# Fourth order PDE solution for vibrating beam rigidly fastened at one end and simply fastened at other end.

I have Partial Differential equation in the form:

$$\frac{\partial^2 y}{\partial t^2} + \frac{\partial^4 y}{\partial x^4} =0, \quad 0<x<1$$

Vibrating Beam

Boundary Conditions: $$y(0,t) = \frac{\partial y}{\partial x}(0,t) = 0 \\ y(1,t) = \frac{\partial^2 y}{\partial x^2}(1,t) = 0$$

Initial conditions:

$$y(x,0) = 0.1\sin(\pi x), \ \frac{\partial y}{\partial t}(x,0) = 0$$

I have come up to the point where I have solved the boundary conditions and got the four equations:

$A+C=0$

$B+D=0$

$A\sinh(\beta) + B\cosh(\beta) + C\cos(\beta) + D\sin(\beta) = 0 \\$

$A\sinh(\beta)+B\cosh(\beta)-C\sin(\beta)-D\cos(\beta) = 0$

Not sure what to do next. I am supposed to find the Eigenvalues and Eigenfunctions.

We have $$C=-A$$ and $$D=-B$$, so

$$A(\sinh \beta - \cos \beta) + B(\cosh \beta - \sin \beta) = 0$$

$$A(\sinh \beta + \sin \beta) + B(\cosh \beta + \cos \beta) = 0$$

This guarantees a solution if the determinant of the coefficient matrix is zero

$$\left|\begin{matrix} \sinh\beta - \cos\beta && \cosh\beta -\sin\beta \\ \sinh\beta + \sin\beta && \cosh\beta + \cos\beta \end{matrix}\right| = 0$$

which, after some manipulation, gives $$e^{-\beta} = \sin\beta - \cos \beta = \sqrt{2}\sin\left(\beta - \frac{\pi}{4}\right)$$

You can solve this graphically as the intersection of the two functions. The first eigenvalue is $$\beta_0 \approx 1.038415$$. Since $$e^{-\beta}$$ decays rapidly, the larger solutions are approximately $$\beta_n \sim (n+1/4)\pi$$

EDIT: I'm not sure if your equations are actually correct. I tried solving for the eigenfunction and got $$X(x) = A\cosh \beta x + B\sinh \beta x + C\cos\beta x + D\sin\beta x$$

And the four boundary conditions give $$X(0) = A + C = 0$$ $$X'(0) = B + D = 0$$ $$X(1) = A\cosh\beta + B\sinh\beta + C\cos\beta + D\sin\beta = 0$$ $$X''(1) = A\cosh\beta + B\sinh\beta - C\cos\beta - D\sin\beta = 0$$

In this case, the eigenvalues satisfy $$\left|\begin{matrix} \cosh\beta - \cos\beta && \sinh\beta -\sin\beta \\ \cosh\beta + \cos\beta && \sinh\beta + \sin\beta \end{matrix}\right| = 0$$ or $$\tanh \beta = \tan\beta$$

Since $$\tanh\beta$$ very quickly goes to $$1$$, $$\beta_n \sim (n + 1/4)\pi, \ n \ge 1$$. This is a very good approximation; the first zero is $$\beta_1 \approx 3.9266 = 5\pi/4 - 0.0004$$

EDIT 2: For the eigenfunctions, we have $$X(x) = A(\cosh\beta x - \cos\beta x) + B(\sinh\beta x - \sin\beta x)$$

where $$(A,B)$$ satisfy $$A(\cosh\beta-\cos\beta) + B(\sinh\beta-\sin\beta) = 0$$ $$A(\cosh\beta+\cos\beta) + B(\sinh\beta+\sin\beta) = 0$$

Note that these are actually the same equation (since the determinant is zero), so we can just pick any arbitrary pair that satisfies one of the above. For example, let $$A = \sinh\beta-\sin\beta$$ and $$-B = \cosh\beta-\cos\beta$$, then the eigenfunction is (up to a constant) $$X_n(x) = (\sinh\beta_n-\sin\beta_n)(\cosh\beta_n x - \cos\beta_n x) - (\cosh\beta_n-\cos\beta_n)(\sinh\beta_n x - \sin\beta_n x)$$

To show these functions are mutually orthogonal, we use the definition of orthogonality $$\int_0^1 X_n(x) X_m(x)\ dx = 0, \ m \ne n$$

The integral is a bit messy, but it should reduce down to some expression in terms of the determinant.

EDIT 3: The B.C.s are $$X(0) = X(1) = X'(0) = X''(1) = 0$$. Using integration by parts, we can show that

\begin{align} \int_0^1 {X_m}^{(4)} X_n &= {X_m}'''X_n\Bigg|_0^1 - \int_0^1 {X_m}'''{X_n}', && X_n(0)=X_n(1)=0 \\ &= -{X_m}''{X_n}'\Bigg|_0^1 + \int_0^1 {X_m}''{X_n}'', && {X_m}''(1) = {X_n}'(0) = 0 \\ &= {X_m}'{X_n}''\Bigg|_0^1 - \int_0^1 {X_m}'{X_n}''', && {X_m}'(0) = {X_n}''(1) = 0 \\ &= -X_m{X_n}^{(4)}\Bigg|_0^1 + \int_0^1 X_m{X_n}^{(4)}, && X_m(0) = X_n(1) = 0 \end{align}

$$\implies \int_0^1 {X_m}^{(4)}X_n - \int_0^1 X_m{X_n}^{(4)} = 0 \implies (\beta_m^4 - \beta_n^4) \int_0^1 X_mX_n = 0$$

since $$\beta_m \ne \beta_n$$, the integral must be $$0$$

• You're right. I made a mistake. The second solution given is suitable. Thank You Commented Dec 8, 2017 at 21:17
• you have any idea how to find the eigen function? Commented Dec 8, 2017 at 21:18
• You just need to pick a pair $(A,B)$ that satisfies on of the given equations. For example $A = \sinh\beta-\sin\beta,\ -B = \cosh\beta-\cos\beta$. The conditions for $\beta$ ensure they satisfy the other equation as well. Commented Dec 9, 2017 at 3:34
• I got the Eigenfunctions by substituting C = -A and D = -B in the X(x). I got the equation as A(coshb-cosb)+B(sinhb-sinb). How can I prove that these eigenfunctions are orthogonal? Commented Dec 9, 2017 at 4:59
• I suppose my last comment wasn't too clear. I made another edit to the answer above. Commented Dec 9, 2017 at 5:21