PDE Cauchy problem to Laplace equation

Question

I have been given this question in polars and I tried but not sure if I did it right and if my solution is correct.

$$U_{rr}+\frac{1}{r^2}U_{\theta\theta} +\frac{1}{r}U_r=0$$ $$U(1,\theta)=1$$ $$U_r(1,\theta)=1$$

Initial conditons given means $U(1,\theta)=1$ and $U_\theta(1,\theta)=0$, $U_r(1,\theta)=1$.

Subtract both sides by $-\frac{1}{r^2}U_{\theta \theta}(1,\theta)$ $$U_{rr}(1,\theta)+\frac{1}{r}U_r(1,\theta)=-\frac{1}{r^2}U_{\theta \theta}(1,\theta)$$

$$ -r^2U_{rr}(1,\theta)-r=U_{\theta \theta}(1,\theta)$$

$$R =1$$

$$U_{\theta \theta}(1,\theta)=0$$

$$U_{rr}(1,\theta)=-1$$

So using Taylor series, $$U(r,\theta)= u(1,0)+U_r(1,0)(r-1)...... =1+(r-1)-\frac{1}{2}(r-1)^2$$ And this was my full solution and the next terms are all zeroes so this is the actual solution as well I guess. Did I do this right? Any help will be appreciated! Thanks!

Answer 1

You have basically taken the first few Taylor terms of the solution about $r=1$, but what you have does not satisfy the PDE elsewhere. Indeed:

$$\left ( \frac{d^2}{dr^2} + \frac{1}{r} \frac{d}{dr} \right ) \left ( 1 + (r-1) - \frac{1}{2} (r-1)^2 \right ) = -1 + \frac{1}{r} - \frac{r-1}{r}$$

which is not zero anywhere except $r=1$.

You might be able to continue using power series to solve the problem; the technical term for this procedure is the Frobenius method. But I doubt this is what your reference expects you to do. Notice that by the symmetry $U$ is really just a function of $r$, so you can look at the ODE problem

$$u''+u'/x=0,u(1)=1,u'(1)=1$$

which is actually not that hard to solve; consider substituting $v=u'$ to get started. (By the way, the method of solution of this ODE is closely related to writing the original PDE in divergence form...)

Answer 2

$$\left\{ \begin{align} &u_{rr}+\frac{1}{r^2}u_{\theta\theta} +\frac{1}{r}u_r=0\\ \\ &u(1,\theta)=1,\quad u_r(1,\theta)=1 \end{align} \right. $$ Let $r=\operatorname e^t$ so that for $u(r(t),\theta)$ we have $u_t=u_r r_t$ and $u_{tt}=ru_r+r^2u_{rr}$. Thus the PDE becomes $$ u_{tt}+u_{\theta\theta}=0 $$ By symmetry $u(t,\theta)=X(t)$.

So we have $u_{tt}=0$ and then $X(t)=at+b$ that is $u(r,\theta)=a\log r+b$.

From $u_r(1,\theta)=1=a$ and $u(1,\theta)=1=b$, we find

$$ u(r,\theta)=1+\log r\quad \text{for }r>0,\,\forall \theta $$