Equilibrium Soln for $u_{t}=K u_{xx}+\gamma$

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In summary, the conversation discusses finding the equilibrium solution for the partial differential equation $u_t=Ku_{xx}+\gamma$ with given boundary and initial conditions. The solution is found to be $u(x)=-\frac{\gamma}{2K}x^2$ with a constant of integration. The conversation also mentions removing the time parameter and using the initial conditions to determine the constant of integration.
  • #1
Markov2
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Well I don't know if is the correct term for this for here goes:

Let

$\begin{align}
& {{u}_{t}}=K{{u}_{xx}}+\gamma ,\text{ }0<x<L,\text{ }t>0, \\
& u(0,t)=\alpha ,\text{ }u(L,t)=\beta ,\text{ }t>0, \\
& u(x,0)=0,
\end{align}
$

where $\alpha,\beta,\gamma$ are constant, then find the equilibrium solution. I don't know what I need to do.
 
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  • #2
Markov said:
Well I don't know if is the correct term for this for here goes:

Let

$\begin{align}
& {{u}_{t}}=K{{u}_{xx}}+\gamma ,\text{ }0<x<L,\text{ }t>0, \\
& u(0,t)=\alpha ,\text{ }u(L,t)=\beta ,\text{ }t>0, \\
& u(x,0)=0,
\end{align}
$

where $\alpha,\beta,\gamma$ are constant, then find the equilibrium solution. I don't know what I need to do.

To find the equilibrium solution $U(x)$, set

$0=Ku_{xx}+\gamma$
$U(0)=\alpha$
$U(L)=\beta$
 
  • #3
So I get $u(x)=-\dfrac\gamma{2K} x^2,$ does this make sense? Why the boundary conditions aren't something like $u(0,t)$ ?
 
  • #4
Markov said:
Why the boundary conditions aren't something like $u(0,t)$ ?

We removed the t parameter.

---------- Post added at 02:45 PM ---------- Previous post was at 02:43 PM ----------

Markov said:
So I get $u(x)=-\dfrac\gamma{2K} x^2,$ does this make sense?

Shouldn't you have a constant of integration?
 
  • #5
dwsmith said:
Shouldn't you have a constant of integration?
Oh yes, yes, so then I just put the initial conditions and that's it?
 

Related to Equilibrium Soln for $u_{t}=K u_{xx}+\gamma$

1. What is an equilibrium solution for the equation $u_{t}=K u_{xx}+\gamma$?

An equilibrium solution is a solution to the given equation that remains constant over time, meaning that the value of $u$ does not change with respect to time. In other words, the derivative of $u$ with respect to time is equal to zero, making the equation $u_{t}=0$. This is often referred to as a steady state solution.

2. How is equilibrium affected by the values of $K$ and $\gamma$?

The values of $K$ and $\gamma$ determine the stability of the equilibrium solution. If the values of $K$ and $\gamma$ are such that they lead to a positive second derivative of $u$ with respect to $x$, then the equilibrium solution is stable. However, if the values of $K$ and $\gamma$ lead to a negative second derivative, the equilibrium solution is unstable. This means that small perturbations or changes in the initial conditions can cause the solution to deviate significantly from the equilibrium.

3. What are the applications of the equilibrium solution for this equation?

The equation $u_{t}=K u_{xx}+\gamma$ has various applications in fields such as physics, chemistry, and biology. One example is in diffusion processes, where $u$ represents the concentration of a substance and the equation describes the change in concentration over time. It can also be used to model heat transfer, population dynamics, and other phenomena where the change in a quantity depends on its second derivative with respect to space or time.

4. How can the equilibrium solution be solved for analytically?

The equilibrium solution can be solved for analytically by setting the derivative of $u$ with respect to time equal to zero and solving the resulting differential equation. This usually involves finding the general solution and applying any initial or boundary conditions to determine the specific solution. However, in some cases, the equation may not have an analytical solution and numerical methods may be used to approximate the equilibrium solution.

5. Can the equilibrium solution change over time?

No, the equilibrium solution remains constant over time. As mentioned earlier, the derivative of $u$ with respect to time is equal to zero for an equilibrium solution, meaning that the value of $u$ does not change with respect to time. However, the equilibrium solution can shift or change in response to changes in the values of $K$ and $\gamma$ or changes in the initial or boundary conditions of the equation.

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