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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# PDE : Can not solve Helmholtz equation

**Physics Forums | Science Articles, Homework Help, Discussion**