I am currently deriving the solution formula for Dirichlet problem of a boundary source on the half-line:
$$v_t-kv_{xx}=f(x, t), \quad 0<x<\infty, \quad t>0$$ given that $v(0, t)=h(t)$, and $v(x, 0)=\phi(x)$.
I know the general solution formula for the homogeneous boundary condition. Following the hints given in the lecture notes, I substituted as $V(x, t)=v(x, t)-h(t)$. Then, $$V_t-kV_{xx}=f(x, t)-h'(t), \quad 0<x<\infty, \quad t>0$$ with the boundary/initial conditions as $V(0, t)\equiv0$, $V(x, 0)=\phi(x)-h(0)$. I then used the odd extension as $$\tilde{\phi}(x)=\phi(x)-h(0) \quad \text{if } x>0$$ $$\tilde{\phi}(x)=-\phi(-x)+h(0) \quad \text{if } x<0$$ and $\tilde{\phi}(0)=0$, and consider the Cauchy problem on the whole real line: $$u_t-ku_{xx}=f(x, t)-h'(t), \quad -\infty<x<\infty, \quad t>0$$ with the boundary/initial conditions as $u(x, 0)=\tilde{\phi}(x)$. We see that $V(x, t):= u(x, t)|_{x\geq 0}$ must be the unique solution by the maximum principle, and it satisfies the initial/boundary conditions, too. By the solution formula to the IVP of $u(x, t)$, we get $$V(x, t)=\int_{-\infty}^{\infty}S(x-y, t)\tilde{\phi}(y)dy+\int_{0}^{\infty}\int_{-\infty}^{\infty}S(x-y, t-s)(f(y, s)-h'(s))dyds \quad (x\geq0) \\=\int_{-\infty}^{0}S(x-y, t)(-\phi(-y)+h(0))dy+\int_{0}^{\infty}S(x-y, t)(\phi(y)-h(0))dy+\int_{0}^{\infty}\int_{-\infty}^{\infty}S(x-y, t-s)(f(y, s)-h'(s))dyds \quad \\ = -\int_{0}^{\infty}S(x+y, t)(\phi(y)-h(0))dy + \int_{0}^{\infty}S(x-y, t)(\phi(y)-h(0))dy+\int_{0}^{\infty}\int_{-\infty}^{\infty}S(x-y, t-s)(f(y, s)-h'(s))dyds$$ We can retrieve $v(x, t)$ back by $v(x, t)=V(x, t)+h(t)$. I have several questions about this derivation.
- Is this correct?
- Can some simplifications be made to the solution?
- Is there another approach? (I think I can decompose the system into two systems(a system with homogeneous boundary condition and no source, and a system with nonhomogeneous boundary condition and a source, but I'm not sure what each system should look like.)