MHB Initial and boundary value problem

[evinda]

Hello! (Wave)

I want to find the solution of the following initial and boundary value problem:

$$u_t(x,t)-u_{xx}(x,t)=0, x>0, t>0 \\ u_x(0,t)=0, t>0, \\ u(x,0)=x^2, x>0.$$I have done the following so far:

$$u(x,t)=X(x) T(t)$$

$$u_t(x,t)=u_{xx}(x,t) \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$

$$u_x(0,t)=0 \Rightarrow X'(0)=0$$

Then we have the following two problems:$$\left\{\begin{matrix}
X''(x)+\lambda X(x)=0\\
X'(0)=0
\end{matrix}\right.$$

and

$$\left\{\begin{matrix}
T'(t)+\lambda T(t)=0
\end{matrix}\right.$$We have that $\lambda=\beta^2$ for some $\beta>0$.

So $X''(x)+\beta^2 X(x)=0 \Rightarrow X(x)=C \sin{(\beta x)}+D \cos{(\beta x)}$.

$X'(0)=0 \Rightarrow C=0$.

So $X(x)=D \cos{(\beta x)}$.

But don't we have to write $\beta$ in respect to $n$ ? (Thinking)

How could we do so? Or have I done something wrong?

 

[I like Serena]

But don't we have to write $\beta$ in respect to $n$ ?

How could we do so? Or have I done something wrong?

Hey evinda!

What is $n$?
Doesn't the solution for X hold for any $\beta$?
Perhaps we should just continue? (Wondering)