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Hello everyone I was doing a bit of homework when I ran into the following problem: PDE: $$u_t= \alpha^2 u_{xx}, \text{for } 0\le x \lt \infty$$ With boundary condition of:

BC: $$u_x(0,t) = 0, \text{for } 0 \le t \lt \infty$$ And Initial Condition: $$ u(x,0) = H(1-x) =\begin{cases} 0, & x \gt 1 \\ 1, & x \le 1 \end{cases}$$

To begin with,I transformed the whole function by applying the Fourier cosine transform, which produce $$\frac{dU}{dt} = \alpha^2 [\frac{2}{\pi}u'(0,t)-\omega^2 U]$$ Next I plugged in the boundary conditions to obtain the final form,: $$\frac{dU}{dt} = -(\alpha \omega)^2U$$ This is of course a simple separable equation which I separated and solved and got: $$U=Ce^{-(\alpha \omega)^2t}$$ Then I also transformed the Initial Condition, $$ \mathscr{F}[H(1-x)] = \frac{2}{\pi}[\int_0^1 (1)cos(\omega x)dx + \int_0^\infty (0)cos(\omega x)dx] = \frac{2}{\pi} \frac{sin(\omega)}{\omega}$$ Putting it all together I get, $$U = \frac{2sin(\omega)}{\omega}e^{-(\alpha \omega)^2t}$$ Finally I apply the inverse cosine transform, although I do not evaluate it and as such, $$u(x,t) = \mathscr{F}[U] = \int_0^\infty \frac{2sin(\omega)}{\omega}e^{-(\alpha \omega)^2t} cos(\omega x)dx$$

Here's my question, how would this problem be done using the Fourier Sine Transform. I checked the answer in the book and it was a bit different but that was because they used the sine transform. Also if anybody can spot errors in the way I solved the problem please let me know. Thank you.

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    $\begingroup$ You wouldn't use the sine transform because of the condition $u_x(0,t)=0$. If you had $u(0,t)=0$, then the sine transform would be appropriate. $\endgroup$ – DisintegratingByParts Feb 22 at 21:24
  • $\begingroup$ I think I may be wrong think about it this way, but what if I apply the Fourier Sine transform to $u_x$ and then from that get $-\omega \mathscr_c {f}$, then I can input the condition that $x=0$. Obviously $\omega \neq 0$ is false. However that must meant that $\mathscr_c{f(0)} = 0$. Plugging this in I should get something similar shouldn't i? $\endgroup$ – Jonathan Aguilera Feb 23 at 22:34

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