I'm afraid you take me beyond my province there. You see how little I know about pde's!
I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.
Solving the original PDE I get
. Taking the positive solution of y
which gives
The IC's are u(x,0)=0, therefore
. Let therefore
I am not sure of this is right either...
Referring to post #15
IC given is
Using polar coordinates, when y is 0 (hence along the x line) theta is 0 therefore r Cos 0 =1 implies the f(r) =0. therefore
for
Not sure how this domain of influence came about though.
regarding post #16, Iv no idea how to solve it. I will try dx/dy instead of dy/dx and see where that gets me.
What you have here is fine except replace the with and simplify your final answer.
You have a solution that satisifes the PDE and BC.
Note: You can transform to polar coordinates if you like - the PDE becomes simple but you must interpret the BC, i.e.
when , for all .
Thus, for all (This means that )