Solution of hyperbolic equation

[mathmari]

Hey!

I have to solve the following pde:

$$u_{tt}(x, t)-u_{xt}(x, t)=0, x \in \mathbb{R}, t>0 \\ u(x, 0)=f(x), x \in \mathbb{R} \\ u_t(x, 0)=g(x), x \in \mathbb{R}$$

I have done the following:

$$\Delta(x, t)=1>0$$

so it is an hyperbolic equation.

We want to write the equation into the canonical form, which is of the form $u_{\xi \eta}=D(\xi, \eta, u, u_{\xi}, u_{\eta})$.

$u(x, t)=U(\xi, \eta)$

$\xi=at+bx \\ \eta=ct+dx$

$$u_t=aU_{\xi}+cU_{\eta} \\ u_{tt}=a^2U_{\xi \xi}+2acU_{\xi \eta}+c^2U_{\eta \eta} \\ u_{xt}=u_{tx}abU_{\xi \xi}+adU_{\xi \eta}++cbU_{\eta \xi}+cdU_{\eta \eta}$$

$$u_{tt}-u_{xt}=0 \Rightarrow (a^2-ab)U_{\xi \xi}+(2ac-ad-cb)U_{\xi \eta}+(c^2-cd)U_{\eta \eta}=0$$

Since it is hyperbolic, it has to stand that $$a^2-ab=0 \text{ and } c^2-cd=0 \\ \Rightarrow \left (a=0 \text{ or } a=b\right ) \text{ and } \left (c=0 \text{ or } c=d\right )$$

For these values we get $0 \cdot U_{\xi \eta }=0$, or not?? (Wondering)

Have I done something wrong?? (Wondering)

 

[chisigma]

Hey!

I have to solve the following pde:

$$u_{tt}(x, t)-u_{xt}(x, t)=0, x \in \mathbb{R}, t>0 \\ u(x, 0)=f(x), x \in \mathbb{R} \\ u_t(x, 0)=g(x), x \in \mathbb{R}$$

I have done the following:

$$\Delta(x, t)=1>0$$

so it is an hyperbolic equation.

We want to write the equation into the canonical form, which is of the form $u_{\xi \eta}=D(\xi, \eta, u, u_{\xi}, u_{\eta})$.

$u(x, t)=U(\xi, \eta)$

$\xi=at+bx \\ \eta=ct+dx$

$$u_t=aU_{\xi}+cU_{\eta} \\ u_{tt}=a^2U_{\xi \xi}+2acU_{\xi \eta}+c^2U_{\eta \eta} \\ u_{xt}=u_{tx}abU_{\xi \xi}+adU_{\xi \eta}++cbU_{\eta \xi}+cdU_{\eta \eta}$$

$$u_{tt}-u_{xt}=0 \Rightarrow (a^2-ab)U_{\xi \xi}+(2ac-ad-cb)U_{\xi \eta}+(c^2-cd)U_{\eta \eta}=0$$

Since it is hyperbolic, it has to stand that $$a^2-ab=0 \text{ and } c^2-cd=0 \\ \Rightarrow \left (a=0 \text{ or } a=b\right ) \text{ and } \left (c=0 \text{ or } c=d\right )$$

For these values we get $0 \cdot U_{\xi \eta }=0$, or not?? (Wondering)

Have I done something wrong?? (Wondering)

A possible approach to the second order PDE...

$$u_{tt}(x, t)-u_{xt}(x, t)=0, x \in \mathbb{R}, t>0 \\ u(x, 0)=f(x), x \in \mathbb{R} \\ u_t(x, 0)=g(x), x \in \mathbb{R}\ (1)$$

... may be the following: setting $\displaystyle u_{t} (x,t) = v (x,t)$ You arrive to write...

$$v_{t} - v_{x}=0, x \in \mathbb{R}, t>0 \\ v(x, 0)=g(x), x \in \mathbb{R}\ (2)$$

... the solution of which [though method of characteristics...] is easy enough...

$\displaystyle \frac{d t}{d x} = 1 \implies t = c_{1} - x \implies c_{1} = t + x \implies v = g(t + x)\ (3)$

At this point You have to solve the first order PDE...

$$u_{t} (x,t) = g(t + x), x \in \mathbb{R}, t>0 \\ u(x, 0)=f(x), x \in \mathbb{R}\ (4)$$

Kind regards

$\chi$ $\sigma$