yy}} - k^2 \hat{u} = 0 , \hat{u(k,0)} = 0 , (\hat{u_y}(k,1) + \alpha \hat{u}(k,1)) = \hat{g}(k)$$

Now, the fourier transform g(x) into $\hat{g}(k)$ is simply:

$\hat{g}(k) = \frac{1}{2\pi} \int_{-\infty}^{infty} e^{-ikx} g(x) dx = \frac{1}{2\pi} \int_{-1}^{1} e^{-ikx} dx = \frac{1}{2\pi} \frac{-e^{-ikx}}{ik} |_{-1}^{1} = \frac{1}{2\pi} \frac{e^{ik} - e^{-ik}}{ik} = \frac{1}{\pi k} sin(k)$

The solution to the differential equation from the fourier transform of (1) is:

$\hat{u} = A(k)e^{|k|y} +B(k)e^{-|k|y}$ ,  applying (2), we need: $-B(k) = A(k).$

$\hat{u} = A(k)e^{|k|y} -A(k)e^{-|k|y} = A(k)\sinh(|k|y)$

Finally, applying the condition (3):

$(A(k)|k| \cosh(|k|) + \alpha A(k)\\sinh(|k|) = \frac{sin(k)}{\pi k}$

So that: $A(k) = \frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1}$

Thus, our solution in x basis is given as:

$f(x) = \int_{-\infty}^{\infty} \hat{f(k)} e^{ikx} dk = \int_{-\infty}^{\infty} (\frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1})\sinh(|k|y)e^{ikx}dk$

« Last Edit: March 25, 2018, 09:16:46 PM by Tristan Fraser »