# Show that this is a solution of Laplace's eqn

1. Sep 13, 2009

### Prologue

1. The problem statement, all variables and given/known data

Suppose that u(x,y) is a solution of Laplace's equation. If theta is a fixed real number and $$v(x,y)=u(xcos\theta-ysin\theta,xsin\theta+ycos\theta)$$
show that v is a solution also.

2. Relevant equations

$$\nabla^{2}u(x,y)=0$$

3. The attempt at a solution

To be honest I haven't gotten anywhere. I have tried just taking the derivatives and seeing what would happen but that didn't get me anywhere (turned ugly fast in other words). I know that Laplace's equation is linear and I bet this has something to do with it but I can't find the way to separate them. I am guessing this problem is supposed to be a quick exercise but it is not clicking with me.

2. Sep 13, 2009

### Office_Shredder

Staff Emeritus
Taking derivatives sounds like a fine idea to me. Multi variable chain rules can be messy, but just slog through it and get the final answer. It's good for you

3. Sep 13, 2009

### Gregg

4. Sep 13, 2009

### Prologue

Will do, I just figured there would be a linearity trick (or change of variables) that does it in 0.73 seconds. But if not I'll just do the multivariable chain rule stuff. I'll report back.

5. Sep 13, 2009

### Prologue

Ok, I think I arrived at something, just one more logical hurdle for me.

I set

$$s=xcos\theta-ysin\theta$$

and

$$t=xsin\theta+ycos\theta$$

Then

$$\nabla^{2}v(x,y)=\nabla^{2}u(s,t)$$

After that I differentiated using many a chain rule to arrive at this

$$\nabla^{2}u(s,t)=\frac{\partial ^{2} u(s,t)}{\partial s^{2}}+\frac{\partial^{2}u(s,t)}{\partial t ^{2}}$$

I am a little iffy on how to interpret this. My apprehension is because s and t are not independent variables as far as I can see, they both depend on things (implicitly) that the other one depends on (x and y). If they were independent I would say, there, it is shown since this

$$\frac{\partial ^{2} u(x,y)}{\partial x ^{2}}+\frac{\partial ^{2} u(x,y)}{\partial y ^{2}}=0$$

is true by definition and then why would it be any different for switching the variables to s and t? But if s and t aren't independent then it seems to me that we have a problem. I have always had this type of apprehension in these types of problems, can someone clear this up once and for all? When does the implicitness matter and when doesn't it?

Last edited: Sep 14, 2009
6. Sep 15, 2009

### Prologue

Can anyone offer any insight into this?

7. Sep 15, 2009

### Gregg

$$\left[ \begin{array}{cc} \text{cos}(\theta ) & -\text{sin}(\theta ) \\ \text{sin}(\theta ) & \text{cos}(\theta ) \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right]=\left[ \begin{array}{c} s \\ t \end{array} \right]$$

The matrix is non singular doesn't that imply independence?