MHB Solving the DE for $\phi$: Find $A,B,\phi$.

[evinda]

Hello! (Wave)
Let $I=(0,1)$. Find solution $\phi$ that has a continuous derivative in $\mathbb{R}$ and satisfies:

$$y''=0 \text{ in } I \\ y''+k^2y=0 \text{ apart from I , where k>0}$$

and furthermore $\phi$ has the form:$\left\{\begin{matrix}
e^{ikx}+ A e^{-ikx} &, x \leq 0 \\
B e^{ikx} &, x \geq 1
\end{matrix}\right.$

i.e. find $\phi$ computing the constants $A,B$ and computing $\phi$ in the interval $I$.I have tried the following:

$\phi$ satisfies the differential equation $y''=0$ in $I$.

So $\phi''(x)=0 \Rightarrow \phi(x)= c_1 x + c_2, c_1, c_2 \in \mathbb{R}, \forall x \in I$.

$\phi$ has a continuous derivative in $\mathbb{R}$, so:

$$\lim_{x \to 0^{-1}} \phi(x)= \lim_{x \to 0^{+}} \phi(x) \\ \lim_{x \to 1^{-1}} \phi(x)= \lim_{x \to 1^{+}} \phi(x) \\ \lim_{x \to 0^{-1}} \phi'(x)= \lim_{x \to 0^{+}} \phi'(x) \\ \lim_{x \to 1^{-1}} \phi'(x)= \lim_{x \to 1^{+}} \phi'(x) $$
$\lim_{x \to 0^{-1}} \phi(x)= \lim_{x \to 0^{+}} \phi(x) \Rightarrow 1+A=c_2 (\star)$

$\lim_{x \to 1^{-1}} \phi(x)= \lim_{x \to 1^{+}} \phi(x) \Rightarrow Be^{ik}=c_1+c_2 (\star \star)$

$\lim_{x \to 0^{-1}} \phi'(x)= \lim_{x \to 0^{+}} \phi'(x) \Rightarrow ik-ikA=c_1 (1)$

$\lim_{x \to 1^{-1}} \phi'(x)= \lim_{x \to 1^{+}} \phi'(x) \Rightarrow ik e^{ik}-ik A e^{-ik}=c_1 (2)$

$(1), (2) \Rightarrow ike^{ik}-ikAe^{-ik}=ik-ikA \Rightarrow A= \frac{e^{ik}-1}{e^{-ik}-1}$

Thus, $c_1=ik \frac{e^{-ik}-e^{ik}}{e^{-ik}-1}$

$(\star) \Rightarrow c_2=\frac{e^{-ik}+e^{ik}-2}{e^{-ik}-1}$

$(\star \star) \Rightarrow B=\frac{e^{-2ik}(ik+1)+(1-ik)-2}{e^{-ik}-1}$.Therefore,$\phi(x)=\left\{\begin{matrix}
e^{ikx}+ \frac{e^{ik}-1}{e^{-ik}-1} e^{-ikx} &, x \leq 0 \\
\frac{e^{-2ik}(ik+1)+(1-ik)-2}{e^{-ik}-1} e^{ikx} &, x \geq 1
\end{matrix}\right.$

and $\phi(x)=ik \frac{e^{-ik}-e^{ik}}{e^{-ik}-1} x+\frac{e^{-ik}+e^{ik}-2}{e^{-ik}-1} $ in $I$.

Is it right or have I done something wrong? (Thinking)

 

[I like Serena]

Hey! (Evilgrin)

$\lim_{x \to 1^{-1}} \phi'(x)= \lim_{x \to 1^{+}} \phi'(x) \Rightarrow ik e^{ik}-ik A e^{-ik}=c_1 (2)$

Shouldn't that be:
$$\lim_{x \to 1^{-1}} \phi'(x)= \lim_{x \to 1^{+}} \phi'(x) \Rightarrow c_1 = B e^{ik} \qquad\qquad(2)$$
? (Wasntme)

 

[evinda]

Hey! (Evilgrin)
Shouldn't that be:
$$\lim_{x \to 1^{-1}} \phi'(x)= \lim_{x \to 1^{+}} \phi'(x) \Rightarrow c_1 = B e^{ik} \qquad\qquad(2)$$
? (Wasntme)

I retried it now... Don't we get $c_1= B ik e^{ik}$ ? Or am I wrong? (Thinking)

 

[I like Serena]

I retried it now... Don't we get $c_1= B ik e^{ik}$ ? Or am I wrong? (Thinking)

You're right... I forgot to take the derivative... (Blush)
It's a bicycle! (Mmm)

 

[evinda]

You're right... I forgot to take the derivative... (Blush)
It's a bicycle! (Mmm)

So will we have the followng values? (Thinking)

$$B=\frac{2e^{-ik}}{2-ik} \\ c_1= \frac{2ik}{2-ik} \\ c_2=\frac{2i+2k}{2i+k} \\ A=\frac{k}{2i+k}$$

 

[I like Serena]

So will we have the followng values? (Thinking)

$$B=\frac{2e^{-ik}}{2-ik} \\ c_1= \frac{2ik}{2-ik} \\ c_2=\frac{2i+2k}{2i+k} \\ A=\frac{k}{2i+k}$$

That's what I get as well. (Nod)

 

[evinda]

That's what I get as well. (Nod)

Great... Thanks a lot! (Whew)