# solution of Given boundary value problem

For $$\lambda\in \mathbb{R}$$ consider the boundary value problem

$$x^2 \frac{d^2y}{dx^2}+2x\frac{dy}{dx}+\lambda y=0\;\;\;,y(1)=y(2)=0$$ called as $$(P_{\lambda})$$

Then i Have to prove which of following is right and which is wrong and why?

$$1).$$For any continuous function $$f:[1,2]\to \mathbb{R}\;\;$$with $$f(x)\neq 0\;\;$$for some $$x \in[1,2]$$ there $$\\$$ exist a solution $$u$$ of ($$P_{\lambda}$$)

$$2).$$There exist a $$\lambda_0 \in \mathbb{R}$$ such that ($$P_{\lambda}$$) has a nontrivial solution for any $$\lambda>\lambda_0$$

The solution i tried- we can write the given differntial equation in form

$$\theta(\theta-1)+2\theta+\lambda=0\\ \theta^2+\theta+\lambda=0\\$$ which is quadratic in $$\theta$$

after solving this i get

$$\theta =\frac{-1 \pm \sqrt {1-4\lambda}}{2}$$

after that i solved further i get

$$\displaystyle y=c_1x^{\frac{-1 + \sqrt {1-4\lambda}}{2}}+c_2x^{\frac{-1 -\sqrt {1-4\lambda}}{2}}$$

further applying the boundary conditions i get two conditions

$$c_1+c_2=0\;\; and\;\; c_1 2^{\frac{-1 + \sqrt {1-4\lambda}}{2}}+c_2 2^{\frac{-1 - \sqrt {1-4\lambda}}{2}}$$

for non trivial solution there Cofficents matrix determinent should not be zero. in end after all calculation i get

$$\lambda \neq \frac{1}{4}$$

after that i have no clue how to get to these options? how can i check which is true and why ?

Thankyou.

• For question (1), what is $f$ doing there? The question is about a solution in $P_\lambda$, no? Oct 15, 2019 at 10:27
• Don't know in option it is given Oct 15, 2019 at 10:33

Let $$t:=\ln(x)$$. Then, $$\frac{\text{d}y}{\text{d}x}=\exp(-t)\,\frac{\text{d}y}{\text{d}t}\,.$$ That is, $$x^2\,\frac{\text{d}^2y}{\text{d}x^2}=\exp(t)\,\frac{\text{d}}{\text{d}t}\,\left(\exp(-t)\,\frac{\text{d}y}{\text{d}t}\right)=\frac{\text{d}^2y}{\text{d}t^2}-\frac{\text{d}y}{\text{d}t}\,.$$ Therefore, the given differential equation now becomes $$\frac{\text{d}^2y}{\text{d}t^2}+\frac{\text{d}y}{\text{d}t}+\lambda\,y=0\,.$$ If $$\lambda<\dfrac14$$, then the solutions are given by \begin{align}y&=A_-\,\exp\left(\frac{-1-\sqrt{1-4\lambda}}{2}\,t\right)+A_+\,\exp\left(\frac{-1-\sqrt{1+4\lambda}}{2}\,t\right)\\&=A_-\,x^{\frac{-1-\sqrt{1-4\lambda}}{2}}+A_+\,x^{\frac{-1+\sqrt{1-4\lambda}}{2}}\,,\end{align} where $$A_-$$ and $$A_+$$ are constants. If $$\lambda=\dfrac14$$, then the solutions are given by $$y=(a+b\,t)\,\exp\left(-\frac{1}{2}\,t\right)=\big(a+b\,\ln(x)\big)\,x^{-\frac12}\,,$$ where $$a$$ and $$b$$ are constants. If $$\lambda>\dfrac14$$, then the solutions are given by \begin{align}y&=\exp\left(-\frac{1}{2}\,t\right)\,\Biggl(C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,t\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,t\right)\Biggr) \\&=x^{-\frac{1}{2}}\,\Biggl(C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(x)\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(x)\right)\Biggr)\,,\end{align} where $$C$$ and $$S$$ are constants.
For $$\lambda<\dfrac14$$, if $$y(x=1)=0$$ and $$y(x=2)=0$$, then $$A_-+A_+=0\text{ and }A_-+A_+\,2^{\sqrt{1-4\lambda}}=0\,.$$ It follows that $$A_-=A_+=0$$, so $$y$$ is identically zero. For $$\lambda=\dfrac14$$, if $$y(x=1)=0$$ and $$y(x=2)=0$$, then $$a=0\text{ and }a+b\,\ln(2)=0\,,$$ whence $$a=b=0$$ and so $$y$$ is again identically zero. For $$\lambda>\dfrac14$$, if $$y(x=1)=0$$ and $$y(x=2)=0$$, then $$C=0\text{ and }C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)\right)=0\,.$$ This shows that $$C=0$$ and $$S=0$$, or $$C=0$$ and $$\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)=k\pi$$ for some positive integer $$k$$. Thus, $$\lambda=\frac{k^2\pi^2}{\big(\ln(2)\big)^2}+\frac{1}{4}\tag{*}$$ for some positive integer $$k$$.
That is, there exists a nonzero solution to $$P_\lambda$$ if and only if $$\lambda$$ is given by (*). In that case, all solutions take the form $$y=S\,x^{-\frac12}\,\sin\big(k\pi\,\log_2(x)\big)$$ where $$S$$ is a constant. Hence, Option (2) is false. I do not understand Option (1).