Verify or refute the function is a solution to a PDE

[docnet]

Homework Statement

Verify or refute the functions are solutions to a PDE

Relevant Equations

$u(x)=\frac{1}{2}(x_1^2+x_2^2+...+x_n^2)$

Solution attempt:

We first write $u(x)=\frac{1}{2}||x||^2$ as $u(x)=\frac{1}{2}(x_1^2+x_2^2+...+x_n^2)$

Operating on $u(x)$ with $\Delta$, we have $u(x)=\frac{1}{2}(2+2+...+2)$ adding 2 to itself $n$ times.

So $\Delta u(x)=n$ and the function satisfies the first condition.

The outward unit normal field on the boundary of unit ball denoted $\partial B_1(0)$ is $\nu=\left<x_1, x_2,...,x_n\right>$

The derivative of $u$ in the direction $\nu$ is given by $D_{\nu}u=\frac{1}{2}(2x_1+2x_2+...+2x_n)=(x_1+x_2+...+x_n)$

So the function u does not satisfy the Neumann boundary conditions.

For the condition $u=1$ on the boundary of the unit ball, we consider the euclidian norm to be always equal to 1 by definition.

So the function $u$ satisfies the Dirichlet boundary conditions.

 

[Office_Shredder]

For the second one, u isn't equal to the euclidean norm.

For the first one, how do you know that the thing you wrote down doesn't add up to 1? If n=1, I think it does. Edit to add: actually I think you have computed the derivative in part 1 incorrectly.

 

[docnet]

For the second one, u isn't equal to the euclidean norm.

I am not sure why $u$ does not equal the euclidian norm? $u$ is defined as the magnitude of vector $\left<x_i\right>$ squared. The boundary of a unit ball is an $n$ dimensional sphere, defined roughly as collection of points distance 1 from the origin. So every point on the boundary of the unit ball is a vector of length 1.

For the first one, how do you know that the thing you wrote down doesn't add up to 1? If n=1, I think it does.

I am not sure what you mean.. I added 2 to itself n times because there are n identical terms in the form of $x_i^2$ in $u(x)$ then i factored out $2$ which cancels with $\frac{1}{2}$ leaving n.

Edit to add: actually I think you have computed the derivative in part 1 incorrectly.

I just used the chain rule. most terms cancel out giving $(x_1+...+x_n)$

 

[Office_Shredder]

I am not sure why $u$ does not equal the euclidian norm? $u$ is defined as the magnitude of vector $\left<x_i\right>$ squared. The boundary of a unit ball is an $n$ dimensional sphere, defined roughly as collection of points distance 1 from the origin. So every point on the boundary of the unit ball is a vector of length 1.

But isn't u equal to one half the norm squared?

I am not sure what you mean.. I added 2 to itself n times because there are n identical terms in the form of $x_i^2$ in $u(x)$ then i factored out $2$ which cancels with $\frac{1}{2}$ leaving n.

To compute the directional derivative, you have to compute the total derivative (which depends on the coordinates of the point you are at), and then apply that matrix to the direction you care about.

 

[docnet]

But isn't u equal to one half the norm squared?

Yes, it is. It just occurred to me I did not account for the factor of $\frac{1}{2}$. thank you.

To compute the directional derivative, you have to compute the total derivative (which depends on the coordinates of the point you are at), and then apply that matrix to the direction you care about.

total derivative? i thought it was partial. i computed

$D_{\left<x_1+x_2+...+x_n\right>}(x_1^2+x_2^2+...+x_n^2)=$
$\partial_{x_1}\frac{1}{2}(x_1^2+x_2^2+...+x_n^2)+...+\partial_{x_n}\frac{1}{2}(x_1^2+x_2^2+...+x_n^2)=$
$x_1+...+x_n$

 

[pasmith]

You have <br /> \frac{\partial u}{\partial \nu} = \boldsymbol{\nu} \cdot \nabla u<br /> where \boldsymbol{\nu} is the outward normal, which here is \mathbf{x}/\|\mathbf{x}\|. Thus <br /> \frac{\partial u}{\partial \nu} = \sum_{i=1}^n x_i \frac{\partial u}{\partial x_i} since \|\mathbf{x}\| = 1.

 

[docnet]

You have <br /> \frac{\partial u}{\partial \nu} = \boldsymbol{\nu} \cdot \nabla u<br /> where \boldsymbol{\nu} is the outward normal, which here is \mathbf{x}/\|\mathbf{x}\|. Thus <br /> \frac{\partial u}{\partial \nu} = \sum_{i=1}^n x_i \frac{\partial u}{\partial x_i} since \|\mathbf{x}\| = 1.

Thank you, that makes sense
<br /> \frac{\partial u}{\partial \nu} = \sum_{i=1}^n x_i \frac{\partial u}{\partial x_i}
<br /> \frac{\partial u}{\partial \nu} = x_1^2+...+x_n^2=1
which makes u a solution of the PDE problem.