Solve the boundary value problem

Question

Let $U(x,t)$ be the solution of $$U_{tt} - U_{xx} = 1$$ $x \in\mathbb R, t> 0$ with \begin{cases} U(x,0) = 0\\ U_t(x,0) =0 \end{cases} $x\in\mathbb R$.

Then $U(\frac{1}{2},\frac{1}{2})$ is equal to

1)$\frac{1}{8}$

2)$\frac{-1}{8}$

3)$\frac{1}{4}$

4)$\frac{-1}{4}$

I have tried by laplace transformation, there is some probem

Answer 1

Using Laplace transformation we have: $$U_{tt} - U_{xx} = 1$$

$$→\frac{\partial^2 U(x,t)}{\partial t^2} - \frac{\partial^2 U(x,t)}{\partial x^2} = 1$$ $$ →(s^2U(x,t)-sU(x,0)-U_t(x,0))-\frac{\partial^2 U(x,t)}{\partial x^2} = \frac{1}{s}$$ $$→(s^2U(x,t)-0-0)-\frac{\partial^2 U(x,t)}{\partial x^2} = \frac{1}{s}$$ $$→s^2U(x,t)-\frac{\partial^2 U(x,t)}{\partial x^2} = \frac{1}{s}$$ Form the characteristic equation:$$s^2-λ^2 = 0 → λ= +s,-s, $$ Homogenouse solution:$$U_h(x,s)= A(s)e^{sx} + B(s)e^{-sx}$$ Particular solution:$$U_p(x,s)= (\frac{1}{s^2-D^2})\frac{1}{s}e^{0x}=\frac{1}{s^3}$$ General solution:$$U(x,s)= A(s)e^{sx} + B(s)e^{-sx}+\frac{1}{s^3}$$

Answer 2

Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,

Then $\dfrac{\partial U}{\partial x}=\dfrac{\partial U}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial U}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial U}{\partial p}+\dfrac{\partial U}{\partial q}$

$\dfrac{\partial^2U}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial U}{\partial p}+\dfrac{\partial U}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial U}{\partial p}+\dfrac{\partial U}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial U}{\partial p}+\dfrac{\partial U}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2U}{\partial p^2}+\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}=\dfrac{\partial^2U}{\partial p^2}+2\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}$

$\dfrac{\partial U}{\partial t}=\dfrac{\partial U}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial U}{\partial q}\dfrac{\partial q}{\partial t}=\dfrac{\partial U}{\partial p}-\dfrac{\partial U}{\partial q}$

$\dfrac{\partial^2U}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{\partial U}{\partial p}-\dfrac{\partial U}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial U}{\partial p}-\dfrac{\partial U}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial U}{\partial p}-\dfrac{\partial U}{\partial q}\right)\dfrac{\partial q}{\partial t}=\dfrac{\partial^2U}{\partial p^2}-\dfrac{\partial^2U}{\partial pq}-\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}=\dfrac{\partial^2U}{\partial p^2}-2\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}$

$\therefore\dfrac{\partial^2U}{\partial p^2}-2\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}-\left(\dfrac{\partial^2U}{\partial p^2}+2\dfrac{\partial^2U}{\partial pq}+\dfrac{\partial^2U}{\partial q^2}\right)=1$

$-4\dfrac{\partial^2U}{\partial pq}=1$

$\dfrac{\partial^2U}{\partial pq}=-\dfrac{1}{4}$

$U(p,q)=f(p)+g(q)-\dfrac{pq}{4}$

$U(x,t)=f(x+t)+g(x-t)-\dfrac{(x+t)(x-t)}{4}=f(x+t)+g(x-t)-\dfrac{x^2-t^2}{4}$

$U(x,0)=0$ :

$f(x)+g(x)-\dfrac{x^2}{4}=0$

$f(x)+g(x)=\dfrac{x^2}{4}......(1)$

$U_t(x,t)=f_t(x+t)+g_t(x-t)+\dfrac{t}{2}=f_x(x+t)-g_x(x-t)+\dfrac{t}{2}$

$U_t(x,0)=0$ :

$f_x(x)-g_x(x)=0$

$f(x)-g(x)=c~......(2)$

$\therefore f(x)=\dfrac{x^2}{8}+\dfrac{c}{2}$ , $g(x)=\dfrac{x^2}{8}-\dfrac{c}{2}$

$\therefore U(x,t)=\dfrac{(x+t)^2}{8}+\dfrac{c}{2}+\dfrac{(x-t)^2}{8}-\dfrac{c}{2}-\dfrac{x^2-t^2}{4}=\dfrac{t^2}{2}$

Hence $U\left(\dfrac{1}{2},\dfrac{1}{2}\right)=\dfrac{1}{8}$