Question

Define the formal solution of $$ \begin{array}{c} u_{r r}+\frac{1}{r} u_{r}+\frac{1}{r^{2}} u_{\theta \theta}=0, \quad \rho_{0}

Short answer

Question: Find the general solution for the function $u(r,\theta )$ that satisfies the given partial differential equation and boundary conditions. Solution: The general solution $u(r,\theta )$ is obtained by summing the separated solutions: $$u(r,\theta )=\sum _{n=1}^{\mathrm{\infty }}[A_n\cos \left(\frac{n\pi }{\gamma }\theta \right)]\left[D_n(\frac{1}{{\rho }_0^{2a}}{r}^a-r^{-a})\right]$$ where $a=\frac{n\pi }{\gamma }$, and $A_n$ and $D_n$ are coefficients that need to be determined using the final boundary condition $u_r(\rho ,\theta )=g(\theta )$.

Step-by-step solution

Separate variables

To begin, we will attempt to separate variables. Assume that the solution can be expressed in the form $u(r,\theta )=R(r)\mathrm{\Theta }(\theta )$. Then, the partial derivatives are given by: $$u_r=R^{\prime }(r)\mathrm{\Theta }(\theta ),u_{rr}=R^{″}(r)\mathrm{\Theta }(\theta ),u_{\theta }=R(r){\mathrm{\Theta }}^{\prime }(\theta ),u_{\theta \theta }=R(r){\mathrm{\Theta }}^{″}(\theta )$$ Inserting these expressions into the given PDE, we obtain: $$R^{″}(r)\mathrm{\Theta }(\theta )+\frac{1}{r}{R}^{\prime }(r)\mathrm{\Theta }(\theta )+\frac{1}{r^2}R(r){\mathrm{\Theta }}^{″}(\theta )=0$$

Separate the equation

Now, we will separate the equation into two parts: one for $R(r)$ and one for $\mathrm{\Theta }(\theta )$. Divide both sides of the equation by $R(r)\mathrm{\Theta }(\theta )$: $$\frac{R^{″}(r)}{R(r)}+\frac{1}{r}\frac{R^{\prime }(r)}{R(r)}+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{″}(\theta )}{\mathrm{\Theta }(\theta )}=0$$ The left-hand side of the equation consists of two terms, one that depends on $r$ only and one that depends on $\theta$ only. Therefore, each term must be equal to a constant, say $-k^2$: $$\frac{R^{″}(r)}{R(r)}+\frac{1}{r}\frac{R^{\prime }(r)}{R(r)}=-k^2,\frac{{\mathrm{\Theta }}^{″}(\theta )}{\mathrm{\Theta }(\theta )}=k^2$$

Solve the ODE for Θ

First, we will solve the ODE for $\mathrm{\Theta }(\theta )$: $${\mathrm{\Theta }}^{″}(\theta )-k^2\mathrm{\Theta }(\theta )=0$$ This is a second-order linear homogeneous ODE with constant coefficients. The general solution is given by: $$\mathrm{\Theta }(\theta )=A\cos (k\theta )+B\sin (k\theta )$$ Now, we need to consider the boundary conditions for $\mathrm{\Theta }(\theta )$: $u_{\theta }(r,0)=0$ and $u_{\theta }(r,\gamma )=0$. So, $${\mathrm{\Theta }}^{\prime }(0)=-Ak\sin (0)+Bk\cos (0)=0\Rightarrow B=0$$ and $${\mathrm{\Theta }}^{\prime }(\gamma )=-Ak\sin (k\gamma )=0$$ Since $A$ cannot be zero (otherwise it would imply the trivial solution), we have that $\sin (k\gamma )=0$, which means $k\gamma =n\pi$ for $n\in \mathbb{Z}$. Thus, $k=\frac{n\pi }{\gamma }$, and the solution for $\mathrm{\Theta }(\theta )$ is given by: $${\mathrm{\Theta }}_n(\theta )=A_n\cos \left(\frac{n\pi }{\gamma }\theta \right)$$

Solve the ODE for R

Now, we will solve the ODE for $R(r)$: $$r^2{R}^{″}(r)+r{R}^{\prime }(r)+k^2{r}^{2}R(r)=0$$ With $k=\frac{n\pi }{\gamma }$, the equation becomes: $$r^2{R}^{″}(r)+r{R}^{\prime }(r)+\frac{n^2{\pi }^2}{{\gamma }^2}{r}^{2}R(r)=0$$ This is a second-order Euler equation. The general solution is given by: $$R_n(r)=C_n{r}^a+D_n{r}^{-a},\text{with}a=\frac{n\pi }{\gamma }$$ Now, apply the given boundary condition $u({\rho }_0,\theta )=0$. Since ${\mathrm{\Theta }}_n(\theta )$ cannot be equal to zero for all $\theta$, $R_n({\rho }_0)$ must be zero: $$R_n({\rho }_0)=C_n{\rho }_0^a+D_n{\rho }_0^{-a}=0$$ We can use this equation to eliminate either $C_n$ or $D_n$ from the expression for $R_n(r)$. Eliminating $C_n$: $$C_n=-\frac{D_n}{{\rho }_0^{2a}}\Rightarrow R_n(r)=D_n(\frac{1}{{\rho }_0^{2a}}{r}^a-r^{-a})$$ Finally, the separated solution is given by the product $u_n(r,\theta )=R_n(r){\mathrm{\Theta }}_n(\theta )$: $$u_n(r,\theta )=[A_n\cos \left(\frac{n\pi }{\gamma }\theta \right)]\left[D_n(\frac{1}{{\rho }_0^{2a}}{r}^a-r^{-a})\right]$$ We still need to find the coefficients $A_n$ and $D_n$ and the sum of the separated solutions that satisfies the boundary conditions.

Key concepts

Separation of Variables

The technique known as separation of variables is a fundamental method to solve partial differential equations (PDEs), especially when dealing with problems that possess some sort of symmetry. The essence of this method is to represent the solution as a product of functions, each dependent on a single independent variable. For example, a solution to a PDE in two variables, say r and θ, would be represented as u(r, θ) = R(r)Θ(θ).

In the given exercise, the separation of variables is utilized to transform a PDE into two ordinary differential equations (ODEs), each in a single variable. This is done by assuming a solution of the form u(r, θ) = R(r)Θ(θ) and substituting this into the original PDE. The resulting equation can then be organized such that one side depends only on r, while the other depends only on θ, leading to the conclusion that both sides must be constant since r and θ can vary independently.

The process simplifies the complex PDE into simpler, more manageable ODEs that can be solved using standard techniques. Once the individual functions R(r) and Θ(θ) are determined, the product of these functions gives the solution to the original PDE. It is important to note that, after solving for the functions, one must apply the boundary conditions to find specific solutions and determine any arbitrary constants involved.

Boundary Value Problems

A boundary value problem (BVP) is a differential equation coupled with a set of additional restraints called boundary conditions. These problems arise in situations where you know the behavior of a solution at certain points and are looking for a solution that fulfills the differential equation as well as respects the boundary conditions.

In our example, the boundary conditions are given for the function u(r, θ) at specific points. The conditions u(ρ₀, θ) = 0 and u_r(ρ, θ) = g(θ) specify the values of the function and its radial derivative at the internal and external boundaries, respectively, while u_θ(r, 0) = 0 and u_θ(r, γ) = 0 set constraints on the angular derivative at the edges of the angular domain.

Solving a BVP involves not only finding the general solution to the ODEs but also adjusting the arbitrary constants so that the boundary conditions are satisfied. This often requires careful consideration and may involve the use of various mathematical techniques, from straightforward substitution to the more advanced methods such as eigenvalue problems and Fourier series expansion. Ultimately, the solution to a BVP is a synthesis of the theory of ODEs and the physical or geometric constraints imposed by the problem.

Euler's Differential Equation

A distinct class of ODEs is represented by Euler's differential equation, which appears in the form ax²y'' + bxy' + cy = 0, where a, b, and c are constants, and the primes denote derivatives with respect to x. These equations are named after the Swiss mathematician Leonhard Euler and are characterized by the variable coefficients being powers of the independent variable.

In this particular differential equation, the presence of terms like r²R''(r) and rR'(r) signifies an Euler equation. The solutions to Euler equations can often be found by trying a solution of the form r^a, where a is a constant. This leads to an algebraic characteristic equation that determines the possible values of a and, consequently, the form of the general solution.

For the exercise at hand, after determining the appropriate form of k using the boundary conditions for Θ(θ), the resulting Euler equation for R(r) presents a solution that involves powers of r. By applying the remaining boundary conditions, one can find expressions for the coefficients involved in the final solution. Euler's equations are particularly useful in problems with radial symmetry and appear frequently in physics and engineering, often in the study of phenomena such as heat conduction, wave propagation, and quantum mechanics.