I need some help with the following questions:
If $γ>0$ and $λ ≤ 0$.
1) I have to find all solutions to the differential equation
$γv'' + λv = 0$.
2) Then, I have to prove that the only solution that satisfies the boundary conditions $v(0) = 0$, $v(l)=0$ , is the zero solution $v(x) ≡ 0$.
I tried to find the answer to the first question and I eventually found:
$v = A + Be^{-λt/γ}$.
I am for sure that this is NOT right. But, I cannot find any other solution. So any help would be grateful
On
Assume the solution takes the general form
$$v(t) = Ae^{at}+Be^{-at}$$
Plug into $γv'' + λv = 0$ to get
$$a^2 γ + λ =0$$
Solve for the exponential coefficient $a$
$$a=\sqrt{ \frac{|λ|}{γ}}$$
Then, with the boundary conditions,
$$v(0)=A+B=0$$ $$v(l)=Ae^{al}+Be^{-al}=0$$
Solve for $A$ and $B$,
$$A=B=0$$
Thus, $v(t) = 0$.
Rewrite the equation as $$v′′+\alpha v=0$$ where $\alpha = \frac{\lambda}{\gamma}\leq 0$ is well defined as $\gamma > 0$. Then analyze the characteristic equation $$r^2+\alpha=0$$ which as $\alpha \le 0$ gives two roots $$r_1=\sqrt{\alpha},~~r_2=-\sqrt{\alpha}$$ so the general solution is $$v(t)=c_1e^{\sqrt{\alpha}t}+c_2e^{-\sqrt{\alpha}t}\tag{1}$$ which after substituting $\alpha = \frac{\lambda}{\gamma}$ forms $$v(t)=c_1e^{\sqrt{\frac{\lambda}{\gamma}}t}+c_2e^{-\sqrt{\frac{\lambda}{\gamma}}t}\tag{2}$$ then evaluate the boundary conditions using $(1)$ $$ v(0)=0 \implies c_1+c_2=0 $$ $$ v(l)=0 \implies c_1e^{\sqrt{\alpha}l}+c_2e^{-\sqrt{\alpha}l}=0$$ so that $$c_1+c_2=0 \implies c_2=-c_1$$ and $$c_1e^{\sqrt{\alpha}l}+c_2e^{-\sqrt{\alpha}l}=0 \implies c_1e^{\sqrt{\alpha}l}= c_1e^{-\sqrt{\alpha}l} \iff c_1 = 0$$ then by our first condition $$c_2=0$$ therefore $$v(t)\equiv 0$$