Solving U_x+U_y=1 with BC U(y,y/2)=y

  • Thread starter Thread starter xdrgnh
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves solving the partial differential equation U_x + U_y = 1 with the boundary condition U(y, y/2) = y. The discussion centers around the methods of finding the solution and verifying its correctness.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of solving the homogeneous and non-homogeneous parts of the equation. There are questions about the validity of the derived solution and its alignment with the boundary condition.

Discussion Status

Some participants express confidence in the solution derived, while others remain cautious about the simplicity of the result. There is a mix of verification and reassurance regarding the correctness of the solution.

Contextual Notes

Participants note the boundary condition and its implications for the solution, as well as the nature of the PDE being solved. There is an underlying concern about the elegance of the solution and its potential implications for correctness.

xdrgnh
Messages
415
Reaction score
0

Homework Statement



U_x+U_y=1 with boundary condition U(y,y/2)=y

Homework Equations

The Attempt at a Solution



Well first I solved the homogeneous solution and got ax-ay where a is just a constant. Then for the non homogeneous solution I got ax+(1-a)y. After adding them both together and pluggin in the boundary condition I got U=x+(0)y? Is that right?
 
Physics news on Phys.org
xdrgnh said:

Homework Statement



U_x+U_y=1 with boundary condition U(y,y/2)=y



Homework Equations




The Attempt at a Solution



Well first I solved the homogeneous solution and got ax-ay where a is just a constant. Then for the non homogeneous solution I got ax+(1-a)y. After adding them both together and pluggin in the boundary condition I got U=x+(0)y? Is that right?

It's easy enough to check it yourself. Does your solution ##U(x,y)=x## satisfy ##U_x+U_y = 1## and ##U(y,\frac y 2) = y##?
 
It does. But I'm suspicious whenever a solution to a PDE comes out that nice.
 
Hey, if you have the solution you have the solution. There's nothing left worry about. Good going.
 

Similar threads

Use Wronskian method in solving the given second order differential equation
  • · Replies 7 ·
Replies
7
Views
2K
Conceptual question with first-order linear PDEs.
  • · Replies 3 ·
Replies
3
Views
2K
Solving differential equation (t^2-1)y''-6y=1
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Helmholtz Equation in Cartesian Coordinates
  • · Replies 11 ·
Replies
11
Views
4K
How Do You Solve the Differential Equation dy/dx = 1 - y^2?
  • · Replies 12 ·
Replies
12
Views
3K
Solving Partial Differential Equation
  • · Replies 2 ·
Replies
2
Views
2K
Solve (t^2-1)y'' +4ty'+2y=6t, given two particular solutions
  • · Replies 2 ·
Replies
2
Views
1K
Mr. Tenenbaum's and Prof. Mattuck's advice not working (ODE)
  • · Replies 3 ·
Replies
3
Views
2K
Solving Harmonic Oscillation w/ BC y(1)=B
  • · Replies 2 ·
Replies
2
Views
1K
How should I show that ## B ## is given by the solution of this?
  • · Replies 2 ·
Replies
2
Views
1K