Determine the normalised eigenfunctions for the BVP: $y''+λy=0, y(0)=0, y(1)=0$
$begingroup$
Solving it I get:
$y(x)=c_1 cos(x sqrt{lambda}) + c_2 sin (x sqrt{lambda})$
$y(0)= C1 + 0 = 0, C1=0$
$y(1)=0+C2sin(sqrt{lambda})=0$
So, $(sqrt{lambda})=npi$, $({lambda})=(npi)^2$
So, $ψ_n= sin(nxpi)$
However, the solution shows that the answer is:
$ψ_n=(sqrt{2})sin(nxpi)$
Where have I gone wrong in this?
fourier-series eigenfunctions sturm-liouville
edited Dec 23 '18 at 11:28
Bernard
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
$endgroup$
add a comment |
$begingroup$
Solving it I get:
$y(x)=c_1 cos(x sqrt{lambda}) + c_2 sin (x sqrt{lambda})$
$y(0)= C1 + 0 = 0, C1=0$
$y(1)=0+C2sin(sqrt{lambda})=0$
So, $(sqrt{lambda})=npi$, $({lambda})=(npi)^2$
So, $ψ_n= sin(nxpi)$
However, the solution shows that the answer is:
$ψ_n=(sqrt{2})sin(nxpi)$
Where have I gone wrong in this?
fourier-series eigenfunctions sturm-liouville
edited Dec 23 '18 at 11:28
Bernard
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
$endgroup$
add a comment |
$begingroup$
Solving it I get:
$y(x)=c_1 cos(x sqrt{lambda}) + c_2 sin (x sqrt{lambda})$
$y(0)= C1 + 0 = 0, C1=0$
$y(1)=0+C2sin(sqrt{lambda})=0$
So, $(sqrt{lambda})=npi$, $({lambda})=(npi)^2$
So, $ψ_n= sin(nxpi)$
However, the solution shows that the answer is:
$ψ_n=(sqrt{2})sin(nxpi)$
Where have I gone wrong in this?
fourier-series eigenfunctions sturm-liouville
edited Dec 23 '18 at 11:28
Bernard
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
$endgroup$
Solving it I get:
$y(x)=c_1 cos(x sqrt{lambda}) + c_2 sin (x sqrt{lambda})$
$y(0)= C1 + 0 = 0, C1=0$
$y(1)=0+C2sin(sqrt{lambda})=0$
So, $(sqrt{lambda})=npi$, $({lambda})=(npi)^2$
So, $ψ_n= sin(nxpi)$
However, the solution shows that the answer is:
$ψ_n=(sqrt{2})sin(nxpi)$
Where have I gone wrong in this?
fourier-series eigenfunctions sturm-liouville
fourier-series eigenfunctions sturm-liouville
edited Dec 23 '18 at 11:28
Bernard
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
edited Dec 23 '18 at 11:28
Bernard
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
edited Dec 23 '18 at 11:28
Bernard
124k741118
edited Dec 23 '18 at 11:28
Bernard
124k741118
edited Dec 23 '18 at 11:28
Bernard
124k741118
124k741118
asked Dec 23 '18 at 11:24
luffyluffy
92
asked Dec 23 '18 at 11:24
luffyluffy
92
asked Dec 23 '18 at 11:24
luffyluffy
92
92
add a comment |
add a comment |
1 Answer
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$begingroup$
The solution is $C_2sin (npi x).$ It is normalized when $$1=int_{0}^{1} (C_2 sin(npi x))^2 dx=frac{(C_2)^2}{2}.$$ From this follows $C_2=sqrt 2.$
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
The solution is $C_2sin (npi x).$ It is normalized when $$1=int_{0}^{1} (C_2 sin(npi x))^2 dx=frac{(C_2)^2}{2}.$$ From this follows $C_2=sqrt 2.$
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
The solution is $C_2sin (npi x).$ It is normalized when $$1=int_{0}^{1} (C_2 sin(npi x))^2 dx=frac{(C_2)^2}{2}.$$ From this follows $C_2=sqrt 2.$
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
The solution is $C_2sin (npi x).$ It is normalized when $$1=int_{0}^{1} (C_2 sin(npi x))^2 dx=frac{(C_2)^2}{2}.$$ From this follows $C_2=sqrt 2.$
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
$endgroup$
The solution is $C_2sin (npi x).$ It is normalized when $$1=int_{0}^{1} (C_2 sin(npi x))^2 dx=frac{(C_2)^2}{2}.$$ From this follows $C_2=sqrt 2.$
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
answered Dec 23 '18 at 13:08
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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