I am asked to use the Laplace transform to solve the initial value problem for the advection-diffusion equation:
$$u_{t}(x,t)-u_{x}(x,t)=u_{xx}(x,t)$$ with $u(x,0)=\cos(2x)$
Taking the Laplace transform of this I got:
$$s\widetilde{u}(x,s)-u(x,0)-\widetilde{u}_{x}(x,s) = \widetilde{u}_{xx}(x,s)$$
$$ \implies \widetilde{u}_{xx}(x,s) + \widetilde{u}_{x}(x,s) - s\widetilde{u}(x,s) = -\cos(2x) $$
And I found the auxiliary equation to be: $λ^2+λ-s$
and by completing the square I got the general solution to be:
$$\widetilde{u}(x,s)= A(s)e^{[(s+1/4)^{1/2}-1/2]x} + B(s)e^{[-(s+1/4)^{1/2}-1/2]x} +\cos(2x)$$
I'm not sure about this but I think we assume that the equation is bounded so $A$ and $B$ are equivalent to $0$. It does not say this in the question so I may be wrong.
I'm not sure what to do after this point, I'm assuming we have to use the Laplace transform but I think I'm missing a step or 2. Would really appreciate your help.
Note that
$$ \widetilde{u}_{xx}(x,s) + \widetilde{u}_{x}(x,s) - s\widetilde{u}(x,s) = -\cos(2x) $$
is a linear non homogeneous DE which solution can be represented as
$$ \widetilde{u}(x,s) = \widetilde{u}_h(x,s)+\widetilde{u}_p(x,s) $$
The homogeneous solution is
$$ \widetilde{u}_h(x,s)=A(s)e^{[(s+1/4)^{1/2}-1/2]x} + B(s)e^{[-(s+1/4)^{1/2}-1/2]x} $$
and the particular is
$$ \widetilde{u}_p(x,s) = \frac{4 ((s + 4)\cos (2 x)-2 \sin (2 x))}{\left((9+2s)^2-(4s+1)\right)} $$
obtained after the substitution of $\widetilde{u}_p(x,s) = C_1\cos(2x)+C_2\sin(2x)$ into the DE.