Verifying the Minimum Principle for $u(x,y)$

In summary, the conversation discusses a solution, $u(x,y)$, for a mathematical equation and proves that the minimum principle holds true for this solution. This is shown by demonstrating that the minimum value of $u(x,y)$ within the region $x^2+y^2 \leq 1$ is equal to the minimum value on the boundary, $x^2+y^2=1$.
  • #1
evinda
Gold Member
MHB
3,836
0
Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$ a solution of $u_{xx}(x,y)+u_{yy}(x,y)+(1+x^2) e^{-u(x,y)}=0, x^2+y^2 \leq 1$.
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Is the following right?

$u_{xx}(x,y)+u_{yy}(x,y)=-(1+x^2) e^{-u(x,y)}$

$(1+x^2) e^{-u(x,y)}>0 \ \ \ \ \ \ \forall x,y \text{ with } x^2+y^2 \leq 1$.

So $u_{xx}(x,y)+u_{yy}(x,y) \leq 0$.

Thus the Minimum principe is satisfied for $u$ and so we have: $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.
 
Physics news on Phys.org
  • #2
evinda said:
Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$ a solution of $u_{xx}(x,y)+u_{yy}(x,y)+(1+x^2) e^{-u(x,y)}=0, x^2+y^2 \leq 1$.
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Is the following right?

$u_{xx}(x,y)+u_{yy}(x,y)=-(1+x^2) e^{-u(x,y)}$

$(1+x^2) e^{-u(x,y)}>0 \ \ \ \ \ \ \forall x,y \text{ with } x^2+y^2 \leq 1$.

So $u_{xx}(x,y)+u_{yy}(x,y) \leq 0$.

Thus the Minimum principe is satisfied for $u$ and so we have: $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Sounds good to me! (Nod)
 

FAQ: Verifying the Minimum Principle for $u(x,y)$

1. What is the Minimum Principle for $u(x,y)$?

The Minimum Principle for $u(x,y)$ states that the minimum value of a harmonic function $u(x,y)$ in a given domain is attained on the boundary of the domain or at its critical points.

2. How do you verify the Minimum Principle for $u(x,y)$?

To verify the Minimum Principle for $u(x,y)$, you must first show that the function is harmonic in the given domain. Then, you can use various techniques such as the Mean Value Property or the Maximum Principle to show that the minimum value is indeed attained on the boundary or at critical points.

3. What happens if the Minimum Principle for $u(x,y)$ is not satisfied?

If the Minimum Principle for $u(x,y)$ is not satisfied, it means that there exists a minimum value of $u(x,y)$ in the given domain that is not attained on the boundary or at critical points. This suggests a violation of the assumptions required for the Minimum Principle to hold.

4. Can the Minimum Principle be extended to functions of more than two variables?

Yes, the Minimum Principle can be extended to functions of more than two variables. In this case, the function must be harmonic in the given domain and the minimum value will be attained on the boundary or at critical points of the function.

5. What are the practical applications of the Minimum Principle for $u(x,y)$?

The Minimum Principle for $u(x,y)$ has various practical applications in fields such as physics, engineering, and finance. It can be used to analyze and solve problems involving heat flow, fluid dynamics, electrostatics, and optimization, among others.

Similar threads

MHB Why Does the Minimum Value of u(x,y) Occur on the Boundary of the Unit Disk?
Replies
8
Views
2K
I Solving BVP with Analytical Solution in (0,1)^2
Replies
5
Views
1K
MHB I.7 verify that the given function is a solution
Replies
5
Views
2K
I Alternating Direction Implicit method for solving 2D Heat diffusion
Replies
1
Views
2K
I On sub and super solutions: Teschl and others
Replies
5
Views
1K
MHB Verifying Solution: U(x, y) = cos(x) cosh(y) for PDE U_xx + U_yy = 0
Replies
5
Views
2K
MHB Method of separation of variables
Replies
3
Views
2K
MHB Laplace equation and Median Value Property
Replies
1
Views
2K
A How Does the Derivation of Harmonic Function Derivatives Work in Evans' PDE?
Replies
1
Views
1K
I On vibrating string and differentiating infinite sum
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top