Now \(v(r,\theta)=R(r)\Theta(\theta)\), where, \[\label{eq:12.4.11} r^2R''+rR'-\lambda R=0\], \[\label{eq:12.4.12} \Theta''+\lambda\Theta=0,\quad \Theta(0)=0,\quad \Theta(\gamma)=0.\], From Theorem 11.1.2, the eigenvalues of Equation \ref{eq:12.4.12} are \(\lambda_n=n^2\pi^2/\gamma^2\), with associated eigenfunction \(\Theta_n=\sin n\pi\theta/\gamma\), \(n=1\), \(2\), \(3\),…. where \(0<\rho_0<\rho\) (Figure \(\PageIndex{2}\)). LECTURE 11. The indicial polynomial of this equation is, so the general solution of Equation \ref{eq:12.4.6} is, \[\label{eq:12.4.7} R_n=c_1r^n+c_2r^{-n},\], by Theorem 7.4.3. I have derived the Laplace Equation in 2 variables in cartesian form already. Solve this functional equation Solving this differential equation Geometric Series of nr^n Recent Insights. Consistent with our previous assumption on \(R_0\), we now require \(R_n\) to be bounded as \(r\to0+\). I will do it for a 2-dimensional case: [math]\dfrac{\partial^2u}{\partial x^2} + \dfrac{\partial^2u}{\partial y^2} = 0[/math]. It includes the boundary value conditions of 3 types which is used to simplify the equation. Recall that (from 1st year Calculis) polar coordinates are connected with Cartesian coordinates by , and inverselysurely the second formula is not exactly correct as changing does not change it ratio but replaces by (or as is defined modulo with . is the Fourier sine expansion of \(f\) on \([0,\gamma]\); that is, \[\alpha_n=\frac{2}{\gamma}\int_0^\gamma f(\theta)\sin\frac{n\pi\theta}{\gamma}\,d\theta. Laplace’s Equation 3 Idea for solution - divide and conquer We want to use separation of variables so we need homogeneous boundary conditions. Fluid Dynamics Dr. A.J. Discrete mathematics, Math 209 class taught by Professor Branko Curgus, Mathematics department, Western Washington University 3 Laplace’s Equation We now turn to studying Laplace’s equation ∆u = 0 and its inhomogeneous version, Poisson’s equation, ¡∆u = f: We say a function u satisfying Laplace’s equation is a harmonic function. This implies that \(c_2=0\), and we choose \(c_1=\rho^{-n}\). Because of the Dirichlet condition at \(r=\rho\), it is convenient to have \(r(\rho)=1\); therefore we take \(c_1=\rho^{-n\pi/\gamma}\), so, \[R_n(r)=\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}}.\nonumber\], \[v_n(r,\theta)=R_n(r)\Theta_n(\theta)=\frac{r^{n\pi/\gamma}} {\rho^{n\pi/\gamma}}\sin\frac{n\pi\theta}{\gamma}\nonumber\], \[f(\theta)=\sin\frac{n\pi\theta}{\gamma}.\nonumber\], More generally, if \(\alpha_1\), \(\alpha_2\), …, \(\alpha_m\) and are arbitrary constants then, \[u_m(r,\theta)=\sum_{n=1}^m\alpha_n\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}} \sin\frac{n\pi\theta}{\gamma}\nonumber\], \[f(\theta) =\sum_{n=1}^m\alpha_n\sin\frac{n\pi\theta}{\gamma}.\nonumber\], This motivates us to define the bounded formal solution of Equation \ref{eq:12.4.10} to be, \[u_m(r,\theta)=\sum_{n=1}^\infty\alpha_n\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}} \sin\frac{n\pi\theta}{\gamma},\nonumber\], \[S(\theta)=\sum_{n=1}^\infty\alpha_n \sin\frac{n\pi\theta}{\gamma}\nonumber\]. is the Fourier series of \(f\) on \([-\pi,\pi]\); that is, \[\alpha_0=\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)\,d\theta,\nonumber\], \[\alpha_n=\frac{1}{\pi}\int_{-\pi}^\pi f(\theta)\cos n\theta\,d\theta \quad \text{and} \quad \beta_n=\frac{1}{\pi}\int_{-\pi}^\pi f(\theta)\sin n\theta\,d\theta, \quad n=1,2,3,\dots.\nonumber\], Since \(\sum_{n=0}^\infty n^k(r/\rho)^n\) converges for every \(k\) if \(0< r<\rho\), Theorem 12.1.2 can be used to show that if \(0< r<\rho\) then Equation \ref{eq:12.4.8} can be differentiated term by term any number of times with respect to both \(r\) and \(\theta\). For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. From Theorem 11.1.6, the eigenvalues of Equation \ref{eq:12.4.4} are \(\lambda_0=0\) with associated eigenfunctions \(\Theta_0=1\) and, for \(n=1,2,3,\dots,\) \(\lambda_n=n^2\), with associated eigenfunction \(\cos n\theta\) and \(\sin n\theta\) therefore, \[\Theta_n=\alpha_n\cos n\theta+\beta_n\sin n\theta \nonumber\]. Since the equation is linear we can break the problem into simpler problems which do have suﬃcient homogeneous BC and use superposition to obtain the solution to (24.8). In the next several lectures we are going to consider Laplace equation in the disk and similar domains and separate variables there but for this purpose we need to express Laplace operator in polar coordinates. Now examining the potential inside the sphere, the potential must have a term of order r 2 to give a constant on the left side of the equation, so the solution is of the form. If u is a solution of the Laplace equation Δu = 0, then the value of u at any point is just the average values of u on a circle centered on that point. so the solution to LaPlace's law outside the sphere is . From Theorem 11.2.4, this is true if \(f\) is continuous and piecewise smooth on \([-\pi,\pi]\) and \(f(-\pi)=f(\pi)\). Laplace’s Equation in Cylindrical Coordinates and Bessel’s Equation (I) 1 Solution by separation of variables Laplace’s equation is a key equation in Mathematical Physics. Solutions of Laplace’s equation in 3d Motivation The general form of Laplace’s equation is: ∇=2Ψ 0; it contains the laplacian, and nothing else. This is the desired expression of the Laplacian in polar coordinates. Legal. Thus the equation takes the form; ∂2V ∂x 2 2V ∂y + ∂ V ∂z2 = 0 and we assume that we can ﬁnd a solution using the method of separation of variables. 5.7 Solutions to Laplace's Equation in Polar Coordinates. time independent) for the two dimensional heat equation with no sources. Missed the LibreFest? Insights How Bayesian Inference Works in the Context of Science We will also convert Laplace’s equation to polar coordinates and solve it on a disk of radius a. where \(\alpha_n\) and \(\beta_n\) are arbitrary constants. Ð!£ Q²¿,v +¶Te{Qé2ÏmÏÂÅ«d>óö7áù>5_Ç¨qµUDç7²¥ÛÍ\'¬`0B©ÁApBTêË@² µ%»«)Ý,ê:ÖaX+©atL¥ÎPu. To obtain a solution that remains bounded as \(r\to0+\) we let \(c_2=0\). §T !â=XÌéqÕã#=&²`4e¢+4cò>lO6»;ÐävgÂM â»¢À`éíï¦ìyîEÁK'ÍïTä¸ÐüÎMó÷²ù©a~bWf~¶Ë~2ÿFØÞkÐ%Íÿ¿0>å.oâéCÏM+SyNð¯HÕ3Äá5ºqfb:e°`%ñ8öt¹ You can extend the argument for 3-dimensional Laplace’s equation on your own. Solution to Laplace’s Equation In Cartesian Coordinates Lecture 6 1 Introduction We wish to solve the 2nd order, linear partial diﬀerential equation; ∇2V(x,y,z) = 0 We ﬁrst do this in Cartesian coordinates. 27 0. For this function, \[v_{rr}+\frac{1}{r}v_r+\frac{1}{r^2}v_{\theta\theta}= R''\Theta+\frac{1}{r}R'\Theta +\frac{1}{r^2}R\Theta''= 0 \nonumber\], \[\frac{r^2R''+rR'}{R}=-\frac{\Theta''}{\Theta}=\lambda, \nonumber\], where \(\lambda\) is a separation constant. \[\label{eq:12.4.10} \begin{array}{c}{u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta\theta}=0,\quad 0