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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