Symmetric Boundary Conditions/Eigenvalues (PDEs)












0














Consider the following eigenvalue problem for the Laplacian



$-Delta u = lambda u$ in $U$



$u + a left(frac{partial u}{partial v}right)$ on $partial U$



where $v$ is the outward unit normal to the boundary. Suppose that $a(x) geq 0$ for any $x in partial U$. Show that the boundary condition given in this system is symmetric.






pde eigenvalues-eigenvectors boundary-value-problem laplacian







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  • I can't discern what you boundary condition is. Please write it as an equation.
    – DisintegratingByParts
    Dec 11 '18 at 23:54
















0














Consider the following eigenvalue problem for the Laplacian



$-Delta u = lambda u$ in $U$



$u + a left(frac{partial u}{partial v}right)$ on $partial U$



where $v$ is the outward unit normal to the boundary. Suppose that $a(x) geq 0$ for any $x in partial U$. Show that the boundary condition given in this system is symmetric.






pde eigenvalues-eigenvectors boundary-value-problem laplacian







share|cite|improve this question
























  • I can't discern what you boundary condition is. Please write it as an equation.
    – DisintegratingByParts
    Dec 11 '18 at 23:54














0












0








0


1





Consider the following eigenvalue problem for the Laplacian



$-Delta u = lambda u$ in $U$



$u + a left(frac{partial u}{partial v}right)$ on $partial U$



where $v$ is the outward unit normal to the boundary. Suppose that $a(x) geq 0$ for any $x in partial U$. Show that the boundary condition given in this system is symmetric.






pde eigenvalues-eigenvectors boundary-value-problem laplacian







share|cite|improve this question















Consider the following eigenvalue problem for the Laplacian



$-Delta u = lambda u$ in $U$



$u + a left(frac{partial u}{partial v}right)$ on $partial U$



where $v$ is the outward unit normal to the boundary. Suppose that $a(x) geq 0$ for any $x in partial U$. Show that the boundary condition given in this system is symmetric.






pde eigenvalues-eigenvectors boundary-value-problem laplacian




pde eigenvalues-eigenvectors boundary-value-problem laplacian






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edited Dec 13 '18 at 4:24









Dylan

12.4k31026




12.4k31026










asked Dec 11 '18 at 21:33









Tim

1




1












  • I can't discern what you boundary condition is. Please write it as an equation.
    – DisintegratingByParts
    Dec 11 '18 at 23:54


















  • I can't discern what you boundary condition is. Please write it as an equation.
    – DisintegratingByParts
    Dec 11 '18 at 23:54
















I can't discern what you boundary condition is. Please write it as an equation.
– DisintegratingByParts
Dec 11 '18 at 23:54




I can't discern what you boundary condition is. Please write it as an equation.
– DisintegratingByParts
Dec 11 '18 at 23:54










1 Answer
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Recall that a symmetric boundary condition is one that satisfies
$$<u, Delta v > = <Delta u, v >$$
where $< cdot, cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity:



$$<u, Delta v> - <Delta u, v> = int_{U} (v Delta u - u Delta v) dsigma = int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma $$
Now we break up the boundary into cases,
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma + int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma$$



By the round-robin condition, if $$a(x) = 0 implies u = 0 = v implies int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = 0$$
So then consider:
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U} frac{-vu}{a} + frac{uv}{a} dsigma = 0$$
Therefore,
$$<u, Delta v> - <Delta u, v> = 0 implies <u, Delta v> = <Delta u, v>$$
Which proves the boundary condition is symmetric!






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  • use langle and rangle for angle brackets: $langle$ $rangle$
    – Dylan
    Dec 13 '18 at 4:24











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Recall that a symmetric boundary condition is one that satisfies
$$<u, Delta v > = <Delta u, v >$$
where $< cdot, cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity:



$$<u, Delta v> - <Delta u, v> = int_{U} (v Delta u - u Delta v) dsigma = int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma $$
Now we break up the boundary into cases,
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma + int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma$$



By the round-robin condition, if $$a(x) = 0 implies u = 0 = v implies int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = 0$$
So then consider:
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U} frac{-vu}{a} + frac{uv}{a} dsigma = 0$$
Therefore,
$$<u, Delta v> - <Delta u, v> = 0 implies <u, Delta v> = <Delta u, v>$$
Which proves the boundary condition is symmetric!






share|cite|improve this answer























  • use langle and rangle for angle brackets: $langle$ $rangle$
    – Dylan
    Dec 13 '18 at 4:24
















0














Recall that a symmetric boundary condition is one that satisfies
$$<u, Delta v > = <Delta u, v >$$
where $< cdot, cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity:



$$<u, Delta v> - <Delta u, v> = int_{U} (v Delta u - u Delta v) dsigma = int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma $$
Now we break up the boundary into cases,
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma + int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma$$



By the round-robin condition, if $$a(x) = 0 implies u = 0 = v implies int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = 0$$
So then consider:
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U} frac{-vu}{a} + frac{uv}{a} dsigma = 0$$
Therefore,
$$<u, Delta v> - <Delta u, v> = 0 implies <u, Delta v> = <Delta u, v>$$
Which proves the boundary condition is symmetric!






share|cite|improve this answer























  • use langle and rangle for angle brackets: $langle$ $rangle$
    – Dylan
    Dec 13 '18 at 4:24














0












0








0






Recall that a symmetric boundary condition is one that satisfies
$$<u, Delta v > = <Delta u, v >$$
where $< cdot, cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity:



$$<u, Delta v> - <Delta u, v> = int_{U} (v Delta u - u Delta v) dsigma = int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma $$
Now we break up the boundary into cases,
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma + int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma$$



By the round-robin condition, if $$a(x) = 0 implies u = 0 = v implies int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = 0$$
So then consider:
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U} frac{-vu}{a} + frac{uv}{a} dsigma = 0$$
Therefore,
$$<u, Delta v> - <Delta u, v> = 0 implies <u, Delta v> = <Delta u, v>$$
Which proves the boundary condition is symmetric!






share|cite|improve this answer














Recall that a symmetric boundary condition is one that satisfies
$$<u, Delta v > = <Delta u, v >$$
where $< cdot, cdot >$ is the $L_2$ inner product. Observe, by applying Green's identity:



$$<u, Delta v> - <Delta u, v> = int_{U} (v Delta u - u Delta v) dsigma = int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma $$
Now we break up the boundary into cases,
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma + int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma$$



By the round-robin condition, if $$a(x) = 0 implies u = 0 = v implies int_{partial U text{and } a(x)= 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = 0$$
So then consider:
$$int_{partial U} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U text{and } a(x) not = 0} (v frac{partial u}{partial nu} - u frac{partial v}{partial nu} ) dsigma = int_{partial U} frac{-vu}{a} + frac{uv}{a} dsigma = 0$$
Therefore,
$$<u, Delta v> - <Delta u, v> = 0 implies <u, Delta v> = <Delta u, v>$$
Which proves the boundary condition is symmetric!







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edited Dec 12 '18 at 0:49

























answered Dec 12 '18 at 0:44









gdepaul

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  • use langle and rangle for angle brackets: $langle$ $rangle$
    – Dylan
    Dec 13 '18 at 4:24


















  • use langle and rangle for angle brackets: $langle$ $rangle$
    – Dylan
    Dec 13 '18 at 4:24
















use langle and rangle for angle brackets: $langle$ $rangle$
– Dylan
Dec 13 '18 at 4:24




use langle and rangle for angle brackets: $langle$ $rangle$
– Dylan
Dec 13 '18 at 4:24


















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