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sompongt
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PDE : Can not solve Helmholtz equation
(This is not a homework. I doing my research on numerical boundary integral. I need the analytical solution to compare the results with my computer program. I try to solve this equation, but it not success. I need urgent help.)
I working on anti-plane elasticity problem. The physical problem is described on a rectangular domain,
[itex] 0 \leq x \leq a [/itex] and [itex]0 \leq y \leq b.[/itex]
Let [itex]u_z = u(x,y;t)[/itex] is the displacement function in [itex]z[/itex] direction.
[tex]u_z = u(x,y;t)[/tex] is the displacement function in [itex]z[/itex] direction.
The boundary condition are specified by:
(1) [itex]u = 0[/itex] on the lower edge ([itex]y=0[/itex])
(2) [itex]\text{Traction} = 0[/itex] on the left edge ([itex]x=0[/itex])
(3) [itex]\text{Traction} = 0[/itex] on the right edge ([itex]x=a[/itex])
(4) [itex]\text{Traction} = P e^{i \omega t}[/itex] on the top edge ([itex]y=b[/itex]) in the [itex]z[/itex] direction
This problem can be formulate in the frequency domain to obtain the Helmholtz equation, as below.
The problem seem not difficult. But I can not solve it. Could anybody help me?
Any suggestion are welcome.
Mathematical derivation start here. The problem statement describe as,
\begin{equation}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial x^2} + k^2 u = 0,
0 \leq x \leq a, 0 \leq y \leq b \text{...(1)}
\end{equation}
where
\begin{equation}
k^2 = \frac{\omega^2}{c^2}
\end{equation}
[itex]\omega[/itex] = angular frequency of the excited force
[itex]c[/itex] = speed of wave = function of material properties
subjected by the following boundary conditions,
\begin{align}
\frac{\partial u}{\partial x}(x=0, y) &= 0 \text{...(2)}\label{bc1}\\
\frac{\partial u}{\partial x}(x=a, y) &= 0 \text{...(3)}\label{bc2} \\
u(x, y = 0) &= 0 \text{...(4)}\label{bc3}\\
\frac{\partial u}{\partial y}(x, y=b) &= \text{constant}\label{bc4} = P \text{...(5)}
\end{align}
Solution:
By separation of variables, let
\begin{equation}
u(x,y) = X(x)Y(y)
\end{equation}
Then, working on the standard process,
\begin{align}
X'' Y + X Y'' + k^2 X Y = 0 \\
\frac{X''}{X} + \frac{Y''}{Y} + k^2 = 0
\end{align}
let
\begin{align}
\frac{X''}{X} &= -\alpha^2 &\rightarrow X''+ \alpha^2 X &= 0 \\
\frac{Y''}{Y} &= -\beta^2 &\rightarrow Y''+ \beta^2 Y &= 0
\end{align}
Relation between [itex]\alpha[/itex] and [itex]\beta[/itex],
\begin{equation}
\alpha ^2 + \beta^2 = k^2 \label{abrelation} \text{...(6)}
\end{equation}
Solve for [itex]X(x)[/itex] and [itex]Y(y)[/itex]
\begin{align}
X(x) &= C_1 \cos \alpha x + C_2 \sin \alpha x \\
Y(y) &= C_3 \cos \beta y + C_4 \sin \beta y
\end{align}
So that,
\begin{align}
u(x,y) &= X(x)Y(y) \\
&= (C_1 \cos \alpha x + C_2 \sin \alpha x)(C_3 \cos \beta y + C_4 \sin \beta y) \label{reduce1}
\end{align}
Determine coefficients [itex]C_1, C_2, C_3, C_4[/itex] by using boundary conditions (BC) eq(2-5),
First, use boundary condition in eq(4):
\begin{align}
u(x,y=0) &= 0 \\
(C_1 \cos \alpha x + C_2 \sin \alpha x)(C_3 \cdot 1 + C_4 \cdot 0) &= 0 \\
C_3(C_1 \cos \alpha x + C_2 \sin \alpha x) &= 0
\end{align}
We obtain,
\begin{equation}
C_3 = 0
\end{equation}
So that, [itex]u(x,y)[/itex] reduce to
\begin{align}
u(x,y) &= C_4 \sin \beta y (C_1 \cos \alpha x + C_2 \sin \alpha x) \\
&= \sin \beta y (C_4C_1 \cos \alpha x + C_4C_2 \sin \alpha x) \\
&= \sin \beta y (C_5 \cos \alpha x + C_6 \sin \alpha x) \label{reduce2} \\
\frac{\partial u}{\partial x} (x,y) &= \alpha \sin \beta y (-C_5 \sin \alpha x + C_6 \cos \alpha x)
\end{align}
Second, use boundary condition in eq(2):
\begin{align}
\frac{\partial u}{\partial x}(x=0,y) &= 0 \\
\alpha \sin \beta y (-C_5 \cdot 0 + C_6 \cdot 1) &= 0 \\
C_6 \cdot \alpha \sin \beta y &= 0
\end{align}
We obtain,
\begin{equation}
C_6 = 0
\end{equation}
So that, [itex]u(x,y)[/itex] reduce to
\begin{align}
u(x,y) &= C_5 \cos \alpha x \sin \beta y \\
\frac{\partial u}{\partial x} (x,y) &= - \alpha C_5 \sin \alpha x \sin \beta y
\end{align}
Third step, use boundary condition in eq(3):
\begin{align}
\frac{\partial u}{\partial x}(x=a,y) &= 0 \\
-\alpha C_5 \sin \alpha a \sin \beta y = 0
\end{align}
Which can be conclude that,
\begin{align}
\sin \alpha a &= 0 \\
\alpha a &= n \pi \\
\alpha_n &= \frac{n \pi}{a}
\end{align}
After we find [itex]\alpha[/itex], we can determine [itex]\beta[/itex] from eq(6)
\begin{align}
\beta_n^2 = k^2 - \alpha_n^2
\end{align}
At this point, we can represent [itex]u(x,y)[/itex] as infinite series by using principle of superposition
\begin{align}
u(x,y) &= \sum_{n = 0}^{\infty} C_n \cos \alpha_n x \sin \beta_n y \\
\frac{\partial u}{\partial y} (x,y) &= \sum_{n = 0}^{\infty} \beta_n C_n \cos \alpha_n x \cos \beta_n y
\end{align}
We can determine the unknowns [itex]C_n[/itex] by using the last boundary condition in eq(5)
\begin{align}
P &= \frac{\partial u}{\partial y} (x,y = b) \\
&= \sum_{n = 0}^{\infty} \beta_n C_n \cos \alpha_n x \cos \beta_n b \\
&= \sum_{n = 0}^{\infty} \underbrace{\left(C_n\beta_n \cos \beta_n b \right)}_{\text{constant} = \bar{C_n}} \cos \alpha_n x \\
&= \sum_{n = 0}^{\infty} \bar{C_n}\cos \alpha_n x \text{...(7)}
\end{align}
Consider [itex]\bar{C_n}[/itex] as coefficient of Fourier cosine series,
\begin{align}
P &= \frac{c_0}{2} + \sum_{ n = 1 }^{\infty} c_n \cos \frac{n\pi x}{a} \\
c_n &= \frac{2}{a} \int_{0}^{a} P \cdot \cos \frac{n\pi x}{a} dx
\end{align}
It is not so hard to find that,
\begin{align}
c_0 &= 2P \\
c_n &= 0
\end{align}
For me, the problem arise here. How can I find [itex]\bar{C_n}[/itex] in eq(7)? Could anybody help me? What point that I am wrong? I feel very headache. Please, please, please help me.
(This is not a homework. I doing my research on numerical boundary integral. I need the analytical solution to compare the results with my computer program. I try to solve this equation, but it not success. I need urgent help.)
I working on anti-plane elasticity problem. The physical problem is described on a rectangular domain,
[itex] 0 \leq x \leq a [/itex] and [itex]0 \leq y \leq b.[/itex]
Let [itex]u_z = u(x,y;t)[/itex] is the displacement function in [itex]z[/itex] direction.
[tex]u_z = u(x,y;t)[/tex] is the displacement function in [itex]z[/itex] direction.
The boundary condition are specified by:
(1) [itex]u = 0[/itex] on the lower edge ([itex]y=0[/itex])
(2) [itex]\text{Traction} = 0[/itex] on the left edge ([itex]x=0[/itex])
(3) [itex]\text{Traction} = 0[/itex] on the right edge ([itex]x=a[/itex])
(4) [itex]\text{Traction} = P e^{i \omega t}[/itex] on the top edge ([itex]y=b[/itex]) in the [itex]z[/itex] direction
This problem can be formulate in the frequency domain to obtain the Helmholtz equation, as below.
The problem seem not difficult. But I can not solve it. Could anybody help me?
Any suggestion are welcome.
Mathematical derivation start here. The problem statement describe as,
\begin{equation}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial x^2} + k^2 u = 0,
0 \leq x \leq a, 0 \leq y \leq b \text{...(1)}
\end{equation}
where
\begin{equation}
k^2 = \frac{\omega^2}{c^2}
\end{equation}
[itex]\omega[/itex] = angular frequency of the excited force
[itex]c[/itex] = speed of wave = function of material properties
subjected by the following boundary conditions,
\begin{align}
\frac{\partial u}{\partial x}(x=0, y) &= 0 \text{...(2)}\label{bc1}\\
\frac{\partial u}{\partial x}(x=a, y) &= 0 \text{...(3)}\label{bc2} \\
u(x, y = 0) &= 0 \text{...(4)}\label{bc3}\\
\frac{\partial u}{\partial y}(x, y=b) &= \text{constant}\label{bc4} = P \text{...(5)}
\end{align}
Solution:
By separation of variables, let
\begin{equation}
u(x,y) = X(x)Y(y)
\end{equation}
Then, working on the standard process,
\begin{align}
X'' Y + X Y'' + k^2 X Y = 0 \\
\frac{X''}{X} + \frac{Y''}{Y} + k^2 = 0
\end{align}
let
\begin{align}
\frac{X''}{X} &= -\alpha^2 &\rightarrow X''+ \alpha^2 X &= 0 \\
\frac{Y''}{Y} &= -\beta^2 &\rightarrow Y''+ \beta^2 Y &= 0
\end{align}
Relation between [itex]\alpha[/itex] and [itex]\beta[/itex],
\begin{equation}
\alpha ^2 + \beta^2 = k^2 \label{abrelation} \text{...(6)}
\end{equation}
Solve for [itex]X(x)[/itex] and [itex]Y(y)[/itex]
\begin{align}
X(x) &= C_1 \cos \alpha x + C_2 \sin \alpha x \\
Y(y) &= C_3 \cos \beta y + C_4 \sin \beta y
\end{align}
So that,
\begin{align}
u(x,y) &= X(x)Y(y) \\
&= (C_1 \cos \alpha x + C_2 \sin \alpha x)(C_3 \cos \beta y + C_4 \sin \beta y) \label{reduce1}
\end{align}
Determine coefficients [itex]C_1, C_2, C_3, C_4[/itex] by using boundary conditions (BC) eq(2-5),
First, use boundary condition in eq(4):
\begin{align}
u(x,y=0) &= 0 \\
(C_1 \cos \alpha x + C_2 \sin \alpha x)(C_3 \cdot 1 + C_4 \cdot 0) &= 0 \\
C_3(C_1 \cos \alpha x + C_2 \sin \alpha x) &= 0
\end{align}
We obtain,
\begin{equation}
C_3 = 0
\end{equation}
So that, [itex]u(x,y)[/itex] reduce to
\begin{align}
u(x,y) &= C_4 \sin \beta y (C_1 \cos \alpha x + C_2 \sin \alpha x) \\
&= \sin \beta y (C_4C_1 \cos \alpha x + C_4C_2 \sin \alpha x) \\
&= \sin \beta y (C_5 \cos \alpha x + C_6 \sin \alpha x) \label{reduce2} \\
\frac{\partial u}{\partial x} (x,y) &= \alpha \sin \beta y (-C_5 \sin \alpha x + C_6 \cos \alpha x)
\end{align}
Second, use boundary condition in eq(2):
\begin{align}
\frac{\partial u}{\partial x}(x=0,y) &= 0 \\
\alpha \sin \beta y (-C_5 \cdot 0 + C_6 \cdot 1) &= 0 \\
C_6 \cdot \alpha \sin \beta y &= 0
\end{align}
We obtain,
\begin{equation}
C_6 = 0
\end{equation}
So that, [itex]u(x,y)[/itex] reduce to
\begin{align}
u(x,y) &= C_5 \cos \alpha x \sin \beta y \\
\frac{\partial u}{\partial x} (x,y) &= - \alpha C_5 \sin \alpha x \sin \beta y
\end{align}
Third step, use boundary condition in eq(3):
\begin{align}
\frac{\partial u}{\partial x}(x=a,y) &= 0 \\
-\alpha C_5 \sin \alpha a \sin \beta y = 0
\end{align}
Which can be conclude that,
\begin{align}
\sin \alpha a &= 0 \\
\alpha a &= n \pi \\
\alpha_n &= \frac{n \pi}{a}
\end{align}
After we find [itex]\alpha[/itex], we can determine [itex]\beta[/itex] from eq(6)
\begin{align}
\beta_n^2 = k^2 - \alpha_n^2
\end{align}
At this point, we can represent [itex]u(x,y)[/itex] as infinite series by using principle of superposition
\begin{align}
u(x,y) &= \sum_{n = 0}^{\infty} C_n \cos \alpha_n x \sin \beta_n y \\
\frac{\partial u}{\partial y} (x,y) &= \sum_{n = 0}^{\infty} \beta_n C_n \cos \alpha_n x \cos \beta_n y
\end{align}
We can determine the unknowns [itex]C_n[/itex] by using the last boundary condition in eq(5)
\begin{align}
P &= \frac{\partial u}{\partial y} (x,y = b) \\
&= \sum_{n = 0}^{\infty} \beta_n C_n \cos \alpha_n x \cos \beta_n b \\
&= \sum_{n = 0}^{\infty} \underbrace{\left(C_n\beta_n \cos \beta_n b \right)}_{\text{constant} = \bar{C_n}} \cos \alpha_n x \\
&= \sum_{n = 0}^{\infty} \bar{C_n}\cos \alpha_n x \text{...(7)}
\end{align}
Consider [itex]\bar{C_n}[/itex] as coefficient of Fourier cosine series,
\begin{align}
P &= \frac{c_0}{2} + \sum_{ n = 1 }^{\infty} c_n \cos \frac{n\pi x}{a} \\
c_n &= \frac{2}{a} \int_{0}^{a} P \cdot \cos \frac{n\pi x}{a} dx
\end{align}
It is not so hard to find that,
\begin{align}
c_0 &= 2P \\
c_n &= 0
\end{align}
For me, the problem arise here. How can I find [itex]\bar{C_n}[/itex] in eq(7)? Could anybody help me? What point that I am wrong? I feel very headache. Please, please, please help me.