PDE : Can not solve Helmholtz equation(adsbygoogle = window.adsbygoogle || []).push({});

(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.

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# PDE : Can not solve Helmholtz equation

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