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To use the method of separation of variables, we write as
![{\displaystyle u(r,\theta )=R(r)\Theta (\theta ).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f379d643e190a2769a99c0d1f37c9e519534f878)
We compute that
![{\displaystyle u_{r}(r,\theta )=R'(r)\Theta (\theta ),\;u_{rr}(r,\theta )=R''(r)\Theta (\theta ),}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4a72d5e4b5da8c41de9434e9e6d8b381fd095b9a)
and
![{\displaystyle u_{\theta \theta }(r,\theta )=R(r)\Theta ''(\theta ).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/97fec9a436f3f8cdbc2515eb6f62d407b48a221e)
Substitute the above into our differential equation
![{\displaystyle u_{rr}+{\frac {1}{r}}u_{r}+{\frac {1}{r^{2}}}u_{\theta \theta }=0,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6cf988e5caee5682a44a80a29f9f35fade9f357c)
we get that
![{\displaystyle R''\Theta +{\frac {1}{r}}R'\Theta +{\frac {1}{r^{2}}}R\Theta ''=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/566f85c841be1560f3f97d56ff4311ac1993f7eb)
Next, we would like to group all the terms onto one side and all the terms onto another, so divide the entire equation by to get
![{\displaystyle {\frac {R''\Theta }{R\Theta }}+{\frac {1}{r}}{\frac {R'\Theta }{R\Theta }}+{\frac {1}{r^{2}}}{\frac {R\Theta ''}{R\Theta }}=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d718c07e56d97c1683ca51044004e7bbfc9c13e1)
which gives
![{\displaystyle {\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cf78193d4ffc28ea248fe736ec68d84e9004a45b)
To get rid of the factor of in the last term on the left, we will multiply the equation by to get
![{\displaystyle r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}+{\frac {\Theta ''}{\Theta }}=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9bbec6f79ddeaca9d8cc377c70ecc6178b25a5eb)
which gives
![{\displaystyle r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}=-{\frac {\Theta ''}{\Theta }}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/70aec786ffb739b6c43a57ad9585f25fd08c55ec)
Here, and are independent of each other. However, the right hand side of the equation is a function of while the right hand side a function of , so both sides of the equation must be equal to a constant which we will denote . In other words, we have
![{\displaystyle r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}=-{\frac {\Theta ''}{\Theta }}=\mu .}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/12616d77a4607a00faf56d76e5bdf39eee0fd9e7)
This gives us two equations
![{\displaystyle r^{2}R''+rR'=\mu R\;\;{\text{and}}\;\;-\Theta ''=\mu \Theta .}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f752a0a26fa256611fd51664fa908f2eb631e097)
We will first solve the equation for .
Before we go ahead to do so, we would like to figure out the boundary conditions for . The boundary conditions
![{\displaystyle u(1,\theta )=0\;\;{\text{and}}\;\;u(2,\theta )=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1168f885d12a30c7d85d6747ba315ca667985edb)
imply that
![{\displaystyle R(1)\Theta (\theta )=0\;\;{\text{and}}\;\;R(2)\Theta (\theta )=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c15eca8537b7ac565e12a9dce97e585c42cea65a)
Since we do not want to be since this will give the trivial solution (i.e. ) which is not of interest to us. So the boundary conditions give us
![{\displaystyle R(1)=0\;\;{\text{and}}\;\;R(2)=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/78122d79e2d1933832fe169693ca058101570636)
We will now solve the equation for which says
![{\displaystyle r^{2}R''+rR'-\mu R=0.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/eb18e8f09be01ff4f3e9c083dcb3bf9562000203)
This is the Euler's equation and we recall that we should look for solutions in the form
![{\displaystyle R(r)=r^{\alpha }\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9885cf96cd0b745d619a1722df5993fdec78668c)
for an unknown constant to be determined.
We compute that
![{\displaystyle R'=\alpha r^{\alpha -1}\;\;{\text{and}}\;\;R''=\alpha (\alpha -1)r^{\alpha -2}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/eac74bca81b6289de9881e588b9774726cc1993d)
Substituting these into
![{\displaystyle r^{2}R''+rR'-\mu R=0\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/daa565c6b5b1c9ee3540f940957b0e91093b9935)
we get that
![{\displaystyle \alpha (\alpha -1)+\alpha =\mu \,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/15418f87ad3fb4454955dffb361ef00dc117d38b)
which after simplifying gives
![{\displaystyle \alpha ^{2}=\mu .\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8ddc6b2f638431e9f1f612b26d2f81bafaef735f)
Here, the sign of will affect the form of the solution, so we have three cases to consider. They are
- case 1:
![{\displaystyle \mu =-\lambda ^{2}<0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ee9bd86cc5e5fa633ca8daaaed8c9dda6e135950)
- case 2:
![{\displaystyle \mu =0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3753282c0ad2ea1e7d63f39425efd13c37da3169)
- case 3:
![{\displaystyle \mu =\lambda ^{2}>0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/306224f953eb43d24320c0c35708883bd161c480)
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
![{\displaystyle \alpha ^{2}=\mu =-\lambda ^{2}.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/99cc061bd0d4fd79d17f26a273eae7ae1dab18ec)
This gives
![{\displaystyle \alpha =\pm \lambda i.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e71bd43d7dbcbd03ff96c95ac3c07555fda353c6)
Recall that for the Euler's equation, if is in the form of , the solution of the Euler's equation can be written as
![{\displaystyle R(r)=C_{1}r^{a}\cos(b\ln(r))+C_{2}r^{a}\sin(b\ln(r))\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/20229f73492805fb305b066270fbd2520ef0420a)
for some arbitrary constants and .
For us, we have and , so our solution for is
![{\displaystyle R(r)=C_{1}\cos(\lambda \ln(r))+C_{2}\sin(\lambda \ln(r)).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3c5e99d69bbec66baff3cec37d51a61fa58f9f62)
Next, we need to match the boundary conditions
![{\displaystyle R(1)=0\Rightarrow 0=C_{1}\cos(\lambda \underbrace {\ln(1)} _{=0})+C_{2}\sin(\lambda \underbrace {\ln(1)} _{=0})=C_{1}\underbrace {\cos(0)} _{=1}+C_{2}\underbrace {\sin(0)} _{=0}=C_{1}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c9e46f8a284c0eb070109b4fa5dd58095b2b3013)
Hence, and
![{\displaystyle R(r)=C_{2}\sin(\lambda \ln(r)).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/903bc4c3e8fb41b16433a487617c12d20707f9f0)
Now,
![{\displaystyle R(2)=0\Rightarrow C_{2}\sin(\lambda \ln(2))=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cb6e0206f6bafcb186909898abe981f25589ce04)
If , we will have the trivial solution. To get non-trivial solutions, we need
![{\displaystyle \sin(\lambda \ln(2))=0\Rightarrow \lambda \ln(2)=n\pi \Rightarrow \lambda ={\frac {n\pi }{\ln(2)}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a40e3325bb40fec322b430b073cbc5c6c04597e2)
In other words, we have an infinite numbers of eigenvalues give by
![{\displaystyle \lambda _{n}={\frac {n\pi }{\ln(2)}}\;\;{\text{for}}\;\;n={\color {red}1},2,3,\cdots }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c9a3b061acddb778c99b026a4d134d55b9743d89)
with the corresponding eigenfunctions
![{\displaystyle R_{n}(r)=C_{n}\sin(\lambda _{n}\ln(r)).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/85657a9ef4ac36ed65007cde504f3c560783b453)
Now, we will go back to solve the equation.
Recall that we have
so the equation
becomes
![{\displaystyle {\begin{aligned}-\Theta ''&=-\lambda ^{2}\Theta \\\Theta ''&=\lambda ^{2}\Theta \end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/26f4ee63fa7302c13b175e320adc3065780a9d05)
which gives solutions of the form
![{\displaystyle \Theta =A\cosh(\lambda \theta )+B\sinh(\lambda \theta )\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d2e0a6c54451561e9a803b97ad82d13d44d9ce37)
for some arbitrary constants and .
Since we found that an infinite number of , we have an infinite number of given by
![{\displaystyle \Theta _{n}=A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta ).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/50ca39adda97d21b71c0ce1469cb1f1593835336)
Putting everything together, we can express as
![{\displaystyle u(r,t)=\sum _{n=1}^{\infty }C_{n}\sin(\lambda _{n}\ln(r))\left(A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta )\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/14ba248064f3ba84f04381bebf110d4a7166a34f)
We can absorb the arbitrary constant into and to get
![{\displaystyle u(r,t)=\sum _{n=1}^{\infty }\sin(\lambda _{n}\ln(r))\left(A_{n}\cosh(\lambda _{n}\theta )+B_{n}\sinh(\lambda _{n}\theta )\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ac7fd067a6f4d1e4db3d083ce83ad6fd88a1131d)
Finally, we will use the other two boundary conditions and to find and .
![{\displaystyle {\begin{aligned}&u(r,0)=0\,\\&\sum _{n=1}^{\infty }\sin(\lambda _{n}\ln(r))(A_{n}\underbrace {\cosh(0)} _{=1}+B_{n}\underbrace {\sinh(0)} _{=0})=0\\&\sum _{n=1}^{\infty }A_{n}\sin(\lambda _{n}\ln(r))=0.\\&A_{n}=0\;\;{\text{for each}}\;\;n=1,2,3,\cdots \end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3823fcbef9cf6e67b28afec9ad68a7bd68fc1540)
Hence
![{\displaystyle {\begin{aligned}u(r,t)&=\sum _{n=1}^{\infty }B_{n}\sin(\lambda _{n}\ln(r))\sinh(\lambda _{n}\theta )\\&=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}\theta \right)\end{aligned}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b6308200e26634e941c7b60f9fc37a04d57c49ac)
Finally, to find , we use the last boundary condition
.
This boundary condition gives us that
![{\displaystyle u\left(r,{\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }B_{n}\sin \left({\frac {n\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/77abc8178a1dab4589723acdaf857885b2593261)
If we expand out the summation, we see that
![{\displaystyle {\begin{aligned}&B_{1}\sin \left({\frac {\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{1}{\frac {\pi }{2}}\right)+{\color {blue}B_{2}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{2}{\frac {\pi }{2}}\right)}\\&+B_{3}\sin \left({\frac {3\pi }{\ln(2)}}\ln(r)\right)\sinh \left(\lambda _{3}{\frac {\pi }{2}}\right)+\cdots ={\color {blue}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)}\end{aligned}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7ede6594d59183bc6e55905de3ac82393de5e0fe)
Here, we see that is the only term on the left hand side that matches with the right hand side. Hence, we have
![{\displaystyle B_{n}\sinh \left(\lambda _{n}{\frac {\pi }{2}}\right)=0\;\;{\text{if}}\;\;n\not =2}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/366aec0af668c4c0e9a787c249f17432a29286a9)
and
![{\displaystyle B_{2}\sinh \left(\lambda _{2}{\frac {\pi }{2}}\right)=1}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4d41db58f4fca8d410baf94d1347f7d1208fdb1e)
![{\displaystyle \Rightarrow B_{2}={\frac {1}{\sinh(\lambda _{2}{\frac {\pi }{2}})}}={\frac {1}{\sinh({\frac {2\pi }{\ln(2)}}{\frac {\pi }{2}})}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0feb32ffd8b2dabbed0949c3db8f5c0c132c3836)
In other words, the solution is
![{\displaystyle u(r,\theta )={\frac {1}{\sinh \left({\frac {2\pi }{\ln(2)}}{\frac {\pi }{2}}\right)}}\sin \left({\frac {2\pi }{\ln(2)}}\ln(r)\right)\sinh \left({\frac {2\pi }{\ln(2)}}\theta \right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/bd44754828c025c9571374c3ff1daee436a8aac9)
Appendix:
We will address two final questions in this appendix.
Question 1: Refer to the solution above, why doesn't lead to non-trivial solutions for the Euler's equation?
Answer to question 1: Suppose , then
![{\displaystyle \alpha ^{2}=\mu \Rightarrow \alpha =0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/413d9a88cbe287c25a462d67792ae86dea01c997)
The solution to the Euler's equation is
![{\displaystyle R(r)=C_{1}\ln(r)+C_{2}\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b30b27e9dce59d691f1cdb7708e6e55199971087)
for some arbitrary constants and .
The boundary condition implies
![{\displaystyle R(1)=C_{1}\underbrace {\ln(1)} _{=0}+C_{2}=0\Rightarrow C_{2}=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fc5ce24163cdcb1aa4fef720fbda16e522023726)
So
![{\displaystyle R(r)=C_{1}\ln(r).\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e1c70827acad8157810d606c5988cc70cd38f88c)
The boundary condition implies
![{\displaystyle R(2)=C_{1}\ln(2)=0\Rightarrow C_{1}=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e1a51536a5d68e32d07831937b105df691374c5a)
Hence, we only get the trivial solution.
Question 2: Refer to the solution above, why doesn't lead to non-trivial solutions for the Euler's equation?
Answer to question 2: Suppose , then
![{\displaystyle \alpha ^{2}=\mu \Rightarrow \alpha =\pm \lambda .}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/74bf7d775f200ea286680223350fa869a0ca2d32)
The solution to the Euler's equation is
![{\displaystyle R(r)=C_{1}r^{\lambda }+C_{2}r^{-\lambda }\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/89b1dcb5b6df2591c470e1a1820db7d6dd9efffe)
for some arbitrary constants and .
The boundary condition implies
![{\displaystyle R(1)=C_{1}+C_{2}=0\Rightarrow C_{1}=-C_{2}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/30c8407ee3b1cbf861c143c467007d407ba06c08)
So
![{\displaystyle R(r)=C_{1}r^{\lambda }-C_{1}r^{-\lambda }.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c8bc5c3576ddb505f3d85b24a4967527ca6d6c80)
The boundary condition implies
![{\displaystyle R(2)=C_{1}2^{\lambda }-C_{1}2^{-\lambda }=0.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9b81cf107594fdae461a173b03c2eeaa4dc504d0)
If , we get a trivial solution, so suppose , then we must have
![{\displaystyle 2^{\lambda }=2^{-\lambda }\Rightarrow 2^{2\lambda }=1\Rightarrow \lambda =0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7ed2bd67f7ad6718594f7b79382e7c7886c52027)
However, this contradicts with our original assumption that .
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