When is the PDE u_t= \alpha u_{xx}+ \beta u_{xxxx} well posed?

[MathematicalPhysicist]

I have this question I got in the exam, I am pretty much sure I did it right, but I guess something got wrong.

We have the next PDE
$$u_t= \alpha u_{xx}+ \beta u_{xxxx}$$ for t>0 and x is on the whole plane.
the question asks to analyse when the above PDE is well posed (strong or weakly)
for the next cases:
1. alpha and beta >=0
2. alpha <=0 beta >0
3. alpha >0 and beta <=0
4. alpha and beta <0

after using Fourier tranform, I get that the energy functional is of the form:
$$E(t)=\int_{-\infty}^{\infty} |u(w,0)|^2 e^{2t(-\alpha w^2 +\beta w^4} dw$$

Now in 3 when alpha is greater than 0 and beta is negative we have that E(t)<=E(0) which means it's well posed, but what of the other options?

Anyone?

 

[jvicens]

I would first like to clarify that the PDE given in the question is a diffusion equation with fourth-order spatial derivative. This type of PDE is commonly used to model physical phenomena such as heat conduction and diffusion of particles.

Now, let's address the question of when this PDE is well posed. In order for a PDE to be well posed, it must have a unique solution and that solution must depend continuously on the initial and boundary conditions. This means that small changes in the initial and boundary conditions should result in small changes in the solution.

In the case of the given PDE, there are four different scenarios to consider based on the values of alpha and beta.

1. When both alpha and beta are greater than or equal to zero, the PDE is well posed in the strong sense. This means that it has a unique solution and that solution depends continuously on the initial and boundary conditions. This is because the energy functional, as shown in the forum post, is always non-negative and decreases with time.

2. When alpha is less than or equal to zero and beta is greater than zero, the PDE is not well posed in either the strong or weak sense. This is because the energy functional is not bounded from below and can increase with time, leading to a lack of uniqueness in the solution.

3. When alpha is greater than zero and beta is less than or equal to zero, the PDE is well posed in the weak sense. This means that it has a unique solution, but that solution may not depend continuously on the initial and boundary conditions. In this case, the energy functional is bounded from above, which guarantees uniqueness, but it is not necessarily bounded from below, which may lead to a lack of continuity in the solution.

4. When both alpha and beta are less than zero, the PDE is not well posed in either the strong or weak sense. This is because the energy functional is not bounded from above or below, which leads to a lack of uniqueness and continuity in the solution.

In summary, the well-posedness of the given PDE depends on the values of alpha and beta. When both are positive, the PDE is well posed in the strong sense. When alpha is positive and beta is negative, the PDE is well posed in the weak sense. In all other cases, the PDE is not well posed. I hope this clarifies any confusion and helps with your understanding