MHB Can the Method of Characteristics Solve This PDE Problem?

[evinda]

Hello! (Wave)

I want to solve the following problem:

$$u_x(x,y)+(x+y)u_y(x,y)=0 , x+y>1 \\ u(x,1-x)=f(x), x \in \mathbb{R}$$

How could I do it? Could we apply the method of characteristics? In my lecture notes, there is an example on which this method is applied.

This example is of the form $a(t,x,u) u_x+ b(t,x,u)u_t=c(t,x,u)$.

$$x_t(x,t)-u_x(x,t)=0, x \in \mathbb{R}, t>0 \\ u(x,0)=f(x), x \in \mathbb{R}$$Does it make a difference if the variable is $t$ or $y$ ?

Also, at the beginning, they took: $(x(0),t(0))=(x_0,0)$.

What initial value do we take in this case?

Could we pick $(x(0),y(0))=(x_0,1-x_0)$ ? (Thinking)

 

[chisigma]

Hello! (Wave)

I want to solve the following problem:

$$u_x(x,y)+(x+y)u_y(x,y)=0 , x+y>1 \\ u(x,1-x)=f(x), x \in \mathbb{R}$$

How could I do it? Could we apply the method of characteristics? In my lecture notes, there is an example on which this method is applied.

This example is of the form $a(t,x,u) u_x+ b(t,x,u)u_t=c(t,x,u)$.

$$x_t(x,t)-u_x(x,t)=0, x \in \mathbb{R}, t>0 \\ u(x,0)=f(x), x \in \mathbb{R}$$Does it make a difference if the variable is $t$ or $y$ ?

Also, at the beginning, they took: $(x(0),t(0))=(x_0,0)$.

What initial value do we take in this case?

Could we pick $(x(0),y(0))=(x_0,1-x_0)$ ? (Thinking)

If You apply the method of characteristics to the PDE...

$\displaystyle u_{x} + (x + y)\ u_{y} = 0\ (1)$

... You arrive to write...

$\displaystyle \frac{d y}{d x} = y + x\ (2)$

... the solution of which is...

$ y = c_{1}\ e^{x} - x - 1\ (3)$

... so that is...

$\displaystyle u = g \{ (y + x + 1)\ e^{- x}\}\ (4)$

... where g(*) is an arbitrary function...

At this point g(*) can be found solving the functional equation...

$\displaystyle g \{2\ e^{-x}\} = f(x)\ (5)$ ...

Kind regards

$\chi$ $\sigma$

 

[evinda]

In my lecture notes there is the following example on which we have applied the method of characteristics:

$$u_t+2xu_x=x+u, x \in \mathbb{R}, t>0 \\ u(x,0)=1+x^2, x \in \mathbb{R}$$

$$(x(0), t(0))=(x_0,0)$$

We will find a curve $(x(s), t(s)), s \in \mathbb{R}$ such that $\sigma '(s)=\frac{d}{ds}(u(x(s), t(s))=u_x(x(s), t(s))x'(s)+u_t(x(s), t(s))t'(s)$

$$x'(s)=2x(s), x(0)=x_0 \\ t'(s)=1, t(0)=0 \\ \sigma '(s)=2xu_x+u_t=x(s)+u(s), \sigma(0)=u(x(0), t(0))=1+x_0^2$$

$$\dots \dots \dots \dots \dots$$

$$t(s)=s \\ x(s)=x_0e^{2s} \\ \sigma(s)=x_0 e^{2s}+(1+x_0^2-x_0)e^s$$

If $\overline{s}$ is the value of $s$ such that $(x(\overline{s}), t(\overline{s}))=(x_1, t_1)$ then we have $$x_0e^{2\overline{s}}=x_1 \ , \ \overline{s}=t_1 \\ \Rightarrow x_0=x_1e^{-2t_1}, \overline{s}=t_1$$

So for $s=\overline{s}$ we have $$\sigma(\overline{s})=u(x_1, t_1)=x_1 e^{t_1}+x_1^2e^{-3t_1}-x_1e^{t_1}$$ In our case, what initial value do we have to take at the beginning?

$(x(0), y(0))=(x_0, 1-x_0)$ ?

Or something else? (Thinking)

 

[chisigma]

... in our case, what initial value do we have to take at the beginning?...

$(x(0), y(0))=(x_0, 1-x_0)$?...

Or something else? (Thinking)...

Applying the method of characteristics to the PDE...

$\displaystyle u_{x} + (x + y)\ u_{y} = 0\ (1)$

... we arrive to the general solution...

$\displaystyle u = g \{ (1 + x + y)\ e^{- x} \}\ (2)$

... where g(*) is an arbitrary function...

... in this case the 'contour conditions' [necessari for finding g(*)...] are in the 'unusual form'...

$\displaystyle u(x, 1-x) = f(x), x \in \mathbb{R}\ (3)$

Inserting the condition (3) in (2) You obtain...

$\displaystyle g(2\ e^{- x}) = f(x) \implies g(t) = f( - \ln \frac{t}{2})\ (4)$

... so that the solution is...

$\displaystyle u = f \{ \frac{x - \ln (1 + x + y)}{2} \}\ (5)$

Kind regards

$\chi$ $\sigma$