- #1
AntSC
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Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:
[tex] {u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0 [/tex]
Solution:
Try [itex] u\left ( x \right )=Ae^{rx} [/itex], which gives roots [itex] r=-1\pm \sqrt{1-\lambda } [/itex]. Solution is altered with [itex] \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1 [/itex]
For the first case [itex] \lambda <1 [/itex]:
General solution:
[tex] u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x} [/tex]
[tex] u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x [/tex]
Applying boundaries: (this is where my question lies - how to correctly apply BCs)
[itex] u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 [/itex] (some cases I've seen the conclusion that only [itex] C=0 [/itex]). Do i assume that as [itex] \cosh [/itex] is never zero that [itex]C=0[/itex] and therefore it must be that [itex]D=0[/itex]. Or do i only take [itex]C=0[/itex] and then apply the second BC to see what happens to [itex]D[/itex]?
The latter (assuming [itex]C=0[/itex]) gives [itex]D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0[/itex]. So either [itex]D=0[/itex] or [itex]\sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n[/itex].
I'm confused by the rules for the BCs. Can anyone point out how to proceed? Thanks
[tex] {u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0 [/tex]
Solution:
Try [itex] u\left ( x \right )=Ae^{rx} [/itex], which gives roots [itex] r=-1\pm \sqrt{1-\lambda } [/itex]. Solution is altered with [itex] \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1 [/itex]
For the first case [itex] \lambda <1 [/itex]:
General solution:
[tex] u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x} [/tex]
[tex] u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x [/tex]
Applying boundaries: (this is where my question lies - how to correctly apply BCs)
[itex] u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 [/itex] (some cases I've seen the conclusion that only [itex] C=0 [/itex]). Do i assume that as [itex] \cosh [/itex] is never zero that [itex]C=0[/itex] and therefore it must be that [itex]D=0[/itex]. Or do i only take [itex]C=0[/itex] and then apply the second BC to see what happens to [itex]D[/itex]?
The latter (assuming [itex]C=0[/itex]) gives [itex]D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0[/itex]. So either [itex]D=0[/itex] or [itex]\sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n[/itex].
I'm confused by the rules for the BCs. Can anyone point out how to proceed? Thanks
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