Question

Let \(\Omega\) be the disk \(x^2 + y^2<4\) in \(\mathbb{R}^2\) with boundary \(\partial\Omega\). If \(u(x, y)\) is the solution of the Dirichlet problem \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \quad (x, y) \in \Omega, \] \[ u(x, y) = 1 + 2x^2, \quad (x, y) \in \partial\Omega, \] then the value of \(u(0,1)\) is ..................

Hint:

For Dirichlet problems on simple domains like disks or rectangles with polynomial boundary conditions, always try to find a simple polynomial solution first. A general harmonic polynomial of degree 2 is \(a(x^2-y^2) + bxy + cx + dy + e\). Matching this with the boundary conditions is often the quickest method.

Answer 1

Step 1: Understanding the Concept:
The problem is to solve Laplace's equation in a circular domain (a disk of radius 2 centered at the origin) with a given boundary condition. The solution \(u(x, y)\) is a harmonic function. We need to find the value of this function at an interior point (0,1).
Step 2: Key Formula or Approach:
A powerful technique for such problems is to find a simple harmonic function that satisfies the boundary conditions. We can try to express the boundary condition in terms of x and y and see if a simple polynomial in x and y can be constructed.
The boundary is the circle \(x^2 + y^2 = 4\). On this boundary, we can substitute \(x^2 = 4 - y^2\) or \(y^2 = 4 - x^2\) to simplify expressions.
Let's try to find a harmonic function \(u(x,y)\) of the form \(ax^2 + by^2 + cx + dy + e\) that satisfies the boundary condition.
A function \(u(x,y)\) is harmonic if \(u_{xx} + u_{yy} = 0\).
For \(u(x,y) = ax^2 + by^2 + cx + dy + e\), we have \(u_{xx} = 2a\) and \(u_{yy} = 2b\). So we need \(2a + 2b = 0\), which means \(b = -a\).
So, any harmonic function of this form must be \(u(x,y) = a(x^2 - y^2) + cx + dy + e\).
Step 3: Detailed Explanation or Calculation:
We are looking for a harmonic function \(u(x,y)\) such that on the boundary \(x^2+y^2=4\), we have \(u(x,y) = 1 + 2x^2\).
Let's try to construct such a function. We know \(u(x,y) = A(x^2 - y^2) + B\) is a family of harmonic functions (for constants A, B). Let's see if we can match the boundary condition.
On the boundary, \(y^2 = 4 - x^2\). Substituting this into our trial function: \[ u(x,y) = A(x^2 - (4 - x^2)) + B = A(2x^2 - 4) + B = 2Ax^2 - 4A + B \] We want this to be equal to the given boundary condition, \(1 + 2x^2\). \[ 2Ax^2 - 4A + B = 2x^2 + 1 \] By comparing the coefficients of \(x^2\) and the constant terms, we get: Coefficient of \(x^2\): \(2A = 2 \implies A = 1\). Constant term: \(-4A + B = 1\). Substituting \(A=1\), we get \(-4(1) + B = 1 \implies B = 5\). So, the function \(u(x,y) = 1(x^2 - y^2) + 5 = x^2 - y^2 + 5\) is harmonic and satisfies the boundary condition. By the uniqueness theorem for the Dirichlet problem, this must be the solution.
Step 4: Final Answer:
Now we evaluate this solution at the point (0, 1). \[ u(0, 1) = (0)^2 - (1)^2 + 5 = 0 - 1 + 5 = 4 \] Step 5: Why This is Correct:
We found a function \(u(x,y) = x^2 - y^2 + 5\). We verified that it is harmonic (\(u_{xx} + u_{yy} = 2 - 2 = 0\)) and that it satisfies the boundary condition on \(x^2+y^2=4\) (\(u(x,y) = x^2 - (4-x^2) + 5 = 2x^2+1\)). Since the solution to the Dirichlet problem is unique, this is the correct solution. The evaluation at the specified point is a straightforward substitution.