MHB Differential equation - Green's Theorem

[mathmari]

Hey!

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... I found the following example in my notes:
$$u_{tt}-c^2u_{xx}=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$

View attachment 4134

$$\iint_{\Omega}[u_{tt}(x, t)-c^2u_{xx}(x, t)]dxdt=\iint_{\Omega}f(x, t)dxdt=\int_0^{t_0} \left (\int_{x_0-ct_0+ct}^{x_0+ct_0-ct}f(x, t)dx\right )dt \tag 1$$

$$\iint_{\Omega}\left [\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{t}}\right ]dxdt=\int_{\partial{\Omega}}Pdx+Qdt$$

$$Q(x, t)=-c^2u_x \\ P(x, t)=-u_t$$

$$\iint_{\Omega}\left [u_{tt}(x, t)-c^2u_{xx}(x, t)\right ]dxdt=\int_{\partial{\Omega}}\left [-u_t(x, t)dx-c^2u_x(x, t)dt\right ]=\int_{C_1} [ \ \ ]+\int_{C_2} [ \ \ ]+\int_{C_3} [ \ \ ]$$

$(\int_{C_1} [ \ \ ]=cu(x_0, t_0), \int_{C_2} [ \ \ ]=cu(x_0, t_0), \int_{C_3} [ \ \ ]=0)$

View attachment 4135

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

$$u(x_0, t_0)=\frac{1}{2c}\iint_{c(x_0, t_0)}f(x, t)dxdt$$

I got stuck at the following:

Could you explain to me the first graph?? (Wondering)

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

Why does it stand at the relation $(2)$ that $u(x_0+ct_0, 0)=0$ ?? (Wondering)

 

[I like Serena]

Could you explain to me the first graph?? (Wondering)

Hey mathmari! (Mmm)

The first graph shows how the area $\Omega$ is chosen, which is a triangle.

Since we're talking about the wave equation, it makes sense to have 1 point at $(x_0,t_0)$ where a wave would start.
And to let it expand in both directions with speed $c$ for 2 more points at time $t$.
In the negative direction the wave will travel to $x_0 - c(t - t_0)$, while in the positive direction it will travel to $x_0 + c(t - t_0)$. (Thinking)

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

It's how we set up a line integral.
The left side the triangle is given by $x = x_0 - c(t - t_0)$ with $t_0 \le t \le t_1$ for some unspecified $t_1$. (Nerd)

Why does it stand at the relation $(2)$ that $u(x_0+ct_0, 0)=0$ ?? (Wondering)

Because the boundary condition $u(x, 0)=0$ is given for any $x \in \mathbb R$.
That is, at time $t=0$ there is no wave yet. (Wasntme)

 

[mathmari]

The first graph shows how the area $\Omega$ is chosen, which is a triangle.

Since we're talking about the wave equation, it makes sense to have 1 point at $(x_0,t_0)$ where a wave would start.
And to let it expand in both directions with speed $c$ for 2 more points at time $t$.
In the negative direction the wave will travel to $x_0 - c(t - t_0)$, while in the positive direction it will travel to $x_0 + c(t - t_0)$. (Thinking)

When we don't have the wave equation as for example at the following 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 take the points $x_0 - (t - t_0)$ and $x_0 + (t - t_0)$ ?? (Wondering)

Because the boundary condition $u(x, 0)=0$ is given for any $x \in \mathbb R$.
That is, at time $t=0$ there is no wave yet. (Wasntme)

Oh yes, you're right! (Blush)

 

[mathmari]

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?? The two characteristics are $x=x_0$ and $x+t=x_0+t_0$.

The region $\Omega$ is the triangle:

View attachment 4138

$$\iint_{\Omega}[u_{tt}(x, t)-u_{xt}(x, t)]dxdt=\iint_{\Omega}f(x, t)dxdt=\int_0^{t_0} \left (\int_{x_0}^{x_0+t_0-t}f(x, t)dx\right )dt$$

$$\iint_{\Omega}\left [\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{t}}\right ]dxdt=\int_{\partial{\Omega}}Pdx+Qdt$$

$$Q(x, t)=-u_t \\ P(x, t)=-u_t$$

$$\iint_{\Omega}\left [u_{tt}(x, t)-u_{xt}(x, t)\right ]dxdt=\int_{\partial{\Omega}}\left [-u_t(x, t)dx-u_t(x, t)dt\right ]=\int_{C_1} [ \ \ ]+\int_{C_2} [ \ \ ]+\int_{C_3} [ \ \ ]$$

View attachment 4139

$$C_1: x+t=x_0+t_0 \Rightarrow dx+dt=0 \Rightarrow dx=-dt$$
$$\int_{C_1} \left [-u_t(x, t)dx-u_t(x, t)dt\right ]=\int_{C_1} \left [u_t(x, t)dt-u_t(x, t)dt\right ]=0$$

$$C_2: x=x_0 \Rightarrow dx=0$$
$$\int_{C_2} \left [-u_t(x, t)dx-u_t(x, t)dt\right ]= \int_{C_2} \left [-u_t(x, t)dt\right ]=-\int_{C_2} \left [du\right ]=u(x_0, t_0)-u(x_0, 0)=u(x_0, t_0)$$

$$C_3: t=0 \Rightarrow dt=0$$
$$\int_{C_3} \left [-u_t(x, 0)dx-u_t(x, 0)dt\right ]=0$$

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?? (Wondering) Could I improve something?? (Wondering)