Determining whether the Dirichlet problem is nonhomogenous

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In summary, the given pde can be considered homogenous or non-homogenous based on the boundary conditions. To find the solution, we note that since the steady-state is independent of time, the equation reduces to ##u_{xx}=0## and the solution can be expressed as ##u(x)=Mx+N##, where M and N are constants. By using the boundary conditions, the solution can be further simplified to ##u(x)##=##\dfrac{
  • #1
chwala
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TL;DR Summary
See attached
1650601806284.png

I am going through the literature on Dirichlet and Neumann conditions for the heat equation...i need to be certain on the nonhomogenous bit that is indicated on the excerpt. I am aware that for a pde to be considered homogenous, ##Lu=0## and for it to be non homogenous; ##Lu=f##, where ##L## is the given differential operator.

To determine whether the pde is homogenous or not we make use of the boundary conditions, correct?

In our case we have the given pde with two boundary conditions and one initial condition (not considered due to steady-state condition).

I now get it!
thus to find the solution, we note that since the steady-state is not dependant on time, then it follows that, ##u_{t}=0##
##u(x,t)=u(x)## ##Du_{xx}## =##0→u_{xx}=0## on integration we shall have
##u_{x}=M##, where ##M## is a constant. Integrating again yields
##u(x)=Mx+N##, where ##N## is also a constant. Now using the first boundary conditions, ##u(0,t)=c_1##, we shall have
## c_1=N## on using the second boundary condition, we shall have
##c_2=Ml+c_1## it follows that ##u(x)##=##\dfrac {c_2-c_1}{l}##+##c_1##
cheers guys...
 

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  • #2
chwala said:
Summary: See attached

View attachment 300372
I am going through the literature on Dirichlet and Neumann conditions for the heat equation...i need to be certain on the nonhomogenous bit that is indicated on the excerpt. I am aware that for a pde to be considered homogenous, ##Lu=0## and for it to be non homogenous; ##Lu=f##, where ##L## is the given differential operator.

To determine whether the pde is homogenous or not we make use of the boundary conditions, correct?

There are two ways in which a PDE can be non-homogenous.

(1): The right hand side can be non-zero, as in [itex]L(u) = f[/itex].

(2) A boundary condition can be non-homogenous, ie. of the form [tex]
\alpha u + \beta \frac{\partial u}{\partial n} = g[/tex] where [itex]\alpha[/itex] and [itex]\beta[/itex] are functions of position and time satisfying [itex]\alpha^2 + \beta^2 = 1[/itex].

In our case we have the given pde with two boundary conditions and one initial condition (not considered due to steady-state condition).

I now get it!
thus to find the solution, we note that since the steady-state is not dependant on time, then it follows that, ##u_{t}=0##
##u(x,t)=u(x)## ##Du_{xx}## =##0→u_{xx}=0## on integration we shall have
##u_{x}=M##, where ##M## is a constant. Integrating again yields
##u(x)=Mx+N##, where ##N## is also a constant. Now using the first boundary conditions, ##u(0,t)=c_1##, we shall have
## c_1=N## on using the second boundary condition, we shall have
##c_2=Ml+c_1## it follows that ##u(x)##=##\dfrac {c_2-c_1}{l}##+##c_1##
cheers guys...

Your final expression for [itex]u(x)[/itex] does not contain [itex]x[/itex]. It is easiest to set [tex]
\begin{split}
u(x) &= c_2\frac xl + c_1\left(1 - \frac xl\right) + v(x) \\
&= c_1 + (c_2 - c_1)\frac xl + v(x)
\end{split}[/tex] where [itex]v[/itex] satisfies [tex]v_{xx} = -(c_1 + (c_2-c_1)x/l)_{xx} = 0[/tex] subject to the homogenous condition [itex]v(0) = v(l) = 0[/itex]. This forces [itex]v(x) = 0[/itex].
 
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  • #3
chwala said:
TL;DR Summary: See attached

View attachment 300372
I am going through the literature on Dirichlet and Neumann conditions for the heat equation...i need to be certain on the nonhomogenous bit that is indicated on the excerpt. I am aware that for a pde to be considered homogenous, ##Lu=0## and for it to be non homogenous; ##Lu=f##, where ##L## is the given differential operator.

To determine whether the pde is homogenous or not we make use of the boundary conditions, correct?

In our case we have the given pde with two boundary conditions and one initial condition (not considered due to steady-state condition).

I now get it!
thus to find the solution, we note that since the steady-state is not dependant on time, then it follows that, ##u_{t}=0##
##u(x,t)=u(x)## ##Du_{xx}## =##0→u_{xx}=0## on integration we shall have
##u_{x}=M##, where ##M## is a constant. Integrating again yields
##u(x)=Mx+N##, where ##N## is also a constant. Now using the first boundary conditions, ##u(0,t)=c_1##, we shall have
## c_1=N## on using the second boundary condition, we shall have
##c_2=Ml+c_1## it follows that ##u(x)##=##\dfrac {c_2-c_1}{l}##+##c_1##
cheers guys...
Amended...

it follows that ##u(x)##=##\dfrac {c_2-c_1}{l}x##+##c_1##
 

Related to Determining whether the Dirichlet problem is nonhomogenous

1. What is the Dirichlet problem?

The Dirichlet problem is a mathematical problem that involves finding a solution to a partial differential equation (PDE) with specified boundary conditions. It is named after the German mathematician Peter Gustav Lejeune Dirichlet.

2. How do you determine if the Dirichlet problem is nonhomogeneous?

A Dirichlet problem is considered nonhomogeneous if the boundary conditions are not equal to zero. This means that the solution to the PDE is influenced by external factors or sources, rather than being solely determined by the given boundary conditions.

3. What are some examples of nonhomogeneous Dirichlet problems?

Examples of nonhomogeneous Dirichlet problems include heat conduction problems with a heat source or a boundary with varying temperature, and fluid flow problems with a pressure source or a boundary with varying velocity.

4. How do you solve a nonhomogeneous Dirichlet problem?

The solution to a nonhomogeneous Dirichlet problem can be found by using various techniques such as separation of variables, Fourier series, or Green's functions. The specific method used will depend on the type of PDE and the given boundary conditions.

5. Why is it important to determine if the Dirichlet problem is nonhomogeneous?

Knowing whether the Dirichlet problem is nonhomogeneous is important because it affects the approach and techniques used to solve the problem. Nonhomogeneous problems require more complex methods and may have multiple solutions, making it crucial to accurately determine the nature of the problem before attempting to solve it.

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