I know how to solve heat equation where it's like $u_t=k\cdot u_{xx}$ (using Fourier Transform or using Separation of Variables) but this exercise is really difficult for me.
I have this:
$$u_t(x,t)=k \cdot u_{xx}(x,t)-a\cdot k \cdot u(x,t)$$
$$u_x(0,t)=0$$
$$u(x,0) = f(x)$$
with $x>0, t>0$ and $a, k$ are positive constants.
I have to find $u(x,t)$ and propose a possible $f(x)$
Any help? Thanks
I was told I cannot use Fourier Transform, I have to use Fourier Cosine Transform, and I don't know why
Best Answer
Indeed, this problem should be solved with Fourier cosine transform rather than Fourier transform. If you use Fourier transform, you won't be able to take the b.c. at $x=0$ into consideration, and the equation will then be solved in $-\infty<x<\infty$ with implicit b.c.s at $\pm\infty$. (For this part check this post for more information. )
The key point is, when $f(\infty)=0$, Fourier cosine transform
$$ \mathcal{F}_t^{(c)}[f(t)](\omega)=\sqrt{\frac{2}{\pi }}\int _0^{\infty } f(t) \cos (\omega t) d t $$
has the following property:
$$ \mathcal{F}_t^{(c)}\left[f''(t)\right](\omega)=-\omega^2 \mathcal{F}_t^{(c)}[f(t)](\omega)-\sqrt{\frac{2}{\pi }} f'(0) $$
You can easily verify this property using integration by parts.
So, by applying this property on your equation, we have
$$ \mathcal{F}_x^{(c)}\left[u^{(1,0)}(t,x)\right](\omega )=k \left(-\omega ^2 \mathcal{F}_x^{(c)}[u(t,x)](\omega )-\sqrt{\frac{2}{\pi }} u^{(0,1)}(t,0)\right)-a k \mathcal{F}_x^{(c)}[u(t,x)](\omega ) $$
Substitute the b.c. into the equation, we obtain a simple initial value problem (IVP) of linear ODE:
$$U'(t)=-a k U(t)-k \omega ^2 U(t)$$ $$U(0)=F$$
where $U(t)=\mathcal{F}_x^{(c)}[u(t,x)](\omega )$, $F=\mathcal{F}_x^{(c)}[f(x)](\omega)$.
If you have difficulty in solving the IVP, check the wikipedia page). Anyway, we can easily find its solution:
$$U(t)=F e^{-k t \left(a + \omega ^2\right)}$$
The last step is to transform back with inverse Fourier Cosine transform
$${\mathcal{F}_\omega^{(c)}}^{-1}[F(\omega)](t)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } F(\omega ) \cos (t \omega) \, d\omega $$
and the solution is
$$ u(t,x)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } e^{-k t \left(a + \omega ^2\right)} \mathcal{F}_x^{(c)}[f(x)](\omega ) \cos (\omega t) \, d\omega $$
Notice this solution is probably the same as the one given by doraemonpaul. I guess he has just chosen a different convention for Fourier parameters.