Consider the limit that .In this case, according to Equation (), the allowed values of become more and more closely spaced.Consequently, the sum over discrete -values in morphs into an integral over a continuous range of -values.For instance, suppose that we wish to solve Laplace's equation … and our solution is fully determined. Note that while the matrix in Eq. Because Laplace's equation is a linear PDE, we can use the technique of separation of variables in order to convert the PDE into several ordinary differential equations (ODEs) that are easier to solve. This equation also arises in applications to fluid mechanics and potential theory; in fact, it is also called the potential equation . PHY2206 (Electromagnetic Fields) Analytic Solutions to Laplace’s Equation 3 Hence R =γrm +δr−m is the general form for m i≠ i0 and R =α0 lnr +β0 when m i= i0 and the most general form of the solution is φ()r,θ=α0lnr +β0 + γmr m +δ mr ()−m α mcos()mθ+βmsin()mθ m=1 ∞ ∑ including a redundant constant. This is Laplace’s equation. Furthermore, unlike the method of undetermined coefficients, the Laplace transform … Figure 4. Note that there are many functions satisfy this equation. The Laplace transform is an integral transform that is widely used to solve linear differential equations with constant coefficients. The solution is illustrated below. Laplace on a disk Next up is to solve the Laplace equation on a disk with boundary values prescribed on the circle that bounds the disk. Ψ = Ae±imφ The geometry is 3 dimensional so there will be 2 eigenvalue equations… Be-cause the solution is harmonic this means m must be integral forming harmonic eigenvalues and eigenfunctions. The solution for the problem is obtained by addition of solutions of the same form as for Figure 2 above. Such equations can (almost always) be … Laplace’s Equation • Separation of variables – two examples • Laplace’s Equation in Polar Coordinates – Derivation of the explicit form – An example from electrostatics • A surprising application of Laplace’s eqn ... and the final solution to the stress distribution is a y a x a b w LAPLACE’S EQUATION ON A DISC 66 or the following pair of ordinary di erential equations (4a) T00= 2T (4b) r2R00+ rR0= 2R The rst equation (4a) should be quite familiar by now. The fundamental solution of Laplace’s equation Consider Laplace’s equation in R2, ∆u(x) = 0, x ∈ R2, (1) where ∆ = ∂2/∂x2 +∂2/∂y2. It has as its general solution (5) T( ) = Acos( ) + Bsin( ) The second equation (4b) is an Euler type equation. We seek solutions of Equation \ref{eq:12.3.2} in a region \(R\) that satisfy specified conditions – called boundary conditions – on the boundary of \(R\). The behavior of the solution is well expected: Consider the Laplace's equation as the governing equation for the steady state solution of a 2-D heat equation, the "temperature", u, should decrease from the top right corner to lower left corner of the domain. 15. In his case the boundary conditions of the superimposed solution match those of the problem in question. The equation takes the form of an eigenvalue equation with the boundary condition that the function Ψ must repeat as φ circles beyond 2π. Linearity ensures that the solution set consists of an arbitrary linear combination of solutions. When such a differential equation is transformed into Laplace space, the result is an algebraic equation, which is much easier to solve. Laplace’s equation is linear and the sum of two solutions is itself a solution. In particular, all u satisfies this equation is called the harmonic function.

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