Differential equation - Green's Theorem I want to find the solution of the following initial value problem: 
$$u_{tt}(x, t)-u_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$ 
using Green's theorem but I got stuck... 
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I found the following example in my notes: 
$$u_{tt}-c^2u_{xx}=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$ 
 
$$\iint_{\Omega}[u_{tt}(x, t)-c^2u_{xx}(x, t)]dxdt=\iint_{\Omega}f(x, t)dxdt=\int_0^{t_0} \left (\int_{x_0-ct_0+ct}^{x_0+ct_0-ct}f(x, t)dx\right )dt \tag 1$$ 
$$\iint_{\Omega}\left [\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{t}}\right ]dxdt=\int_{\partial{\Omega}}Pdx+Qdt$$ 
$$Q(x, t)=-c^2u_x \\ P(x, t)=-u_t$$ 
$$\iint_{\Omega}\left [u_{tt}(x, t)-c^2u_{xx}(x, t)\right ]dxdt=\int_{\partial{\Omega}}\left [-u_t(x, t)dx-c^2u_x(x, t)dt\right ]=\int_{C_1} [ \ \ ]+\int_{C_2} [ \ \ ]+\int_{C_3} [  \ \ ]$$
$(\int_{C_1} [ \ \ ]=cu(x_0, t_0), \int_{C_2} [ \ \ ]=cu(x_0, t_0), \int_{C_3} [  \ \ ]=0)$ 

$$\int_{C_3}[-u_t(x, 0)dx-c^2u_x(x, 0)dt], \text{ where } u_t(x, 0)=0, u_x(x, 0)=0$$ 
$$C_1: x+ct=x_0+ct_0 \Rightarrow dx+cdt=0$$ 
$$\int_{C_1}(-u_tdx-c^2u_xdt=\int_{C_1}-u_t(-cdt)-c^2u_x\left (-\frac{dx}{c}\right )=\int_{C_1}cu_tdt+cu_xdx=c \int_{C_1}u_tdt+u_xdx=c\int_{C_1}du=c(u(x_0, t_0)-u(x_0+ct_0, 0))\overset{ u(x_0+ct_0, 0)=0 }{ = }cu(x_0, t_0) \ \ \ \ \ (2)$$ 
$$2cu(x_0, t_0)=\int_0^{t_0}\int_{x_0-ct_0+ct}^{x_0+ct_0-ct}f(x, t)dx$$ 
$$u(x_0, t_0)=\frac{1}{2c}\iint_{c(x_0, t_0)}f(x, t)dxdt$$ 
I got stuck at the following: 


*

*Could you explain to me the first graph?? 

*Why are the limits of the integral at the relation $(1)$ the following: $x_0-ct_0+ct$ and $x_0+ct_0-ct$ ??  

*Why does it stand at the relation $(2)$ that $u(x_0+ct_0, 0)=0$ ?? 
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EDIT: 
So, for the problem  $$u_{tt}(x, t)-u_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$ do we have the following?? 
Let $P=(x_0, t_0)$. 
The two characteristics are $x=x_0$ and $x+t=x_0+t_0$. 
The characteristics intersect the line $t=0$ at the points $A(x_0, 0)$ and $B(x_0+t_0, 0)$. 
So, we get the following region of influence: 
 
$$\iint_{\Omega}[u_{tt}(x, t)-u_{xt}(x, t)]dxdt=\iint_{\Omega}f(x, t)dxdt=\int_0^{t_0} \left (\int_{x_0}^{x_0+t_0-t}f(x, t)dx\right )dt$$ 
$$\iint_{\Omega}\left [\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{t}}\right ]dxdt=\int_{\partial{\Omega}}Pdx+Qdt$$ 
$$Q(x, t)=-u_t \\ P(x, t)=-u_t$$ 
$$\iint_{\Omega}\left [u_{tt}(x, t)-u_{xt}(x, t)\right ]dxdt=\int_{\partial{\Omega}}\left [-u_t(x, t)dx-u_t(x, t)dt\right ]=\int_{C_1} [ \ \ ]+\int_{C_2} [ \ \ ]+\int_{C_3} [  \ \ ]$$
 
$$C_1: x+t=x_0+t_0 \Rightarrow dx+dt=0 \Rightarrow dx=-dt$$
$$\int_{C_1} \left [-u_t(x, t)dx-u_t(x, t)dt\right ]=\int_{C_1} \left [u_t(x, t)dt-u_t(x, t)dt\right ]=0$$ 
$$C_2: x=x_0 \Rightarrow dx=0$$ 
$$\int_{C_2} \left [-u_t(x, t)dx-u_t(x, t)dt\right ]= \int_{C_2} \left [-u_t(x, t)dt\right ]=-\int_{C_2} \left [du\right ]=u(x_0, t_0)-u(x_0, 0)=u(x_0, t_0)$$ 
$$C_3: t=0 \Rightarrow dt=0$$ 
$$\int_{C_3} \left [-u_t(x, 0)dx-u_t(x, 0)dt\right ]=0$$ 
So, we have $$u(x_0, t_0)=\int_0^{t_0} \left (\int_{x_0}^{x_0+t_0-t}f(x, t)dx\right )dt$$ 
Is this correct?? Could I improve something?? 
 A: i am going to try. the idea here is pick a point $P = (x_0, t_0)$ and see if you can find the value of $u$ at the point $P.$  you draw the two characteristics $x = \pm c(t-t_0)$ and let them intersect the line $t = 0$ at $A = (x_0 - ct_0, 0), B = (x_0+ct_0, 0).$  the data $u, u_t$ are given on the line $t = 0,$ but only the portion $AB$ has any influence on the value of $u$ at $P.$  the triangular region $ABP$ is called the region of influence i believe.
now you integrate the equation $u_{tt} - c^2 u_{xx} = f$ over $APB$ and use the greens theorem. 
$\bf edit:$
you can factor $u_{tt} - c^2u_{xx}$ in two ways: 
(a) $$\left(\frac{\partial}{\partial t}  - c \frac{\partial}{\partial x}\right) \left(\frac{\partial}{\partial t}  + c \frac{\partial}{\partial x}\right) u = f$$
and (b) switch the order.
you can see that both $f(x - ct)$ and $g(x + ct)$ are  solutions of the homogenous wave equation for any differentiable functions $f$ and $g.$ the lines $x \pm ct = const$ are called characteristics of the wave equation because the information propagates along these lines.
