Consider the following boundary and initial value problem:
$$\frac{1}{\kappa}u_t-u_{xx}=0,\quad t>0, x>0\tag{1}$$ $$u(0,x)=A,\quad t>0\tag{2}$$ $$u(t,0)=f(x),\quad x\ge 0\tag{3}$$ where $A$ is a constant. Show that the solution to this problem is given by $$u(t,x)=\Phi*\widetilde{f}(x)+A[1-\gamma(\frac{x}{\sqrt{4\kappa t}})]$$ where, $$\widetilde{f}(x)=-f(-x),\quad x<0\tag{4}$$ $$\gamma(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-s^2}ds\tag{5}$$ $$\Phi(x)=\frac{1}{\sqrt{4\pi \kappa t}}exp(\frac{-x^2}{4\kappa t})\tag{6}$$
(4) is the odd extension of $f$, (5) is an error function and (6) has the following relation $$\sqrt{2\pi}\cdot \Phi(x)=\phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\widetilde{\phi}(\xi)e^{i\xi x}d\xi\tag{7}$$ where (7) is the antitransform of $\widetilde{\phi}(\xi)=e^{-\kappa \xi^2 t}$

My attempt was:
Let $F_s(\xi)$ be the sine-transform of $f(x)$ given by $$F_s(\xi)=\int_{0}^{\infty}f(x)sin(x\xi)dx\tag{8}$$ using (4) we get $$F_s(\xi)=\int_{-\infty}^{\infty}\widetilde{f}(x)sin(x\xi)dx\tag{9}$$ $$\Rightarrow F_s(\xi)=\int_{-\infty}^{0}\widetilde{f}(x)sin(x\xi)dx +\int_{0}^{\infty}f(x)sin(\xi x)dx$$
I don't know how to use the Fourier tranform on (9) in order to get $f(x)$ and proceed to finding $u(t,x)$. Any tips?


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