Mixing
solve 2nd order PDE
$begingroup$
I have a PDE:
$$
frac{partial^2 u}{partial x^2} + afrac{partial^2 u}{partial y^2} +bu = f(x,y)
$$
where $a$ and $b$ are constants and $b>a>0$.
Also $space u(x,0) = g(x)$, and $space u(0,y) = h(y)$
How do i approach such a problem? I tried finding C.F. by separation of variables technique but got stuck on deciding the value of separation parameter. And how do i find the P.I.?
pde boundary-value-problem
edited Jan 18 at 9:32
Dylan
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
$endgroup$
|
show 8 more comments
$begingroup$
I have a PDE:
$$
frac{partial^2 u}{partial x^2} + afrac{partial^2 u}{partial y^2} +bu = f(x,y)
$$
where $a$ and $b$ are constants and $b>a>0$.
Also $space u(x,0) = g(x)$, and $space u(0,y) = h(y)$
How do i approach such a problem? I tried finding C.F. by separation of variables technique but got stuck on deciding the value of separation parameter. And how do i find the P.I.?
pde boundary-value-problem
edited Jan 18 at 9:32
Dylan
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
$endgroup$
1
$begingroup$
What does "C.F." mean$
$endgroup$
– Robert Lewis
Jan 15 at 19:52
$begingroup$
CF: complementary function
$endgroup$
– titusarmah99
Jan 15 at 20:07
$begingroup$
You haven't got enough boundary conditions.
$endgroup$
– Mattos
Jan 15 at 22:31
$begingroup$
@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
$endgroup$
– titusarmah99
Jan 16 at 2:32
1
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
$endgroup$
– Dylan
Jan 16 at 14:18
|
show 8 more comments
$begingroup$
I have a PDE:
$$
frac{partial^2 u}{partial x^2} + afrac{partial^2 u}{partial y^2} +bu = f(x,y)
$$
where $a$ and $b$ are constants and $b>a>0$.
Also $space u(x,0) = g(x)$, and $space u(0,y) = h(y)$
How do i approach such a problem? I tried finding C.F. by separation of variables technique but got stuck on deciding the value of separation parameter. And how do i find the P.I.?
pde boundary-value-problem
edited Jan 18 at 9:32
Dylan
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
$endgroup$
I have a PDE:
$$
frac{partial^2 u}{partial x^2} + afrac{partial^2 u}{partial y^2} +bu = f(x,y)
$$
where $a$ and $b$ are constants and $b>a>0$.
Also $space u(x,0) = g(x)$, and $space u(0,y) = h(y)$
How do i approach such a problem? I tried finding C.F. by separation of variables technique but got stuck on deciding the value of separation parameter. And how do i find the P.I.?
pde boundary-value-problem
pde boundary-value-problem
edited Jan 18 at 9:32
Dylan
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
edited Jan 18 at 9:32
Dylan
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
edited Jan 18 at 9:32
Dylan
12.9k31027
edited Jan 18 at 9:32
Dylan
12.9k31027
edited Jan 18 at 9:32
Dylan
12.9k31027
12.9k31027
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
asked Jan 15 at 19:37
titusarmah99titusarmah99
105
105
1
$begingroup$
What does "C.F." mean$
$endgroup$
– Robert Lewis
Jan 15 at 19:52
$begingroup$
CF: complementary function
$endgroup$
– titusarmah99
Jan 15 at 20:07
$begingroup$
You haven't got enough boundary conditions.
$endgroup$
– Mattos
Jan 15 at 22:31
$begingroup$
@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
$endgroup$
– titusarmah99
Jan 16 at 2:32
1
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
$endgroup$
– Dylan
Jan 16 at 14:18
|
show 8 more comments
1
$begingroup$
What does "C.F." mean$
$endgroup$
– Robert Lewis
Jan 15 at 19:52
$begingroup$
CF: complementary function
$endgroup$
– titusarmah99
Jan 15 at 20:07
$begingroup$
You haven't got enough boundary conditions.
$endgroup$
– Mattos
Jan 15 at 22:31
$begingroup$
@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
$endgroup$
– titusarmah99
Jan 16 at 2:32
1
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
$endgroup$
– Dylan
Jan 16 at 14:18
1
1
$begingroup$
What does "C.F." mean$
$endgroup$
– Robert Lewis
Jan 15 at 19:52
$begingroup$
What does "C.F." mean$
$endgroup$
– Robert Lewis
Jan 15 at 19:52
$begingroup$
CF: complementary function
$endgroup$
– titusarmah99
Jan 15 at 20:07
$begingroup$
CF: complementary function
$endgroup$
– titusarmah99
Jan 15 at 20:07
$begingroup$
You haven't got enough boundary conditions.
$endgroup$
– Mattos
Jan 15 at 22:31
$begingroup$
You haven't got enough boundary conditions.
$endgroup$
– Mattos
Jan 15 at 22:31
$begingroup$
@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
$endgroup$
– titusarmah99
Jan 16 at 2:32
$begingroup$
@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
$endgroup$
– titusarmah99
Jan 16 at 2:32
1
1
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
$endgroup$
– Dylan
Jan 16 at 14:18
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
$endgroup$
– Dylan
Jan 16 at 14:18
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First use the substitution $y = sqrt{a} cdot bar{y}$ to obtain the inhomogeneous Helmholtz equation
begin{cases} nabla^2 u + k^2 u = f, & (x,bar y) in D \ u = g, & y=0 \ u = h, & x = 0 end{cases}
where $D$ is the quarter plane ${x > 0, bar y > 0}$ on $Bbb R^2$
We can use the same method used for the Laplace equation here, with some modifications.
Step 1. Find the Green's function $G(x,x',bar y,bar y')$ on an open domain such that
$$ (nabla^2 + k^2)G = delta(x-x',bar y-bar y'), quad (x',bar y') in Bbb R^2 $$
There are several options, the most common one you'll find is
$$ G(rho) = frac{i}{4}[J_0(krho) + iY_0(krho)], quad rho = sqrt{(x-x')^2+(bar y-bar y')^2} $$
where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.
Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',bar y,bar y')$ such that
begin{cases}
(nabla^2 + k^2)G_D = delta(rho), & (x',bar y') in D \
G = 0, & bar y' = 0 \
G = 0, & x' = 0
end{cases}
Using the method of images, we find
$$ G_D(x,x',bar y, bar y') = frac14 big[G(x,x',bar y,bar y') - G(x,x',bar y,-bar y') - G(x,-x',bar y, bar y') + G(x,-x',bar y,-bar y')big] $$
Step 3. Integrate the Green's function. Using Green's second identity
begin{align}
iint_D big[u(mathbf{r}')cdot(nabla^2+k^2)G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r}')cdot (nabla^2+k^2)u(mathbf{r}') big]dA' &= int_{partial D} big[u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r'})nabla u(mathbf{r})big]cdot hat{mathbf{n}}ds' \
u(mathbf{r}) - iint_D G_D(mathbf{r},mathbf{r}')f(mathbf{r}') dA' &= int_{partial D} u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}')cdot mathbf{n} ds'
end{align}
$$ implies u(x,bar y) = int_0^inftyint_0^infty G_Df(x',bar y')dx' dbar y' + int_0^infty g(x') frac{partial G_D}{partial bar y'}Biggvert_{y'=0}dx' + int_0^infty h(y') frac{partial G_D}{partial x'}Biggvert_{x'=0} dy' $$
Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions
$$ u_h(x,y) = sin(k_x x)sin(k_y y), quad k_x^2 + k_y^2 = k^2 $$
such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.
answered Jan 18 at 9:01
DylanDylan
12.9k31027
$endgroup$
add a comment |
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$begingroup$
First use the substitution $y = sqrt{a} cdot bar{y}$ to obtain the inhomogeneous Helmholtz equation
begin{cases} nabla^2 u + k^2 u = f, & (x,bar y) in D \ u = g, & y=0 \ u = h, & x = 0 end{cases}
where $D$ is the quarter plane ${x > 0, bar y > 0}$ on $Bbb R^2$
We can use the same method used for the Laplace equation here, with some modifications.
Step 1. Find the Green's function $G(x,x',bar y,bar y')$ on an open domain such that
$$ (nabla^2 + k^2)G = delta(x-x',bar y-bar y'), quad (x',bar y') in Bbb R^2 $$
There are several options, the most common one you'll find is
$$ G(rho) = frac{i}{4}[J_0(krho) + iY_0(krho)], quad rho = sqrt{(x-x')^2+(bar y-bar y')^2} $$
where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.
Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',bar y,bar y')$ such that
begin{cases}
(nabla^2 + k^2)G_D = delta(rho), & (x',bar y') in D \
G = 0, & bar y' = 0 \
G = 0, & x' = 0
end{cases}
Using the method of images, we find
$$ G_D(x,x',bar y, bar y') = frac14 big[G(x,x',bar y,bar y') - G(x,x',bar y,-bar y') - G(x,-x',bar y, bar y') + G(x,-x',bar y,-bar y')big] $$
Step 3. Integrate the Green's function. Using Green's second identity
begin{align}
iint_D big[u(mathbf{r}')cdot(nabla^2+k^2)G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r}')cdot (nabla^2+k^2)u(mathbf{r}') big]dA' &= int_{partial D} big[u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r'})nabla u(mathbf{r})big]cdot hat{mathbf{n}}ds' \
u(mathbf{r}) - iint_D G_D(mathbf{r},mathbf{r}')f(mathbf{r}') dA' &= int_{partial D} u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}')cdot mathbf{n} ds'
end{align}
$$ implies u(x,bar y) = int_0^inftyint_0^infty G_Df(x',bar y')dx' dbar y' + int_0^infty g(x') frac{partial G_D}{partial bar y'}Biggvert_{y'=0}dx' + int_0^infty h(y') frac{partial G_D}{partial x'}Biggvert_{x'=0} dy' $$
Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions
$$ u_h(x,y) = sin(k_x x)sin(k_y y), quad k_x^2 + k_y^2 = k^2 $$
such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.
answered Jan 18 at 9:01
DylanDylan
12.9k31027
$endgroup$
add a comment |
$begingroup$
First use the substitution $y = sqrt{a} cdot bar{y}$ to obtain the inhomogeneous Helmholtz equation
begin{cases} nabla^2 u + k^2 u = f, & (x,bar y) in D \ u = g, & y=0 \ u = h, & x = 0 end{cases}
where $D$ is the quarter plane ${x > 0, bar y > 0}$ on $Bbb R^2$
We can use the same method used for the Laplace equation here, with some modifications.
Step 1. Find the Green's function $G(x,x',bar y,bar y')$ on an open domain such that
$$ (nabla^2 + k^2)G = delta(x-x',bar y-bar y'), quad (x',bar y') in Bbb R^2 $$
There are several options, the most common one you'll find is
$$ G(rho) = frac{i}{4}[J_0(krho) + iY_0(krho)], quad rho = sqrt{(x-x')^2+(bar y-bar y')^2} $$
where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.
Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',bar y,bar y')$ such that
begin{cases}
(nabla^2 + k^2)G_D = delta(rho), & (x',bar y') in D \
G = 0, & bar y' = 0 \
G = 0, & x' = 0
end{cases}
Using the method of images, we find
$$ G_D(x,x',bar y, bar y') = frac14 big[G(x,x',bar y,bar y') - G(x,x',bar y,-bar y') - G(x,-x',bar y, bar y') + G(x,-x',bar y,-bar y')big] $$
Step 3. Integrate the Green's function. Using Green's second identity
begin{align}
iint_D big[u(mathbf{r}')cdot(nabla^2+k^2)G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r}')cdot (nabla^2+k^2)u(mathbf{r}') big]dA' &= int_{partial D} big[u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r'})nabla u(mathbf{r})big]cdot hat{mathbf{n}}ds' \
u(mathbf{r}) - iint_D G_D(mathbf{r},mathbf{r}')f(mathbf{r}') dA' &= int_{partial D} u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}')cdot mathbf{n} ds'
end{align}
$$ implies u(x,bar y) = int_0^inftyint_0^infty G_Df(x',bar y')dx' dbar y' + int_0^infty g(x') frac{partial G_D}{partial bar y'}Biggvert_{y'=0}dx' + int_0^infty h(y') frac{partial G_D}{partial x'}Biggvert_{x'=0} dy' $$
Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions
$$ u_h(x,y) = sin(k_x x)sin(k_y y), quad k_x^2 + k_y^2 = k^2 $$
such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.
answered Jan 18 at 9:01
DylanDylan
12.9k31027
$endgroup$
add a comment |
$begingroup$
First use the substitution $y = sqrt{a} cdot bar{y}$ to obtain the inhomogeneous Helmholtz equation
begin{cases} nabla^2 u + k^2 u = f, & (x,bar y) in D \ u = g, & y=0 \ u = h, & x = 0 end{cases}
where $D$ is the quarter plane ${x > 0, bar y > 0}$ on $Bbb R^2$
We can use the same method used for the Laplace equation here, with some modifications.
Step 1. Find the Green's function $G(x,x',bar y,bar y')$ on an open domain such that
$$ (nabla^2 + k^2)G = delta(x-x',bar y-bar y'), quad (x',bar y') in Bbb R^2 $$
There are several options, the most common one you'll find is
$$ G(rho) = frac{i}{4}[J_0(krho) + iY_0(krho)], quad rho = sqrt{(x-x')^2+(bar y-bar y')^2} $$
where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.
Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',bar y,bar y')$ such that
begin{cases}
(nabla^2 + k^2)G_D = delta(rho), & (x',bar y') in D \
G = 0, & bar y' = 0 \
G = 0, & x' = 0
end{cases}
Using the method of images, we find
$$ G_D(x,x',bar y, bar y') = frac14 big[G(x,x',bar y,bar y') - G(x,x',bar y,-bar y') - G(x,-x',bar y, bar y') + G(x,-x',bar y,-bar y')big] $$
Step 3. Integrate the Green's function. Using Green's second identity
begin{align}
iint_D big[u(mathbf{r}')cdot(nabla^2+k^2)G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r}')cdot (nabla^2+k^2)u(mathbf{r}') big]dA' &= int_{partial D} big[u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r'})nabla u(mathbf{r})big]cdot hat{mathbf{n}}ds' \
u(mathbf{r}) - iint_D G_D(mathbf{r},mathbf{r}')f(mathbf{r}') dA' &= int_{partial D} u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}')cdot mathbf{n} ds'
end{align}
$$ implies u(x,bar y) = int_0^inftyint_0^infty G_Df(x',bar y')dx' dbar y' + int_0^infty g(x') frac{partial G_D}{partial bar y'}Biggvert_{y'=0}dx' + int_0^infty h(y') frac{partial G_D}{partial x'}Biggvert_{x'=0} dy' $$
Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions
$$ u_h(x,y) = sin(k_x x)sin(k_y y), quad k_x^2 + k_y^2 = k^2 $$
such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.
answered Jan 18 at 9:01
DylanDylan
12.9k31027
$endgroup$
First use the substitution $y = sqrt{a} cdot bar{y}$ to obtain the inhomogeneous Helmholtz equation
begin{cases} nabla^2 u + k^2 u = f, & (x,bar y) in D \ u = g, & y=0 \ u = h, & x = 0 end{cases}
where $D$ is the quarter plane ${x > 0, bar y > 0}$ on $Bbb R^2$
We can use the same method used for the Laplace equation here, with some modifications.
Step 1. Find the Green's function $G(x,x',bar y,bar y')$ on an open domain such that
$$ (nabla^2 + k^2)G = delta(x-x',bar y-bar y'), quad (x',bar y') in Bbb R^2 $$
There are several options, the most common one you'll find is
$$ G(rho) = frac{i}{4}[J_0(krho) + iY_0(krho)], quad rho = sqrt{(x-x')^2+(bar y-bar y')^2} $$
where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.
Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',bar y,bar y')$ such that
begin{cases}
(nabla^2 + k^2)G_D = delta(rho), & (x',bar y') in D \
G = 0, & bar y' = 0 \
G = 0, & x' = 0
end{cases}
Using the method of images, we find
$$ G_D(x,x',bar y, bar y') = frac14 big[G(x,x',bar y,bar y') - G(x,x',bar y,-bar y') - G(x,-x',bar y, bar y') + G(x,-x',bar y,-bar y')big] $$
Step 3. Integrate the Green's function. Using Green's second identity
begin{align}
iint_D big[u(mathbf{r}')cdot(nabla^2+k^2)G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r}')cdot (nabla^2+k^2)u(mathbf{r}') big]dA' &= int_{partial D} big[u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}') - G_D(mathbf{r},mathbf{r'})nabla u(mathbf{r})big]cdot hat{mathbf{n}}ds' \
u(mathbf{r}) - iint_D G_D(mathbf{r},mathbf{r}')f(mathbf{r}') dA' &= int_{partial D} u(mathbf{r}')nabla G_D(mathbf{r},mathbf{r}')cdot mathbf{n} ds'
end{align}
$$ implies u(x,bar y) = int_0^inftyint_0^infty G_Df(x',bar y')dx' dbar y' + int_0^infty g(x') frac{partial G_D}{partial bar y'}Biggvert_{y'=0}dx' + int_0^infty h(y') frac{partial G_D}{partial x'}Biggvert_{x'=0} dy' $$
Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions
$$ u_h(x,y) = sin(k_x x)sin(k_y y), quad k_x^2 + k_y^2 = k^2 $$
such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.
answered Jan 18 at 9:01
DylanDylan
12.9k31027
edited Jan 21 at 9:34
answered Jan 18 at 9:01
DylanDylan
12.9k31027
answered Jan 18 at 9:01
DylanDylan
12.9k31027
answered Jan 18 at 9:01
DylanDylan
12.9k31027
12.9k31027
add a comment |
add a comment |
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1
$begingroup$
What does "C.F." mean$
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– Robert Lewis
Jan 15 at 19:52
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CF: complementary function
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– titusarmah99
Jan 15 at 20:07
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You haven't got enough boundary conditions.
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– Mattos
Jan 15 at 22:31
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@Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one
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– titusarmah99
Jan 16 at 2:32
1
$begingroup$
The substitution $y= sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $nabla^2u + k^2u = f$. Then you solve it using Green's function
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– Dylan
Jan 16 at 14:18