Mixing
Analytical solution to the 2D wave equation with Neumann BC's on a square
$begingroup$
I'm struggling to find an exact solution to the 2D wave equation, $u_{tt} = c^2(u_{xx} + u_{yy})$ with the following boundary and initial conditions:
The left edge of the square is equal to 1, while the top, bottom and right are 0.
$u_x(0,y,t) =1$
$u_y(x,0,t) = u_y (x,1,t)= u_x(1,y,0)=0$
Initial conditions
$u(x,y,0) = u_t (x,y,0)=0 $.
By separation of variables, let U(x,y,t) = F(x,y)G(t), and going through the usual process letting $-v^2$ be the constant that lets the two functions equal to each other.
This reduces to two equations requiring to be solved $G'' + lambda^2 G = 0 $ therefore $G(t) = Acos(lambda t) + Bsin(lambda t)$ where $lambda = cv$.
And $F_{xx} + F_{yy} + v^2 F = 0$ (2D Helmholtz equation). However when I apply the conditions all i get is zero, which shouldn't be the case.
pde wave-equation
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
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add a comment |
$begingroup$
I'm struggling to find an exact solution to the 2D wave equation, $u_{tt} = c^2(u_{xx} + u_{yy})$ with the following boundary and initial conditions:
The left edge of the square is equal to 1, while the top, bottom and right are 0.
$u_x(0,y,t) =1$
$u_y(x,0,t) = u_y (x,1,t)= u_x(1,y,0)=0$
Initial conditions
$u(x,y,0) = u_t (x,y,0)=0 $.
By separation of variables, let U(x,y,t) = F(x,y)G(t), and going through the usual process letting $-v^2$ be the constant that lets the two functions equal to each other.
This reduces to two equations requiring to be solved $G'' + lambda^2 G = 0 $ therefore $G(t) = Acos(lambda t) + Bsin(lambda t)$ where $lambda = cv$.
And $F_{xx} + F_{yy} + v^2 F = 0$ (2D Helmholtz equation). However when I apply the conditions all i get is zero, which shouldn't be the case.
pde wave-equation
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
$endgroup$
$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27
add a comment |
$begingroup$
I'm struggling to find an exact solution to the 2D wave equation, $u_{tt} = c^2(u_{xx} + u_{yy})$ with the following boundary and initial conditions:
The left edge of the square is equal to 1, while the top, bottom and right are 0.
$u_x(0,y,t) =1$
$u_y(x,0,t) = u_y (x,1,t)= u_x(1,y,0)=0$
Initial conditions
$u(x,y,0) = u_t (x,y,0)=0 $.
By separation of variables, let U(x,y,t) = F(x,y)G(t), and going through the usual process letting $-v^2$ be the constant that lets the two functions equal to each other.
This reduces to two equations requiring to be solved $G'' + lambda^2 G = 0 $ therefore $G(t) = Acos(lambda t) + Bsin(lambda t)$ where $lambda = cv$.
And $F_{xx} + F_{yy} + v^2 F = 0$ (2D Helmholtz equation). However when I apply the conditions all i get is zero, which shouldn't be the case.
pde wave-equation
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
$endgroup$
I'm struggling to find an exact solution to the 2D wave equation, $u_{tt} = c^2(u_{xx} + u_{yy})$ with the following boundary and initial conditions:
The left edge of the square is equal to 1, while the top, bottom and right are 0.
$u_x(0,y,t) =1$
$u_y(x,0,t) = u_y (x,1,t)= u_x(1,y,0)=0$
Initial conditions
$u(x,y,0) = u_t (x,y,0)=0 $.
By separation of variables, let U(x,y,t) = F(x,y)G(t), and going through the usual process letting $-v^2$ be the constant that lets the two functions equal to each other.
This reduces to two equations requiring to be solved $G'' + lambda^2 G = 0 $ therefore $G(t) = Acos(lambda t) + Bsin(lambda t)$ where $lambda = cv$.
And $F_{xx} + F_{yy} + v^2 F = 0$ (2D Helmholtz equation). However when I apply the conditions all i get is zero, which shouldn't be the case.
pde wave-equation
pde wave-equation
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
asked Jan 22 at 17:18
jake rowbottomjake rowbottom
1
1
$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27
add a comment |
$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27
$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27
$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27
add a comment |
1 Answer
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oldest
votes
$begingroup$
I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.
What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so
$$ w(x,y,t) = -frac{(1-x)^2}{2} $$
Plugigng this into the original equation, we have
begin{align}
v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \
v_y(x,0,t) = v_y(x,1,t) &= 0 \
v_x(0,y,t) = v_x(1,y,t) &= 0 \ v(x,y,0) &= frac{(1-x)^2}{2} \ v_t(x,y,0) &= 0 end{align}
We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving
begin{align}
X'' + lambda^2 X &= 0, quad X'(0) = X'(1) = 0 \
Y'' + mu^2 Y &= 0, quad Y'(0) = Y'(1) = 0
end{align}
which gives
$$ X_n(x) = cos(npi x), quad Y_m(y) = cos(mpi y) $$
However, notice that the solution does not need to depend on $y$, so we're left with
$$ v(x,y,t) = T_0(t) + sum_{n=1}^infty T_n(t)cos(npi x) $$
Plugging this into the equation:
$$ T_0'' + sum_{n=1}^infty [T_n'' + c^2n^2pi^2 T_n]cos(npi x) = -c^2 $$
$$ implies begin{cases} T_0'' = -c^2 \ T_n'' + c^2n^2pi^2 T_n = 0 &&, n ge 1 end{cases} $$
The last step is to break down the initial condition into its Fourier series
$$ v(x,y,0) = frac{(1-x)^2}{2} = T_0(0) + sum_{n=1}^infty T_n(t)cos(npi x) $$
$$ implies begin{cases} T_0(0) = int_0^1 frac{(1-x)^2}{2} dx \ T_n(0) = dfrac{int_0^1 frac{(1-x)^2}{2}cos(npi x) dx}{int_0^1 cos^2 (npi x) dx} &&, n ge 1 end{cases} $$
and $T_n'(0)=0, forall n$.
Solve the remaining IVPs for $T_n(t)$
You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.
answered Jan 23 at 12:53
DylanDylan
13.7k31027
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$begingroup$
I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.
What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so
$$ w(x,y,t) = -frac{(1-x)^2}{2} $$
Plugigng this into the original equation, we have
begin{align}
v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \
v_y(x,0,t) = v_y(x,1,t) &= 0 \
v_x(0,y,t) = v_x(1,y,t) &= 0 \ v(x,y,0) &= frac{(1-x)^2}{2} \ v_t(x,y,0) &= 0 end{align}
We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving
begin{align}
X'' + lambda^2 X &= 0, quad X'(0) = X'(1) = 0 \
Y'' + mu^2 Y &= 0, quad Y'(0) = Y'(1) = 0
end{align}
which gives
$$ X_n(x) = cos(npi x), quad Y_m(y) = cos(mpi y) $$
However, notice that the solution does not need to depend on $y$, so we're left with
$$ v(x,y,t) = T_0(t) + sum_{n=1}^infty T_n(t)cos(npi x) $$
Plugging this into the equation:
$$ T_0'' + sum_{n=1}^infty [T_n'' + c^2n^2pi^2 T_n]cos(npi x) = -c^2 $$
$$ implies begin{cases} T_0'' = -c^2 \ T_n'' + c^2n^2pi^2 T_n = 0 &&, n ge 1 end{cases} $$
The last step is to break down the initial condition into its Fourier series
$$ v(x,y,0) = frac{(1-x)^2}{2} = T_0(0) + sum_{n=1}^infty T_n(t)cos(npi x) $$
$$ implies begin{cases} T_0(0) = int_0^1 frac{(1-x)^2}{2} dx \ T_n(0) = dfrac{int_0^1 frac{(1-x)^2}{2}cos(npi x) dx}{int_0^1 cos^2 (npi x) dx} &&, n ge 1 end{cases} $$
and $T_n'(0)=0, forall n$.
Solve the remaining IVPs for $T_n(t)$
You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.
answered Jan 23 at 12:53
DylanDylan
13.7k31027
$endgroup$
add a comment |
$begingroup$
I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.
What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so
$$ w(x,y,t) = -frac{(1-x)^2}{2} $$
Plugigng this into the original equation, we have
begin{align}
v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \
v_y(x,0,t) = v_y(x,1,t) &= 0 \
v_x(0,y,t) = v_x(1,y,t) &= 0 \ v(x,y,0) &= frac{(1-x)^2}{2} \ v_t(x,y,0) &= 0 end{align}
We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving
begin{align}
X'' + lambda^2 X &= 0, quad X'(0) = X'(1) = 0 \
Y'' + mu^2 Y &= 0, quad Y'(0) = Y'(1) = 0
end{align}
which gives
$$ X_n(x) = cos(npi x), quad Y_m(y) = cos(mpi y) $$
However, notice that the solution does not need to depend on $y$, so we're left with
$$ v(x,y,t) = T_0(t) + sum_{n=1}^infty T_n(t)cos(npi x) $$
Plugging this into the equation:
$$ T_0'' + sum_{n=1}^infty [T_n'' + c^2n^2pi^2 T_n]cos(npi x) = -c^2 $$
$$ implies begin{cases} T_0'' = -c^2 \ T_n'' + c^2n^2pi^2 T_n = 0 &&, n ge 1 end{cases} $$
The last step is to break down the initial condition into its Fourier series
$$ v(x,y,0) = frac{(1-x)^2}{2} = T_0(0) + sum_{n=1}^infty T_n(t)cos(npi x) $$
$$ implies begin{cases} T_0(0) = int_0^1 frac{(1-x)^2}{2} dx \ T_n(0) = dfrac{int_0^1 frac{(1-x)^2}{2}cos(npi x) dx}{int_0^1 cos^2 (npi x) dx} &&, n ge 1 end{cases} $$
and $T_n'(0)=0, forall n$.
Solve the remaining IVPs for $T_n(t)$
You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.
answered Jan 23 at 12:53
DylanDylan
13.7k31027
$endgroup$
add a comment |
$begingroup$
I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.
What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so
$$ w(x,y,t) = -frac{(1-x)^2}{2} $$
Plugigng this into the original equation, we have
begin{align}
v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \
v_y(x,0,t) = v_y(x,1,t) &= 0 \
v_x(0,y,t) = v_x(1,y,t) &= 0 \ v(x,y,0) &= frac{(1-x)^2}{2} \ v_t(x,y,0) &= 0 end{align}
We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving
begin{align}
X'' + lambda^2 X &= 0, quad X'(0) = X'(1) = 0 \
Y'' + mu^2 Y &= 0, quad Y'(0) = Y'(1) = 0
end{align}
which gives
$$ X_n(x) = cos(npi x), quad Y_m(y) = cos(mpi y) $$
However, notice that the solution does not need to depend on $y$, so we're left with
$$ v(x,y,t) = T_0(t) + sum_{n=1}^infty T_n(t)cos(npi x) $$
Plugging this into the equation:
$$ T_0'' + sum_{n=1}^infty [T_n'' + c^2n^2pi^2 T_n]cos(npi x) = -c^2 $$
$$ implies begin{cases} T_0'' = -c^2 \ T_n'' + c^2n^2pi^2 T_n = 0 &&, n ge 1 end{cases} $$
The last step is to break down the initial condition into its Fourier series
$$ v(x,y,0) = frac{(1-x)^2}{2} = T_0(0) + sum_{n=1}^infty T_n(t)cos(npi x) $$
$$ implies begin{cases} T_0(0) = int_0^1 frac{(1-x)^2}{2} dx \ T_n(0) = dfrac{int_0^1 frac{(1-x)^2}{2}cos(npi x) dx}{int_0^1 cos^2 (npi x) dx} &&, n ge 1 end{cases} $$
and $T_n'(0)=0, forall n$.
Solve the remaining IVPs for $T_n(t)$
You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.
answered Jan 23 at 12:53
DylanDylan
13.7k31027
$endgroup$
I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.
What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so
$$ w(x,y,t) = -frac{(1-x)^2}{2} $$
Plugigng this into the original equation, we have
begin{align}
v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \
v_y(x,0,t) = v_y(x,1,t) &= 0 \
v_x(0,y,t) = v_x(1,y,t) &= 0 \ v(x,y,0) &= frac{(1-x)^2}{2} \ v_t(x,y,0) &= 0 end{align}
We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving
begin{align}
X'' + lambda^2 X &= 0, quad X'(0) = X'(1) = 0 \
Y'' + mu^2 Y &= 0, quad Y'(0) = Y'(1) = 0
end{align}
which gives
$$ X_n(x) = cos(npi x), quad Y_m(y) = cos(mpi y) $$
However, notice that the solution does not need to depend on $y$, so we're left with
$$ v(x,y,t) = T_0(t) + sum_{n=1}^infty T_n(t)cos(npi x) $$
Plugging this into the equation:
$$ T_0'' + sum_{n=1}^infty [T_n'' + c^2n^2pi^2 T_n]cos(npi x) = -c^2 $$
$$ implies begin{cases} T_0'' = -c^2 \ T_n'' + c^2n^2pi^2 T_n = 0 &&, n ge 1 end{cases} $$
The last step is to break down the initial condition into its Fourier series
$$ v(x,y,0) = frac{(1-x)^2}{2} = T_0(0) + sum_{n=1}^infty T_n(t)cos(npi x) $$
$$ implies begin{cases} T_0(0) = int_0^1 frac{(1-x)^2}{2} dx \ T_n(0) = dfrac{int_0^1 frac{(1-x)^2}{2}cos(npi x) dx}{int_0^1 cos^2 (npi x) dx} &&, n ge 1 end{cases} $$
and $T_n'(0)=0, forall n$.
Solve the remaining IVPs for $T_n(t)$
You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.
answered Jan 23 at 12:53
DylanDylan
13.7k31027
edited Jan 24 at 8:08
answered Jan 23 at 12:53
DylanDylan
13.7k31027
answered Jan 23 at 12:53
DylanDylan
13.7k31027
answered Jan 23 at 12:53
DylanDylan
13.7k31027
13.7k31027
add a comment |
add a comment |
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$begingroup$
If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake?
$endgroup$
– Vasily Mitch
Jan 22 at 17:27