Science:Math Exam Resources/Courses/MATH257/December 2011/Question 05 (a)

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MATH257 December 2011

  •  Q1  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  

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  •  December 2011  •  

Question 05 (a)
Consider the following problem involving Laplace's equation in an annular region: $u_{rr}+\frac{1}{r}{u}_r+\frac{1}{r^2}{u}_{\theta \theta }=0,1<r<2,0<\theta <\pi /2.$ $u(1,\theta )=0,u(2,\theta )=0,u(r,0)=0,u(r,\pi /2)=f(r).$ Use the method of separation of variables to solve the problem when $f(r)=\sin (\frac{2\pi }{\ln (2)}\ln (r)).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Similar to question 2b of this exam, when using the method of separation of variables, we would like to write $u$ as a product of two solutions of a single variable. In question 2b, we wrote $u$ as $u=\varphi (x)\psi (t)$ but we need something slightly different this time (because there is no $x$ or $t$ anywhere in the equation!). How should we write $u$ as a product of two solutions of a single variable this time?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

To use the method of separation of variables, we write $u$ as $u(r,\theta )=R(r)\mathrm{\Theta }(\theta ).$ We compute that $u_r(r,\theta )=R^{\prime }(r)\mathrm{\Theta }(\theta ),u_{rr}(r,\theta )=R^{″}(r)\mathrm{\Theta }(\theta ),$ and $u_{\theta \theta }(r,\theta )=R(r){\mathrm{\Theta }}^{″}(\theta ).$ Substitute the above into our differential equation $u_{rr}+\frac{1}{r}{u}_r+\frac{1}{r^2}{u}_{\theta \theta }=0,$ we get that $R^{″}\mathrm{\Theta }+\frac{1}{r}{R}^{\prime }\mathrm{\Theta }+\frac{1}{r^2}R{\mathrm{\Theta }}^{″}=0.$ Next, we would like to group all the $R$ terms onto one side and all the $\mathrm{\Theta }$ terms onto another, so divide the entire equation by $R\mathrm{\Theta }$ to get $\frac{R^{″}\mathrm{\Theta }}{R\mathrm{\Theta }}+\frac{1}{r}\frac{R^{\prime }\mathrm{\Theta }}{R\mathrm{\Theta }}+\frac{1}{r^2}\frac{R{\mathrm{\Theta }}^{″}}{R\mathrm{\Theta }}=0$ which gives $\frac{R^{″}}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}=0.$ To get rid of the factor of $\frac{1}{r^2}$ in the last term on the left, we will multiply the equation by $r^2$ to get $r^2\frac{R^{″}}{R}+r\frac{R^{\prime }}{R}+\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}=0$ which gives $r^2\frac{R^{″}}{R}+r\frac{R^{\prime }}{R}=-\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}.$ Here, $r$ and $\theta$ are independent of each other. However, the right hand side of the equation is a function of $r$ while the right hand side a function of $\theta$, so both sides of the equation must be equal to a constant which we will denote $\mu$. In other words, we have $r^2\frac{R^{″}}{R}+r\frac{R^{\prime }}{R}=-\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}=\mu .$ This gives us two equations $r^2{R}^{″}+r{R}^{\prime }=\mu R\text{and}-{\mathrm{\Theta }}^{″}=\mu \mathrm{\Theta }.$ We will first solve the equation for $R$. Before we go ahead to do so, we would like to figure out the boundary conditions for $R$. The boundary conditions $u(1,\theta )=0\text{and}u(2,\theta )=0$ imply that $R(1)\mathrm{\Theta }(\theta )=0\text{and}R(2)\mathrm{\Theta }(\theta )=0.$ Since we do not want $\mathrm{\Theta }$ to be $0$ since this will give the trivial solution (i.e. $u\equiv 0$) which is not of interest to us. So the boundary conditions give us $R(1)=0\text{and}R(2)=0.$ We will now solve the equation for $R$ which says $r^2{R}^{″}+r{R}^{\prime }-\mu R=0.$ This is the Euler's equation and we recall that we should look for solutions in the form $R(r)=r^{\alpha }$ for an unknown constant $\alpha$ to be determined. We compute that $R^{\prime }=\alpha r^{\alpha -1}\text{and}{R}^{″}=\alpha (\alpha -1)r^{\alpha -2}.$ Substituting these into $r^2{R}^{″}+r{R}^{\prime }-\mu R=0$ we get that $\alpha (\alpha -1)+\alpha =\mu$ which after simplifying gives ${\alpha }^2=\mu .$ Here, the sign of $\mu$ will affect the form of the solution, so we have three cases to consider. They are case 1: $\mu =-{\lambda }^2<0$ case 2: $\mu =0$ case 3: $\mu ={\lambda }^2>0$ Further calculations show that case 2 and case 3 only give trivial solutions. If you are not sure, you should try verifying that yourself. For completeness of the solution, I will also include the computation in the appendix of this solution. Now, case 1: says that ${\alpha }^2=\mu =-{\lambda }^2.$ This gives $\alpha =\pm \lambda i.$ Recall that for the Euler's equation, if $\alpha$ is in the form of $\alpha =a\pm bi$, the solution of the Euler's equation can be written as $R(r)=C_1{r}^a\cos (b\ln (r))+C_2{r}^a\sin (b\ln (r))$ for some arbitrary constants $C_1$ and $C_2$. For us, we have $a=0$ and $b=\lambda$, so our solution for $R$ is $R(r)=C_1\cos (\lambda \ln (r))+C_2\sin (\lambda \ln (r)).$ Next, we need to match the boundary conditions $R(1)=0\text{and}R(2)=0.$ $R(1)=0\Rightarrow 0=C_1\cos (\lambda \underset{=0}{\underbrace{\ln (1)}})+C_2\sin (\lambda \underset{=0}{\underbrace{\ln (1)}})=C_1\underset{=1}{\underbrace{\cos (0)}}+C_2\underset{=0}{\underbrace{\sin (0)}}=C_1.$ Hence, $C_1=0$ and $R(r)=C_2\sin (\lambda \ln (r)).$ Now, $R(2)=0\Rightarrow C_2\sin (\lambda \ln (2))=0.$ If $C_2=0$, we will have the trivial solution. To get non-trivial solutions, we need $\sin (\lambda \ln (2))=0\Rightarrow \lambda \ln (2)=n\pi \Rightarrow \lambda =\frac{n\pi }{\ln (2)}.$ In other words, we have an infinite numbers of eigenvalues ${\lambda }_n$ give by ${\lambda }_n=\frac{n\pi }{\ln (2)}\text{for}n=1,2,3,\cdots$ with the corresponding eigenfunctions $R_n(r)=C_n\sin ({\lambda }_n\ln (r)).$ Now, we will go back to solve the $\theta$ equation. Recall that we have $\mu =-{\lambda }^2$ so the $\theta$ equation $-{\mathrm{\Theta }}^{″}=\mu \mathrm{\Theta }$ becomes $\begin{array}{rl}-{\mathrm{\Theta }}^{″}& =-{\lambda }^2\mathrm{\Theta }\\ {\mathrm{\Theta }}^{″}& ={\lambda }^2\mathrm{\Theta }\end{array}$ which gives solutions of the form $\mathrm{\Theta }=A\cosh (\lambda \theta )+B\sinh (\lambda \theta )$ for some arbitrary constants $A$ and $B$. Since we found that an infinite number of ${\lambda }_n=\frac{n\pi }{\ln (2)}$, we have an infinite number of ${\mathrm{\Theta }}_n$ given by ${\mathrm{\Theta }}_n=A_n\cosh ({\lambda }_n\theta )+B_n\sinh ({\lambda }_n\theta ).$ Putting everything together, we can express $u$ as $u(r,t)=\sum _{n=1}^{\mathrm{\infty }}{C}_n\sin ({\lambda }_n\ln (r))(A_n\cosh ({\lambda }_n\theta )+B_n\sinh ({\lambda }_n\theta )).$ We can absorb the arbitrary constant $C_n$ into $A_n$ and $B_n$ to get $u(r,t)=\sum _{n=1}^{\mathrm{\infty }}\sin ({\lambda }_n\ln (r))(A_n\cosh ({\lambda }_n\theta )+B_n\sinh ({\lambda }_n\theta )).$ Finally, we will use the other two boundary conditions $u(r,0)=0$ and $u(r,\frac{\pi }{2})=f(r)$ to find $A_n$ and $B_n$. $\begin{array}{rl}& u(r,0)=0\\ & \sum _{n=1}^{\mathrm{\infty }}\sin ({\lambda }_n\ln (r))(A_n\underset{=1}{\underbrace{\cosh (0)}}+B_n\underset{=0}{\underbrace{\sinh (0)}})=0\\ & \sum _{n=1}^{\mathrm{\infty }}{A}_n\sin ({\lambda }_n\ln (r))=0.\\ & A_n=0\text{for each}n=1,2,3,\cdots \end{array}$ Hence $\begin{array}{rl}u(r,t)& =\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin ({\lambda }_n\ln (r))\sinh ({\lambda }_n\theta )\\ & =\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (\frac{n\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_n\theta \right)\end{array}.$ Finally, to find $B_n$, we use the last boundary condition $u(r,\frac{\pi }{2})=\sin (\frac{2\pi }{\ln (2)}\ln (r))$. This boundary condition gives us that $u(r,\frac{\pi }{2})=\sum _{n=1}^{\mathrm{\infty }}{B}_n\sin (\frac{n\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_n\frac{\pi }{2}\right)=\sin (\frac{2\pi }{\ln (2)}\ln (r)).$ If we expand out the summation, we see that $\begin{array}{rl}& B_1\sin (\frac{\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_1\frac{\pi }{2}\right)+B_2\sin (\frac{2\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_2\frac{\pi }{2}\right)\\ & +B_3\sin (\frac{3\pi }{\ln (2)}\ln (r))\sinh \left({\lambda }_3\frac{\pi }{2}\right)+\cdots =\sin (\frac{2\pi }{\ln (2)}\ln (r))\end{array}.$ Here, we see that $n=2$ is the only term on the left hand side that matches with the right hand side. Hence, we have $B_n\sinh \left({\lambda }_n\frac{\pi }{2}\right)=0\text{if}n=2$ and $B_2\sinh \left({\lambda }_2\frac{\pi }{2}\right)=1$ $\Rightarrow B_2=\frac{1}{\sinh ({\lambda }_2\frac{\pi }{2})}=\frac{1}{\sinh (\frac{2\pi }{\ln (2)}\frac{\pi }{2})}.$ In other words, the solution is $u(r,\theta )=\frac{1}{\sinh \left(\frac{2\pi }{\ln (2)}\frac{\pi }{2}\right)}\sin (\frac{2\pi }{\ln (2)}\ln (r))\sinh \left(\frac{2\pi }{\ln (2)}\theta \right).$ Appendix: We will address two final questions in this appendix. Question 1: Refer to the solution above, why doesn't $\mu =0$ lead to non-trivial solutions for the Euler's equation? Answer to question 1: Suppose $\mu =0$, then ${\alpha }^2=\mu \Rightarrow \alpha =0.$ The solution to the Euler's equation is $R(r)=C_1\ln (r)+C_2$ for some arbitrary constants $C_1$ and $C_2$. The boundary condition $R(1)=0$ implies $R(1)=C_1\underset{=0}{\underbrace{\ln (1)}}+C_2=0\Rightarrow C_2=0.$ So $R(r)=C_1\ln (r).$ The boundary condition $R(2)=0$ implies $R(2)=C_1\ln (2)=0\Rightarrow C_1=0.$ Hence, we only get the trivial solution. Question 2: Refer to the solution above, why doesn't $\mu ={\lambda }^2>0$ lead to non-trivial solutions for the Euler's equation? Answer to question 2: Suppose $\mu ={\lambda }^2>0$, then ${\alpha }^2=\mu \Rightarrow \alpha =\pm \lambda .$ The solution to the Euler's equation is $R(r)=C_1{r}^{\lambda }+C_2{r}^{-\lambda }$ for some arbitrary constants $C_1$ and $C_2$. The boundary condition $R(1)=0$ implies $R(1)=C_1+C_2=0\Rightarrow C_1=-C_2.$ So $R(r)=C_1{r}^{\lambda }-C_1{r}^{-\lambda }.$ The boundary condition $R(2)=0$ implies $R(2)=C_1{2}^{\lambda }-C_1{2}^{-\lambda }=0.$ If $C_1=0$, we get a trivial solution, so suppose $C_1=0$, then we must have $2^{\lambda }=2^{-\lambda }\Rightarrow 2^{2\lambda }=1\Rightarrow \lambda =0.$ However, this contradicts with our original assumption that ${\lambda }^2>0$.

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