2.15.2 \(u_{rr} + u_{\theta \theta } = 0\)

problem number 134

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Second order PDE in Polar coordinates. Solve for \(u(r,\theta )\) \[ u_{rr} + u_{\theta \theta } = 0 \] With boundary conditions \begin {align*} u(2,\theta )&=3 \sin (2 \theta )+1 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, theta], {r, 2}] + D[u[r, theta], {theta, 2}] == 0; 
bc  = u[2, theta] == 3*Sin[2*theta] + 1; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}], 60*10]];
 

\[\left \{\left \{u(r,\theta )\to \fbox {$\begin {array}{cc} \{ & \begin {array}{cc} 6 e^{4-2 r} \cos (\theta ) \sin (\theta )+1 & r\geq 2 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\text { if }r\neq 2$}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(r, theta), r$2)+diff(u(r, theta), theta$2) = 0; 
bc  := u(2, theta) = 3*sin(2*theta)+1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(r,theta),method = Fourier)),output='realtime'));
 

\[u \left (r , \theta \right ) = -\frac {3 i {\mathrm e}^{2 i \theta -2 r +4}}{2}+\frac {3 i {\mathrm e}^{-2 i \theta +2 r -4}}{2}+1\]

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