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**Chapter 5 / Chapter 5.3 Problem 1.1**

« **on:**March 13, 2022, 02:32:37 PM »

For this problem we have the following equations: $u_{xx} + u_{yy} = 0 \space \space (-\infty < x < \infty , y > 0)$ and $u|_{y=0} = f(x)$.

After doing the Fourier Transformation, the equations become $\hat{u}_{yy} - k^2\hat{u} = 0$ and $\hat{u}(k,0) = \hat{f}(k)$.

In my understanding, we should get a general form as $\hat{u}(k,y) = A(k)e^{ky} + B(k)e^{-ky}$ and drop the first term if $k>0$ since as $y \rightarrow \infty$ the term $e^{ky} \rightarrow \infty$, and if $y<0$ we drop the second one.

On the answer provided by Prof. Kennedy it said $\hat{u}(k,y) = \hat{f}(k)e^{-k|y|}$, so I am a little confused since $\hat{u}(k,y) = \hat{f}(k)e^{-|k|y}$ makes more sense to me.

After doing the Fourier Transformation, the equations become $\hat{u}_{yy} - k^2\hat{u} = 0$ and $\hat{u}(k,0) = \hat{f}(k)$.

In my understanding, we should get a general form as $\hat{u}(k,y) = A(k)e^{ky} + B(k)e^{-ky}$ and drop the first term if $k>0$ since as $y \rightarrow \infty$ the term $e^{ky} \rightarrow \infty$, and if $y<0$ we drop the second one.

On the answer provided by Prof. Kennedy it said $\hat{u}(k,y) = \hat{f}(k)e^{-k|y|}$, so I am a little confused since $\hat{u}(k,y) = \hat{f}(k)e^{-|k|y}$ makes more sense to me.