What is the Laplace Tansform of $u_{xt}$?

Question

I need to solve a partial differential equation which includes $u_{xt}$, that is the derivated function of $u(x,t)$ with respect to both $x$ and $t$ but couldn't find any formula about laplace transformation of this function. The question is; $$ u_{xt}=cos(2t)\quad u(x,0)=x, \quad u(0,t)=0 $$ I tried to solve it using Laplace Substitution Method; $$ \frac{\partial^2 u}{\partial x \partial t} =\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}\right) $$ I assumed; $$ \frac{\partial u}{\partial t} =G $$ Now the equation becomes; $$ \frac{\partial G}{\partial x} = cos(2t) $$ By taking laplace transform of the equation with respect to x, I obtained; $$ sG(s,t)-G(0,t)=L_x(cos(2t)) $$ $$ G(s,t)=\frac{G(0,t)}{s}+ \frac{L_x(cos(2t))}{s} $$ $$ G(s,t)=\frac{u_t(0,t)}{s}+ \frac{L_x(cos(2t))}{s} $$ $$ G(s,t)=\frac{u_t(0,t)}{s}+ \frac{\frac{cos(2t)}{s}}{s} $$ $$ G(s,t)=\frac{u_t(0,t)}{s}+ \frac{cos(2t)}{s^2} $$ By taking the inverse laplace transformation, the equation becomes; $$ G(x,t)=L_x^{-1}\left(\frac{u_t(0,t)}{s}\right) + xcos(2t) $$ $$ \frac{\partial u(x,t)}{\partial t}=u_t(0,t)+ xcos(2t) $$ By taking laplace transform of the equation with respect to t, I obtained; $$ sU(x,s)-u(x,0)=L_t\left(u_t(0,t)\right)+\frac{xs}{s^2+4} $$ By putting initial condition of $u(x,0)=x$ in to the equation; $$ sU(x,s)=L_t\left(u_t(0,t)\right)+\frac{xs}{s^2+4}+x $$ $$ U(x,s)=\frac{L_t\left(u_t(0,t)\right)}{s}+\frac{xs}{s^3+4s}+\frac{x}{s} $$ By taking inverse laplace transform, I obtained; $$ u(x,t)=L_t^{-1}\left(\frac{L_t\left(u_t(0,t)\right)}{s}\right)+\frac{xsin(2t)}{2}+x $$ But the boundary condition $u_t(0,t)$ is not given. From the solution by direct integrations, I know that $u_t(0,t)=0$ but I don't know how do I prove that in this problem.

Answer 1

Ok, I think I figured it out. Since $u(0,t)=0$, $u_t(0,t)$ also equals to zero. So, the final answer is; $$ u(x,t)=\frac{xsin(2t)}{2}+x $$