Can the Method of Characteristics Solve This PDE Problem?

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In summary, the conversation discusses solving a partial differential equation using the method of characteristics. The general solution is given as $u=g\{(1+x+y)e^{-x}\}$, where $g(*)$ is an arbitrary function. The contour conditions are in an unusual form, and the solution is found to be $u=f\{\frac{x-\ln(1+x+y)}{2}\}$.
  • #1
evinda
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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)
 
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  • #2
evinda said:
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

$\chi$ $\sigma$
 
  • #3
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)
 
  • #4
evinda said:
... 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

$\chi$ $\sigma$
 

FAQ: Can the Method of Characteristics Solve This PDE Problem?

1. What is the method of characteristics?

The method of characteristics is a mathematical technique used to solve partial differential equations. It involves transforming the original partial differential equation into a system of ordinary differential equations, which can then be solved using standard techniques.

2. When is the method of characteristics used?

The method of characteristics is typically used to solve first-order, linear partial differential equations with constant coefficients. It is also commonly used to solve problems involving conservation laws, such as in fluid dynamics and traffic flow.

3. How does the method of characteristics work?

The method of characteristics works by finding curves, known as characteristic curves, along which the solution to the partial differential equation is constant. By solving for these curves, the original partial differential equation can be transformed into a system of ordinary differential equations, which can then be solved using standard techniques.

4. What are the advantages of using the method of characteristics?

The method of characteristics has several advantages, including its ability to solve first-order, linear partial differential equations with constant coefficients and its applicability to problems involving conservation laws. It also provides a more intuitive understanding of the solution and can be applied to a wide range of physical and engineering problems.

5. Are there any limitations to the method of characteristics?

Yes, the method of characteristics is limited to solving first-order, linear partial differential equations with constant coefficients. It also requires the existence of characteristic curves, which may not always be present in more complex problems. Additionally, it can be computationally expensive for higher-dimensional problems.

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