Linearity solving Laplace equation in rectangular domain with non-homogeneous conditions
$begingroup$
Strauss' book on PDEs explains that to solve Laplace's equation in a 2D rectangular $a < x < b, c < y < d$ domain with non-homogeneous conditions
$$u(a,y) = f$$
$$u(b,y) = g$$
$$u(x,c) = h$$
$$u(x,d) = j$$
Then $u = u_1 + u_2 + u_3 + u_4$
where each $u$ is the solution to the 1D problem with one of the 4 boundary conditions.
The problem I am working on is
$$u(x,0) = x$$
$$u(x,1) = 0$$
$$u_x(0,y) = 0$$
$$u_x(1,y) = y^2$$
for $0<x,y < 1$
The solutions say that it is by solving just 2 different problems instead of the 4 mentioned. Why is this?
pde harmonic-functions
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
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add a comment |
$begingroup$
Strauss' book on PDEs explains that to solve Laplace's equation in a 2D rectangular $a < x < b, c < y < d$ domain with non-homogeneous conditions
$$u(a,y) = f$$
$$u(b,y) = g$$
$$u(x,c) = h$$
$$u(x,d) = j$$
Then $u = u_1 + u_2 + u_3 + u_4$
where each $u$ is the solution to the 1D problem with one of the 4 boundary conditions.
The problem I am working on is
$$u(x,0) = x$$
$$u(x,1) = 0$$
$$u_x(0,y) = 0$$
$$u_x(1,y) = y^2$$
for $0<x,y < 1$
The solutions say that it is by solving just 2 different problems instead of the 4 mentioned. Why is this?
pde harmonic-functions
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
$endgroup$
$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17
add a comment |
$begingroup$
Strauss' book on PDEs explains that to solve Laplace's equation in a 2D rectangular $a < x < b, c < y < d$ domain with non-homogeneous conditions
$$u(a,y) = f$$
$$u(b,y) = g$$
$$u(x,c) = h$$
$$u(x,d) = j$$
Then $u = u_1 + u_2 + u_3 + u_4$
where each $u$ is the solution to the 1D problem with one of the 4 boundary conditions.
The problem I am working on is
$$u(x,0) = x$$
$$u(x,1) = 0$$
$$u_x(0,y) = 0$$
$$u_x(1,y) = y^2$$
for $0<x,y < 1$
The solutions say that it is by solving just 2 different problems instead of the 4 mentioned. Why is this?
pde harmonic-functions
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
$endgroup$
Strauss' book on PDEs explains that to solve Laplace's equation in a 2D rectangular $a < x < b, c < y < d$ domain with non-homogeneous conditions
$$u(a,y) = f$$
$$u(b,y) = g$$
$$u(x,c) = h$$
$$u(x,d) = j$$
Then $u = u_1 + u_2 + u_3 + u_4$
where each $u$ is the solution to the 1D problem with one of the 4 boundary conditions.
The problem I am working on is
$$u(x,0) = x$$
$$u(x,1) = 0$$
$$u_x(0,y) = 0$$
$$u_x(1,y) = y^2$$
for $0<x,y < 1$
The solutions say that it is by solving just 2 different problems instead of the 4 mentioned. Why is this?
pde harmonic-functions
pde harmonic-functions
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
asked Dec 4 '18 at 14:43
The BoscoThe Bosco
547212
547212
$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17
add a comment |
$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17
$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17
$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17
add a comment |
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$begingroup$
Because as you just wrote, you have two non homogeneous BC, so those two problems are to be solved instead of 4, then use the linearity condition and you’re done.. If there is no forcing term then the only solution to the completely homogeneous problem ( $u=0$ on $partial([a,b] times [c,d]$) is the trivial one.
$endgroup$
– DaveNine
Dec 4 '18 at 20:17