- #1
evinda
Gold Member
MHB
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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 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?