Understanding relationship between heat equation & Green's function

  • #1
lriuui0x0
101
25
TL;DR Summary
Trying to understand heat equation general solution through Green's function
Given a 1D heat equation on the entire real line, with initial condition ##u(x, 0) = f(x)##. The general solution to this is:

$$
u(x, t) = \int \phi(x-y, t)f(y)dy
$$

where ##\phi(x, t)## is the heat kernel.

The integral looks a lot similar to using Green's function to solve differential equation. The fact that ##\phi(x, 0) = \delta(x)## also signals something related to Green's function. This wikipedia page talks about Green's function related to heat equation as well.

However after searching on the internet, I don't get how do I understand the Green's function in the context of heat equation. As I understand, Green's function is related to a particular ordinary linear differential operator. What's the differential operator for heat equation?
 

Answers and Replies

  • #2
fresh_42
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Summary:: Trying to understand heat equation general solution through Green's function

Given a 1D heat equation on the entire real line, with initial condition ##u(x, 0) = f(x)##. The general solution to this is:

$$
u(x, t) = \int \phi(x-y, t)f(y)dy
$$

where ##\phi(x, t)## is the heat kernel.

The integral looks a lot similar to using Green's function to solve differential equation. The fact that ##\phi(x, 0) = \delta(x)## also signals something related to Green's function. This wikipedia page talks about Green's function related to heat equation as well.

However after searching on the internet, I don't get how do I understand the Green's function in the context of heat equation. As I understand, Green's function is related to a particular ordinary linear differential operator. What's the differential operator for heat equation?
Have you had a look at
http://web.pdx.edu/~daescu/mth428_528/Green_functions.pdf
 
  • #6
jasonRF
Science Advisor
Gold Member
1,508
576
Summary:: Trying to understand heat equation general solution through Green's function

Given a 1D heat equation on the entire real line, with initial condition ##u(x, 0) = f(x)##. The general solution to this is:

$$
u(x, t) = \int \phi(x-y, t)f(y)dy
$$

where ##\phi(x, t)## is the heat kernel.

The integral looks a lot similar to using Green's function to solve differential equation. The fact that ##\phi(x, 0) = \delta(x)## also signals something related to Green's function. This wikipedia page talks about Green's function related to heat equation as well.

However after searching on the internet, I don't get how do I understand the Green's function in the context of heat equation. As I understand, Green's function is related to a particular ordinary linear differential operator. What's the differential operator for heat equation?
First, as others have noted, Green's functions are used for partial differential equations all the time, not just ordinary differential equations.

Your particular problem is ##u_t - u_{xx} = 0## with ##u(x,t=0) = \delta(x)##, for ##-\infty<x<\infty##.

This is equivalent to the problem ##u_t - u_{xx} = \delta(x) \delta(t) ## with ##u(x,t<0) = 0##, for ##-\infty<x<\infty##. The solution to this problem is the Green's function for the operator ##L(u) = u_t - u_{xx}##.

So in this case the homogeneous initial value problem is equivalent to a non-homogeneous problem with zero initial conditions.

Was that the connection you were looking for?

jason
 
  • #7
lriuui0x0
101
25
First, as others have noted, Green's functions are used for partial differential equations all the time, not just ordinary differential equations.

Your particular problem is ##u_t - u_{xx} = 0## with ##u(x,t=0) = \delta(x)##, for ##-\infty<x<\infty##.

This is equivalent to the problem ##u_t - u_{xx} = \delta(x) \delta(t) ## with ##u(x,t<0) = 0##, for ##-\infty<x<\infty##. The solution to this problem is the Green's function for the operator ##L(u) = u_t - u_{xx}##.

So in this case the homogeneous initial value problem is equivalent to a non-homogeneous problem with zero initial conditions.

Was that the connection you were looking for?

jason
Thanks!
 

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