Method of characteristics for linear PDE's (variable coefficients)

In summary, the conversation revolves around the solution for the equation -yu_x + xu_y = u and the confusion about the solution of the first characteristic being x(t,s) = f_1(s)sin(t) + f_2(s)cos(t). The expert explains that the characteristics are a system of ODEs and that x and y are not constants, leading to a solution of xtt = -x.
  • #1
Defconist
7
0
I was going through an inroductory book on PDE's and at one point they proceed with little show of work. I have problem with equation [itex] -yu_x + xu_y = u [/itex].
Characteristics for this equation are [itex] x_t = -y, y_t = x, u_t = u [/itex].

So far it is clear, but now books states that solution of first characteristic is [itex] x(t,s) = f_1(s)sin(t) + f_2(s)cos(t) [/itex], which is perplexing to me, I would just integrate righthand side treating x or y as constants (we are integrating with respect to t). Any suggestion?
 
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  • #2
Welcome to PF!

Defconist said:
Characteristics for this equation are
[tex] x_t = -y, y_t = x, u_t = u [/tex].

So far it is clear, but now books states that solution of first characteristic is [itex] x(t,s) = f_1(s)sin(t) + f_2(s)cos(t) [/itex], which is perplexing to me, I would just integrate righthand side treating x or y as constants (we are integrating with respect to t). Any suggestion?

Hi Defconist! Welcome to PF! :smile:

Nooo … x and y depend on t, so if you vary t, then you must vary x and y … they're not constants!

Hint: xt = -y, yt = x,

means that xtt = -x. :smile:
 
  • #3
Oh, I get it. It is a system od ODE's because the in y the second equation is the same as y in the first one... It's easy to see why I missed that. It is a possibility I feared from the very beginning. Anyway, thanks for getting me from this predicament. :)
 

1. What is the method of characteristics for linear PDE's (variable coefficients)?

The method of characteristics is a mathematical technique used for solving linear partial differential equations with variable coefficients. It involves transforming the PDE into a system of ordinary differential equations and then using the characteristics of the PDE to solve the system.

2. What types of PDE's can be solved using the method of characteristics?

The method of characteristics can be used to solve first-order linear PDE's with variable coefficients. It is not applicable to higher-order PDE's or nonlinear PDE's.

3. How does the method of characteristics work?

The method of characteristics works by transforming the PDE into a system of ordinary differential equations using a change of variables. The equations are then solved using the characteristics of the PDE, which are curves that represent the evolution of the solution over time.

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

The method of characteristics is particularly useful for solving PDE's with variable coefficients because it simplifies the problem by transforming it into a system of ordinary differential equations. It also allows for the use of initial or boundary conditions to find a specific solution.

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

While the method of characteristics is a powerful tool for solving linear PDE's with variable coefficients, it has its limitations. It is not applicable to nonlinear PDE's or higher-order PDE's. It also requires the PDE to be written in a specific form, which may not always be possible.

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