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The function f(x, y) satisfies the Laplace equation
∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
- a)0
- b)2
- c)3
- d)1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The function f(x, y) satisfies the Laplace equation∇2f(x,y)=0on ...
Understanding the Laplace Equation
The Laplace equation, ∇²f(x, y) = 0, describes a function f that is harmonic within a given domain. In this case, the domain is a circular area with a radius of 1 centered at the origin (0, 0).
Boundary Conditions
- The function f(x, y) is known to take a constant value on the circular boundary defined by the equation x² + y² = 1.
- It is given that f(x, y) = 3 for all points on this boundary.
Properties of Harmonic Functions
- A key property of harmonic functions is that the average value of the function over any closed boundary is equal to its value at the center of that domain.
- Therefore, if the function is constant on the boundary, it must also be constant throughout the entire domain.
Finding the Value at the Center
- Since f(x, y) = 3 on the boundary and the function is harmonic, by the mean value property of harmonic functions, we conclude that the value of the function at the center (0, 0) must also be equal to the boundary value.
Conclusion
- Thus, f(0, 0) = 3, which corresponds to option 'C'.
This demonstrates that within a circular domain, if a harmonic function is constant on the boundary, it will also take that same value at the center.
The Laplace equation, ∇²f(x, y) = 0, describes a function f that is harmonic within a given domain. In this case, the domain is a circular area with a radius of 1 centered at the origin (0, 0).
Boundary Conditions
- The function f(x, y) is known to take a constant value on the circular boundary defined by the equation x² + y² = 1.
- It is given that f(x, y) = 3 for all points on this boundary.
Properties of Harmonic Functions
- A key property of harmonic functions is that the average value of the function over any closed boundary is equal to its value at the center of that domain.
- Therefore, if the function is constant on the boundary, it must also be constant throughout the entire domain.
Finding the Value at the Center
- Since f(x, y) = 3 on the boundary and the function is harmonic, by the mean value property of harmonic functions, we conclude that the value of the function at the center (0, 0) must also be equal to the boundary value.
Conclusion
- Thus, f(0, 0) = 3, which corresponds to option 'C'.
This demonstrates that within a circular domain, if a harmonic function is constant on the boundary, it will also take that same value at the center.
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The function f(x, y) satisfies the Laplace equation∇2f(x,y)=0on ...
Given that,
The function f(x, y) satisfies the Laplace equation ∇2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0.
The value of this function on the circular boundary of this domain is equal to 3.
Here it is given that the value of the function is 3 for its domain, which signifies that it is a constant function whose value is 3.
So the value of the function at (0, 0) is 3.
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The function f(x, y) satisfies the Laplace equation∇2f(x,y)=0on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.The numerical value of f(0, 0) is:a)0b)2c)3d)1Correct answer is option 'C'. Can you explain this answer?
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