I'm trying to solve the following PDE, but I got stuck. So I was hoping for some help.
Solve $u_{tt} = 4u_{xx},\hspace{6px} \hspace{6px} x\geq 0 \hspace{6px} \hspace{6px} t\geq 0$
$u(0,t) = g(t)\\u(x,0) = f(x)\\ u_{t}(x,0) = 0$
Assume that f(0) = g(0) = 0
I've tried introducing v(x,t) = u(x,t) - g(t) to get homogeneous boundary condition. This gives me the equation:
$v_{tt} = 4v_{xx} - g''(t)\\ v(0,t) = 0\\v(x,0) = f(x)\\v_{t}(x,0) = -g'(0)$
Which I'm not sure how to solve. I would appreciate if someone could give me a push in the right direction. If the source term g''(t) wasn't there, I would know how to solve it.
Thanks in advance!
A solution to your problem is to subtract some function other than $g(t)$, chosen in such a way that
One solution is to take $v(x,t) = u(x,t) - g\left(t +\frac{x}{2}\right)$. Then, we get the equation $$\left\{\begin{array}{ll} v_{tt} = 4v_{xx} & 0 < x <+\infty\\ v(0,t) = 0 & \\ v(x,0) = f(x) - g\left(\frac{x}{2}\right)\\ v_t(x,0) = -g'\left(\frac{x}{2}\right) \end{array}\right.$$ Then, the odd extension $\tilde{v}$ of $v$ satisfies $$\left\{\begin{array}{l} \tilde{v}_{tt} = 4\tilde{v}_{xx}\\ \tilde{v}(0,t) = 0 & \\ \tilde{v}(x,0) = \tilde{f}(x) - \tilde{g}\left(\frac{x}{2}\right)\\ v_t(x,0) = -\operatorname{sgn}(x)\tilde{g}'\left(\frac{x}{2}\right) \end{array}\right.$$ Whence, by d'Alembert's formula $$\tilde{v}(x,t) = \frac{\tilde{f}(x+2t) -\tilde{g}(x/2+t)+ \tilde{f}(x-2t)-\tilde{g}(x/2-t)}{2} + \frac{1}{4} \int_{x-2t}^{x+2t} -\operatorname{sgn}(s)\tilde{g}'\left(\frac{s}{2}\right)\;ds$$ $$\tilde{v}(x,t) = \frac{\tilde{f}(x+2t) -\tilde{g}(x/2+t)+ \tilde{f}(x-2t)-\tilde{g}(x/2-t)}{2} - \frac{1}{4} \int_{0}^{x+2t} \tilde{g}'\left(\frac{s}{2}\right)\;ds$$ $$- \frac{1}{4}\operatorname{sgn}(x-2t)\int_{x-2t}^0\tilde{g}'\left(\frac{s}{2}\right)\;ds$$ $$\tilde{v}(x,t) = \frac{\tilde{f}(x+2t) + \tilde{f}(x-2t)}{2} - \tilde{g}(x/2+t) - \frac{1-\operatorname{sgn}(x-2t)}{2}\tilde{g}(x/2-t)$$ Thus, we can obtain the piecewise solution for $u$ in terms of $f$ and $g$ by $$u(x,t) = \begin{cases} \frac{f(x+2t) + f(x-2t)}{2} & x\ge 2t\\ \frac{f(x+2t) - f(2t-x)}{2} - g\left(t-\frac{x}{2}\right) & x< 2t \end{cases}$$