PDE Characteristic Method?

In summary, the conversation discusses solving a differential equation with multiple variables and the use of substitutions to simplify the problem. After trying various substitutions, it is determined that the expression (du/dt) = s^2 can be directly integrated, with s being the other coordinate. However, it is noted that the initial condition is not satisfied.
  • #1
144
2
Homework Statement
u_x + x*u_y = (y-1/2x^2)^2, u(0,y) = e^y
Relevant Equations
du/dt = du/dx*dx/dt + du/dy*dy/dt
dx/dt =1, x(0,s)=0, dy/dt=x, y(0,s) = s, du/dt=(y-1/2x^2)^2, u(0,s)=e^s

I did well at the beginning to get x(t,s) =t and y(t,s)=1/2t^2 + s, but got stuck with the du/dt part.

You can sub in x=t and y=1/2t^2 +s for x and y to get du/dt = s^2. But that's still three variables, and I can't see any obvious substitutions.

This is a regular diffy Q course, so I don't have a lot of knowledge of advanced techniques, and this was just thrown in, so I'm a little unsure how to proceed.
 
Physics news on Phys.org
  • #2
Also, I tried u=e^s => s=ln(u), so du/dt = (ln(u))^2, but that gives me an Ei function, so I feel like that can't be correct.
 
  • #3
The Head said:
You can sub in x=t and y=1/2t^2 +s for x and y to get du/dt = s^2.

The expression
$$
\frac{\partial u}{\partial t} = s^2
$$
can be directly integrated, ##s## is the other coordinate - not an unknown to be solved for.

Edit: Missed the square in the original PDE.
 
Last edited:
  • Like
Likes The Head
  • #4
Orodruin said:
The expression
$$
\frac{\partial u}{\partial t} = s^2
$$
can be directly integrated, ##s## is the other coordinate - not an unknown to be solved for.

Edit: Missed the square in the original PDE.

Ohh, OK gotcha. So it can just be s^2 * t = u(s,t)? Thanks so much for your reply.
 
  • #5
The Head said:
Ohh, OK gotcha. So it can just be s^2 * t = u(s,t)? Thanks so much for your reply.
No, but almost. Your function does not satisfy the initial condition.
 
  • Like
Likes The Head
  • #6
Orodruin said:
No, but almost. Your function does not satisfy the initial condition.
Right, of course! Thank you.
 

What is the PDE Characteristic Method?

The PDE Characteristic Method is a mathematical technique used to solve partial differential equations (PDEs). It involves transforming the PDE into a system of ordinary differential equations (ODEs) by introducing new variables known as characteristics. These characteristics represent the direction in which the solution to the PDE changes most rapidly.

How does the PDE Characteristic Method work?

The PDE Characteristic Method works by first identifying the characteristics of the PDE, which are determined by the coefficients of the highest order derivatives. These characteristics are then used to transform the PDE into a system of ODEs, which can be solved using standard techniques such as separation of variables or integrating factors. The solutions to the ODEs are then combined to obtain the general solution to the PDE.

What types of PDEs can be solved using the PDE Characteristic Method?

The PDE Characteristic Method can be used to solve linear first and second order PDEs with constant coefficients. It is particularly useful for solving hyperbolic and parabolic PDEs, as these types of equations have well-defined characteristics. However, it may not be applicable to all types of PDEs, such as elliptic PDEs.

What are the advantages of using the PDE Characteristic Method?

One advantage of the PDE Characteristic Method is that it can provide an explicit solution to a PDE, which can be easier to interpret and work with compared to implicit solutions. Additionally, it can be used to solve PDEs with non-constant coefficients, as long as the characteristics can still be identified. This method can also be more efficient than other numerical methods for certain types of PDEs.

Are there any limitations to the PDE Characteristic Method?

While the PDE Characteristic Method can be a powerful tool for solving certain types of PDEs, it does have some limitations. It may not be applicable to all types of PDEs, and it may not always produce a unique solution. Additionally, it may not be suitable for solving PDEs with highly irregular or discontinuous solutions. In these cases, other numerical methods may be more appropriate.

Suggested for: PDE Characteristic Method?

Replies
7
Views
467
Replies
2
Views
696
Replies
11
Views
643
Replies
3
Views
543
Replies
1
Views
902
Replies
2
Views
669
Replies
3
Views
1K
Back
Top