Solution given by sum of functions on a PDE

  • Context: MHB 
  • Thread starter Thread starter Markov2
  • Start date Start date
  • Tags Tags
    Functions Pde Sum
Click For Summary
SUMMARY

The discussion focuses on solving the partial differential equation (PDE) given by $u_t + u_x = g(x)$ with initial condition $u(x,0) = f(x)$. The solution is expressed as $u(x,t) = f(x-t) + \sqrt{2\pi}(g*h)(x)$, where $h(x) = \chi_{[0,t]}(x)$. The Fourier transform is applied, leading to the equation $\frac{\partial U}{\partial t} + iwU = U(g)$, with the initial condition $U(u)(w,0) = U(f)$. The main challenge discussed is correctly applying the initial condition to the transformed equation.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with Fourier transforms
  • Knowledge of convolution operations
  • Basic concepts of initial value problems
NEXT STEPS
  • Study the application of Fourier transforms in solving PDEs
  • Learn about convolution and its properties in the context of functions
  • Explore initial value problems and their solutions using Fourier methods
  • Investigate the role of the Dirichlet and Neumann conditions in PDEs
USEFUL FOR

Mathematicians, physicists, and engineers working with partial differential equations, as well as students seeking to deepen their understanding of Fourier analysis and initial value problems.

Markov2
Messages
149
Reaction score
0
Consider $u_t+u_x=g(x),\,x\in\mathbb R,\,t>0$ and $u(x,0)=f(x).$ Given $f,g\in C^1,$ then show that $u(x,t)$ has the form $u(x,t)=f(x-t)+\sqrt{2\pi}(g*h)(x)$ where $h(x)=\chi_{[0,t]}(x).$

So we just apply the Fourier transform to get $\dfrac{{\partial U}}{{\partial t}} + iwU = U(g)$ and $U((x,0))=U(f ),$ so by solving the ODE I get $U(u)(w,t)=ce^{-iwt}+\dfrac{U(g)(w,t)}{iw},$ now I'm quite confusing on placing the initial condition, what's the correct way?
 
Physics news on Phys.org
I'm quite interested on this one, I want to know how to use the initial condition and plug it into $U(u)(w,t),$ please! :D
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K