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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: $$u_{tt}-c^2u_{xx}=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$

enter image description here

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

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

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

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

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

enter image description here

$$\int_{C_3}[-u_t(x, 0)dx-c^2u_x(x, 0)dt], \text{ where } u_t(x, 0)=0, u_x(x, 0)=0$$

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

$$2cu(x_0, t_0)=\int_0^{t_0}\int_{x_0-ct_0+ct}^{x_0+ct_0-ct}f(x, t)dx$$

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

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:

enter image description here

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

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

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

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

enter image description here

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

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

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

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??

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1 Answer 1

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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.

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  • $\begingroup$ How do we get that $f(x-ct)$ and $g(x+ct)$ are the solutions of the homogeneous wave equation?? I got stuck right now... Are the lines $x \pm ct=const$ the characteristics because the solution of the homogeneous wave equation is of the form $f(x-ct)$ and $g(x+ct)$ ?? @abel $\endgroup$
    – Mary Star
    Mar 22, 2015 at 1:45
  • $\begingroup$ @MaryStar, you are right about the reason for the chars. now $\frac{\partial}{\partial t} - c \frac{\partial}{\partial x}u = 0 \implies u = f(x+ ct)$ because you get $f'(x+ct) \times c - c f'(x + ct) = 0$ by chain rule. $\endgroup$
    – abel
    Mar 22, 2015 at 2:01
  • $\begingroup$ How can we find the characteristics at the following problem?? How can we factor the equation?? $$u_{tt}(x, t)-u_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$ Is it maybe as followed?? $$\frac{\partial}{\partial{t}} \left (\frac{\partial}{\partial{t}}-\frac{\partial}{\partial{x}}\right )=f$$ So, are the characteristics $x=t_0$ and $x=t_0-t$ ?? @abel $\endgroup$
    – Mary Star
    Mar 22, 2015 at 11:12
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    $\begingroup$ that is right @MaryStar. $\endgroup$
    – abel
    Mar 22, 2015 at 12:13
  • $\begingroup$ I edited my initial post... I added what I have done for the initial problem... Could you take a look at it and tell me if I have done something wrong?? @abel $\endgroup$
    – Mary Star
    Mar 22, 2015 at 15:16

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