Exploring Parabolic Behavior in PDE Systems: A Mathematical Analysis

[Clausius2]

I am looking for an elegant way of demonstrating the parabolical behavior of the system:

$$\frac{\partial u}{\partial x}+\frac{1}{r}\frac{\partial}{\partial r}(vr)=0$$

$$u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial}{\partial r}\Big(r \frac{\partial u}{\partial r}\Big)$$

Any idea?. I have read some ways of doing so by establishing a complicated matrix of coefficients, but it is only valid for linear equations.

Thanks in Advance!

 

[Aaron Bex]

One way to demonstrate the parabolical nature of the system is to use the method of separation of variables. This involves expressing the equation as a product of functions with respect to each independent variable, i.e.,

u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial}{\partial r}\Big(r \frac{\partial u}{\partial r}\Big)

= A(x)B(r).

By solving for each part of the equation, we can show that the solution is a parabolic function. For example, if we solve for A(x), we find that the solution is of the form A(x) = c1 + c2x, where c1 and c2 are constants. This means that the solution is a parabola in terms of x. Similarly, if we solve for B(r), we find that the solution is of the form B(r) = c3r^2 + c4r + c5, where c3, c4, and c5 are again constants. This also shows that the solution is a parabolic function in terms of r.

 

[tacman]

Thank you for sharing your interest in exploring the parabolic behavior of PDE systems. There are indeed various ways to demonstrate this behavior, but one elegant approach is through the use of the heat equation. The heat equation is a well-known example of a parabolic PDE, and its solution exhibits parabolic behavior.

To demonstrate this, we can rewrite the given PDE in terms of the heat equation by introducing a new variable, t, and setting v = 1. This results in the following equation:

u_t = \frac{1}{r}\frac{\partial}{\partial r}\Big(r\frac{\partial u}{\partial r}\Big)

Now, let's consider the initial condition u(x,0) = f(x), where f(x) is a given function. This represents the initial temperature distribution in our system. We can then solve this PDE using separation of variables, assuming that u can be written as a product of functions of x and r, i.e. u(x,r) = X(x)R(r).

Plugging this into the PDE, we get:

X'R' = \frac{1}{r}\Big(rX''R + X'R'\Big)

Dividing both sides by XR, we get:

\frac{X''}{X} = \frac{R'}{rR}

This is a separated ODE, and its solution is well-known to be of the form X(x) = Ae^{ax} and R(r) = Br^{c}. Substituting these into our original solution, we get:

u(x,r) = \sum_{n=1}^{\infty} A_ne^{anx}B_nr^{cn}

Now, we can use the initial condition to find the coefficients A_n and B_n, which results in the following solution:

u(x,r) = \sum_{n=1}^{\infty} e^{-\lambda_n t} \Big( \int_{0}^{r} f(r')B_nr'^{cn+1} dr' \Big) \sin\Big(\frac{n\pi x}{L}\Big)

where \lambda_n = \frac{n^2 \pi^2}{L^2} is the eigenvalue and c = \frac{1}{2}.

From this solution, we can see that as t increases, the term e^{-\lambda_n t} decreases, resulting