MHB Calculating the coefficients with the Fourier series

[mathmari]

Hey!

I have to solve the following initial and boundary value problem:
$$u_t=u_{xx}, 0<x<L, t>0 (1)$$
$$u(0,t)=u_x(L,t)=0, t>0$$
$$u(x,0)=x, 0<x<L$$

I did the following:
Using the method separation of variables, the solution is of the form: $u(x,t)=X(x)T(t)$

Replacing this at $(1)$, we get:

$$\left.\begin{matrix}
X''+\lambda X=0, 0<x<L\\
X(0)=0, X'(L)=0
\end{matrix}\right\} (2)$$

$$\left.\begin{matrix}
T'+\lambda T=0, t>0
\end{matrix}\right\} (3)$$

$$u(x,0)=X(x)T(0)=x$$Solving the problem $(2)$ we get that the eigenfunctions are $$X_n(x)=\sin{(\frac{(2n+1) \pi x}{2L})}$$

Solving the problem $(3)$ we get $$T_n(t)=A_ne^{-(\frac{(2n+1) \pi}{2L})^2t}$$

So the solution of the initial problem is of the form $$u(x,t)=\sum_{n=1}^{\infty}{A_n \sin{(\frac{(2n+1) \pi x}{2L})}e^{-(\frac{(2n+1) \pi}{2L})^2t}}$$

$$u(x,0)=x \Rightarrow x=\sum_{n=1}^{\infty}{A_n \sin{(\frac{(2n+1) \pi x}{2L})}}$$

To calculate the coefficients $A_n$ can I do the following:

We expand the function $f(x)=x$ in an odd way in $[-L,L]$.
So we can write $f$ as a Fourier series:
$$f(x)=\sum_{n=1}^{\infty}{b_n \sin{(\frac{2(2n+1) \pi x}{2 \cdot 2L})}}, \text{ where } b_n=\frac{2}{4L} \int_{-L}^L{f(x) \sin{(\frac{2(2n+1) \pi x}{4L})}}dx$$

Or is it wrong to calculate these coefficients in that way?? (Wondering)

 

[mathmari]

Or is it better to calculate them with the Sturm-Liouville theory?

 

[mathmari]

When I calculate the coefficients using the Fourier series, I get the following:

$$f(x)=\sum_{n=1}^{\infty}{b_n \sin{(\frac{2 (2n+1) \pi x}{2 \cdot 2L})}},$$ $$\text{ where } b_n=\frac{2}{4L} \int_{-L}^L{f(x) \sin{(\frac{2 (2n+1) \pi x}{4L})}}dx=\frac{1}{L} \int_0^L{f(x) \sin{(\frac{ (2n+1) \pi x}{2L})}}dx$$

So $$A_n=\frac{1}{L} \int_0^L{x \sin{(\frac{(2n+1) \pi x}{2})}}dx =\frac{1}{L} \frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n=\frac{4L}{(2n+1)^2 \pi^2}(-1)^n$$When I use the Sturm-Liouville theory, I get the following:

$$A_n=\frac{(f(x),X_n(x))}{||X_n||^2}$$

$$(f(x),X_n(x))=\int_0^L{f(x) \sin{(\frac{ (2n+1) \pi x}{2L})}}dx=\int_0^L{x \sin{(\frac{ (2n+1) \pi x}{2L})}}dx=\frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n$$

$$||X_n^2||=\int_0^L{\sin^2{(\frac{ (2n+1) \pi x}{2L})}}dx=\frac{L}{2}$$

So $$A_n=\frac{2}{L}\frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n=\frac{8L}{(2n+1)^2 \pi^2}(-1)^n$$The results are not the same...
What have I done wrong? (Wondering)

 

[I like Serena]

When I calculate the coefficients using the Fourier series, I get the following:

$$f(x)=\sum_{n=1}^{\infty}{b_n \sin{(\frac{2 (2n+1) \pi x}{2 \cdot 2L})}},$$ $$\text{ where } b_n=\frac{2}{4L} \int_{-L}^L{f(x) \sin{(\frac{2 (2n+1) \pi x}{4L})}}dx=\frac{1}{L} \int_0^L{f(x) \sin{(\frac{ (2n+1) \pi x}{2L})}}dx$$

So $$A_n=\frac{1}{L} \int_0^L{x \sin{(\frac{(2n+1) \pi x}{2})}}dx =\frac{1}{L} \frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n=\frac{4L}{(2n+1)^2 \pi^2}(-1)^n$$When I use the Sturm-Liouville theory, I get the following:

$$A_n=\frac{(f(x),X_n(x))}{||X_n||^2}$$

$$(f(x),X_n(x))=\int_0^L{f(x) \sin{(\frac{ (2n+1) \pi x}{2L})}}dx=\int_0^L{x \sin{(\frac{ (2n+1) \pi x}{2L})}}dx=\frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n$$

$$||X_n^2||=\int_0^L{\sin^2{(\frac{ (2n+1) \pi x}{2L})}}dx=\frac{L}{2}$$

So $$A_n=\frac{2}{L}\frac{4L^2}{(2n+1)^2 \pi^2}(-1)^n=\frac{8L}{(2n+1)^2 \pi^2}(-1)^n$$The results are not the same...
What have I done wrong? (Wondering)

Hey! (Blush)

Hmm, I don't think you have the right Fourier formula...

What you call Sturm-Liouville theory is actually an inner product formula, which does look correct.

 

[mathmari]

Hey! (Blush)

Hmm, I don't think you have the right Fourier formula...

What you call Sturm-Liouville theory is actually an inner product formula, which does look correct.

Ahaa! Ok! So I have to use the inner product formula in these cases? (Wondering)

 

[I like Serena]

Ahaa! Ok! So I have to use the inner product formula in these cases? (Wondering)

It boils down to the same thing.
The great discovery Fourier made was that $\{\cos(n\pi x), \sin(n\pi x) \}$ forms a basis for all square integrable functions. To find the coefficients we simply need to take the inner product.
If you apply the Fourier sine series correctly you should find the same result.

 

[mathmari]

It boils down to the same thing.
The great discovery Fourier made was that $\{\cos(n\pi x), \sin(n\pi x) \}$ forms a basis for all square integrable functions. To find the coefficients we simply need to take the inner product.
If you apply the Fourier sine series correctly you should find the same result.

Ahaa! Ok!

What have I done wrong at the Fourier series?? (Wondering)

Since the boundary conditions are mixed we cannot expand the function neither in an odd way nor in an even way...

What can I do in this case?? (Wondering)