MHB What is the solution for the given problem using the method of characteristics?

[evinda]

Hello! (Wave)

I want to find the solution of the problem
$$(t+u(x,t))u_x(x,t)+tu_t(x,t)=x-t, x \in \mathbb{R}, t>1,$$
$$u(x,1)=1+x$$
using the method of characteristics.

That's what I have tried:

$$(x(0),t(0))=(x_0,1)$$We are looking for a curve $(x(s),y(s)), s \in \mathbb{R}$ such that $\sigma(s)=u(x(s),t(s))$
$$\sigma'(s)=u_x(x(s),t(s))x'(s)+u_t(x(s),t(s))t'(s)$$

We choose $x'(s)=t+u(x,t)=t+ \sigma(s), s \in \mathbb{R}, x(0)=x_0\\t'(s)=t(s), s \in \mathbb{R}, t(0)=1$

Then $\sigma'(s)=x(s)-t(s), s \in \mathbb{R}\\ \sigma(0)=u(x(0),t(0))=u(x_0,1)=1+x_0$$$t'(s)=t(s) \Rightarrow \frac{dt}{ds}t \Rightarrow \frac{dt}{t}=ds \Rightarrow \ln{|t|}=s+c \Rightarrow t=C e^s$$
$$t(0)=1 \Rightarrow C=1$$
$$\Rightarrow t(s)=e^s$$

$x'(s)=t+ \sigma(s)=e^s+u(x(s),e^s) \Rightarrow x(s)=e^s-1+ \int_0^{s} u(x(\xi), e^{\xi})d \xi$

$\sigma'(s)=e^s-1+ \int_0^{s} u(x(\xi), e^{\xi})d \xi-e^s=\int_0^{s} u(x(\xi), e^{\xi})d \xi-1=\int_0^{s} \sigma(\xi) d\xi-1 $

From the Mean Value theorem, we have that there is a $k \in (0,s)$ such that $\sigma(k)=\frac{\int_0^s \sigma(\xi)d\xi}{s}$

So we have that $\sigma'(s)=s \sigma(k)-1 \Rightarrow \sigma(s)=\frac{s^2}{2} \sigma(k)-s+c$

 

[chisigma]

The PDE is of the type 'quasi-linear' and it is written as...

$\displaystyle a(x,t,u)\ u_{x} + b(x,t,u)\ u_{t} = c(x,t,u)\ (1)$

... with the boundary conditions...

$\displaystyle u=u_{0} (s)\ \text{along}\ x=f(s), y=g(s)$

Under appropriate conditions the solving procedure consists in solving the system of ODE...

$\displaystyle \frac{d\ x}{d\ \sigma} = a, \frac{d\ t}{d\ \sigma} = b, \frac{d\ u}{d\ \sigma}= c\ (3)$

... under the conditions...

$\displaystyle x(\sigma=0,\ s) = f(s), t(\sigma=0, s)= g(s), u(\sigma=0, s)= u_{0} (s)\ (4)$

In thi case is...

$\displaystyle a= t+u,\ b=t,\ c=x-t, f(s)= s, g(s)=1, u_{0} = 1 + s$

... so that You can proceed with the first ODE...

$\displaystyle \frac{d\ t}{d\ \sigma}= t, \ t(0,s) = 1 \implies t = e^{\sigma}\ (5)$

... then, taking into account (5), the second ODE...

$\displaystyle \frac{d\ x}{d\ \sigma} = e^{\sigma} + u,\ x(0,s)= s \implies x = u\ \sigma + e^{\sigma} + s - 1\ (6)$

... and then, taking into account (5) and (6), the third ODE...

$\displaystyle \frac{d\ u}{d\ \sigma} = u\ \sigma + s - 1,\ u(0,1)= 1 + s \implies u = (1 + s)\ e^{\frac{\sigma^{2}}{2}} + \sqrt{\frac{\pi}{2}}\ (s-1)\ e^{\frac{\sigma^{2}}{2}}\ \text{erf}\ (\frac{\sigma}{\sqrt{2}})\ (7)$

Now You have the solution in parametric form...

$\displaystyle x = u\ \sigma + e^{\sigma} + s - 1, t = e^{\sigma}, u = (1 + s)\ e^{\frac{\sigma^{2}}{2}} + \sqrt{\frac{\pi}{2}}\ (s-1)\ e^{\frac{\sigma^{2}}{2}}\ \text{erf}\ (\frac{\sigma}{\sqrt{2}})\ (8)$

... and You can obtain u(x,t) in explicit form eliminating $s$ and $\sigma$ from (8)... non a very comfortable task (Thinking)

Kind regards

$\chi$ $\sigma$