Question

Consider the boundary-value problem $\mathit{y}\mathbf{+}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\left(\pi /2\right)\mathbf{=}\mathbf{0}$. Discuss: Is it possible to determine real values of$\mathit{\lambda }$so that the problem possesses (a) trivial solutions? (b) nontrivial solutions?

Short answer

1. When $\lambda =0$or$\lambda <0$ , the boundary-value problem possesses a trivial solution
2. When $\lambda >0$, the boundary-value problem possesses a nontrivial solution.

Step-by-step solution

General solution

The auxiliary equation of the given boundary-value problem is . To determine if the problem possesses trivial or nontrivial solutions, we consider three different cases.

$\begin{array}{l}\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{x}\\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{ }\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{ }\mathbf{ }\mathbf{a}\mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ }\mathbf{ }\mathbf{a}\mathbf{x}\\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{a}\mathbf{x}}\end{array}$

(a): Repeated real roots                    

Case 1: When$\lambda =0$, then the auxiliary equation becomes$m^2=0$ and has roots$m_1=m_2=0$. Therefore, from Case II: Repeated real roots, the general solution of the differential equation is

$y=c_1+c_{2}x$

Substituting$y\left(0\right)=0$ in the above equation gives

$\begin{array}{l}{c}_1+c_2\left(0\right)=0\\ c_{1 }=0······ \left(1\right)\end{array}$

Substituting$y\left(\pi /2\right)=0$ in the above equation gives

$\begin{array}{l}{c}_1+c_2\left(\frac{\pi }{2}\right)=0\\ 0+\frac{\pi c_2}{2}=0\\ c_2=0 ······ \left(2\right)\end{array}$

Therefore, substituting $c_1\mathrm{and}{c}_2$in the general solution gives y=0 which concludes that the given problem possesses a trivial solution for$\lambda =0$

(b): Conjugate complex roots

Case 2: When$\lambda >0$, i.e., let$\lambda =a^2$for some real number a. Then the auxiliary equation becomes$m^2+a^2=0$and has roots$m_1=ia\mathrm{and}{m}_2=-ia$. Therefore, from Case III: Conjugate complex roots, the general solution of the differential equation$y=c_1 \cos ax+c_2 \sin ax$

Substituting$y\left(0\right)=0$in the above equation gives

$\begin{array}{c}{c}_1 \cos \left(a · 0\right)+c_2 \sin \left(a ·0\right)=0\\ c_1 \cos 0+c_2 \sin 0=0\\ c_1=0······ \left(1\right)\end{array}$

Substituting$y\left(\pi /2\right)=0$in the above equation gives

$\begin{array}{c}{c}_1 \cos \left(a . \frac{\pi }{2}\right)+c_2 \sin \left(a . \frac{\pi }{2}\right)=0\\ 0+c_2 \sin \left(\frac{\pi a}{2}\right)=0\\ c_2=0 \sin \left(\frac{\pi a}{2}\right)=\sin \left(n\pi \right)\\ c_2=0\mathrm{or}a=2n ······\left(2\right)\end{array}$

If $c_2=0$, then the solution from case 1 repeats. On the other hand, substituting$c_1$ and the other value of $c_2$in the general solution gives us $y= c_2 \sin 2nx$where is a nonzero integer. This concludes that the given problem possesses a nontrivial solution for$\lambda >0$

Distinct real roots

Case 3: When$\lambda <0$, i.e., let$\lambda =-a^2$for some real number a. Then the auxiliary equation becomes$m^2-a^2=0$and has roots$m_1=a\mathrm{and}{m}_2=-a$. Therefore, from Case I: Distinct real roots, the general solution of the differential equation is

$y=c_1{e}^{ax}+c_2{e}^{-ax}$

Substituting$y\left(0\right)=0$ in the above equation gives

$\begin{array}{r}{c}_1{e}^{a . 0}+c_2{e}^{-a . 0}=0\\ c_1{e}^0+c_2{e}^0=0\\ c_1+c_2=0\\ c_2=-c_1······\left(1\right)\end{array}$

Substituting$y\left(\pi /2\right)=0$in the above equation gives

$\begin{array}{c}{c}_1{e}^{a . \frac{\pi }{2}}+c_2{e}^{-a . \frac{\pi }{2}}=0\\ c_1{e}^{\frac{\pi a}{2}}-c_1{e}^{-\frac{\pi a}{2}}=0\\ c_1\left(e^{\frac{\pi a}{2}}-e^{-\frac{\pi a}{2}}\right)=0\\ c_1\left(e^{\frac{\pi a}{2}}-\frac{1}{e^{\frac{\pi a}{2}}}\right)=0\\ c_1\left(\frac{e^{\pi a}-1}{e^{\frac{\pi a}{2}}}\right)=0\\ c_1=0\mathrm{or}{e}^{\pi a}=1 ······\left(2\right)\end{array}$

But $e^{\pi a}=1$,only when$a=0$which means$\lambda =0$.But, we have considered$\lambda <0$. So, we ignore this solution. Hence$c_{1 }=0$. Therefore, substituting and in the general solution gives us$y=0$which concludes that the given problem possesses a trivial solution for$\lambda <0$

Therefore, we consider three different cases. When$\lambda =0\mathrm{or}\lambda <0$, the boundary-value problem possesses a trivial solution and when $\lambda >0$, the boundary-value problem possesses a nontrivial solution.