how to solve Sturm Liouville Problem y’’-λy=0, y(0) = 0, y’ (1) = 0 where λ ∈ R

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Question. solve Sturm Liouville Problem y’’-λy=0, y(0) = 0, y’ (1) = 0 where λ ∈ R

Solution:

To solve

\[y’’ – λy = 0\] (1) \[ y(0)\ =\ 0,\ y’ (1) = 0\] This is a Sturm Liouville Problem.

Case(1):

when \lambda=0 Eqn(1) becomes \[y^{\prime\prime}\ =0 \] (2) Auxiliary equation \[m^2=0\] It’s roots are \[m_1=0,\ m_2=0\] \[y\left(x\right)=c_1+c_2x\] Using boundary conditions \[y\left(0\right)=\ 0\Rightarrow c_1+c_2=0\Rightarrow c_2=-c_1\] And \[y\prime\left(1\right)=\ 0\Rightarrow c_2=0\Rightarrow c_1=0\] Thus solution of eqn (1) is \[ y\left(x\right)=0 \] This is a trivial solution in this case

Case(2):

when \[\lambda>0\] Let \[\lambda=\mu^2\] Eqn (1) becomes \[y^{\prime\prime}-\mu^2y=0\] Auxiliary equation \[m^2-\mu^2=0\] It’s roots are \[m_1=\mu,\ m_2=-\mu\] \[y\left(x\right)=c_1e^{\mu x}+c_2e^{-\mu x}\] Using boundary conditions \[y\left(0\right)=\ 0\Rightarrow c_1+c_2=0\Rightarrow c_2=-c_1 \] and \[y^\prime\left(1\right)=\ 0\Rightarrow c_1\mu e^\mu-c_2\mu e^{-\mu}=0\] Since \[c_2=-c_1 \] \[\Rightarrow\ c_1\mu\left(e^\mu-e^{-\mu}\right)=0\] for non trivial solutions \[ c_1\neq0\] hence \[e^\mu-e^{-\mu}=0\ \Rightarrow e^{2\mu}=1\] Which is not possible. Therefore \[y\left(x\right)=0 \] This is a trivial solution in this case also

Case(3):

\[\lambda<0\] Eqn (1) becomes \[y^{\prime\prime}+\mu^2y=0\] Auxiliary equation \[m^2+\mu^2=0\] It’s roots are \[m_1=i\mu,\ m_2=-i\mu\] \[y\left(x\right)=c_1\cos{\left(\mu x\right)}+c_2\sin{\left(\mu x\right)}\] Using boundary conditions \[y\left(0\right)=\ 0\Rightarrow c_1=0 \] and \[y^\prime\left(1\right)=\ 0\Rightarrow c_2\mu\cos{\left(\mu\right)}=0\Rightarrow\ \cos{\left(\mu\right)}=0\Rightarrow\mu=\frac{\left(2n+1\right)\pi}{2}\] Where n is an integer. Therefore eigen values and eigen functions of the given Sturm Liouville problem are \[\lambda_n=\frac{\left(2n+1\right)\pi}{2}\] and eigen functions are \[y_n\left(x\right)=c_2\sin{\left(\frac{\left(2n+1\right)\pi}{2}x\right)}\] Since all the eigenvalues are positive hence the eigen functions corresponding to each eigenvalue form a one dimensional vector space and so the eigen functions are unique upto a constant multiple


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