Exercises — Initial value problems (IVP) vs boundary value problems (BVP)

Dono conditions same point $x=1$ par hain (ek value hai, ek slope hai — dono $x=1$ par). $\Rightarrow$ IVP. Key idea: "initial" ek location ke baare mein hai, number $0$ ke baare mein nahi.

Conditions do alag points $x=0$ aur $x=\pi /2$ par hain. $\Rightarrow$ BVP (ek two-point problem).

Ek condition $x=2$ par, doosri $x=5$ par — alag points. $\Rightarrow$ BVP (ek mixed wala, kyunki yeh ek end par value aur doosre par slope use karta hai, lekin phir bhi BVP hai kyunki locations alag hain).

General solution: $y=c_1\cos x+c_2\sin x$.

• $y(0)=c_1=2$.
• $y^{\prime }(x)=-c_1\sin x+c_2\cos x$, toh $y^{\prime }(0)=c_2=-3$.

Unique: $y(x)=2\cos x-3\sin x.$ Kyun unique hai? Dono conditions ek hi point par hain $\Rightarrow$ hum directly $c_1,c_2$ read off karte hain; yahan koi matrix singular nahi ho sakti.

$y^{\prime \prime }=0$ ko do baar integrate karo aur ek straight line milti hai: $y=c_{1}x+c_2$.

• $y(0)=c_2=10$.
• $y(4)=4c_1+10=2\Rightarrow c_1=\frac{2-10}{4}=-2$.

Unique: $y(x)=10-2x.$ Determinant se check karo $y_1=x,y_2=1$ ke saath: $\det M=y_1(0)y_2(4)-y_2(0)y_1(4)=0\cdot 1-1\cdot 4=-4\ne 0$. Nonzero $\Rightarrow$ exactly ek solution. ✔

$y=c_1\cos x+c_2\sin x$.

• $y(0)=c_1=1$.
• $y(\pi /2)=c_1\cos (\pi /2)+c_2\sin (\pi /2)=0+c_2=5\Rightarrow c_2=5$.

Unique: $y(x)=\cos x+5\sin x.$ Yahan $\det M=y_1(0)y_2(\pi /2)-y_2(0)y_1(\pi /2)=1\cdot 1-0\cdot 0=1\ne 0$, toh endpoints ek bura pair nahi hain — ek clean solution milta hai.

$y=c_1\cos x+c_2\sin x$.

• $y(0)=c_1=0$.
• $y(\pi )=c_1\cos \pi +c_2\sin \pi =-c_1+0=0$ — yeh sirf $c_1=0$ repeat karta hai aur $c_2$ ke baare mein kuch nahi kehta.

$\det M=\cos 0\cdot \sin \pi -\sin 0\cdot \cos \pi =1\cdot 0-0\cdot (-1)=0$. Kyunki right-hand side $(0,0)$ compatible hai, hume infinitely many solutions milte hain: $y(x)=c_2\sin x,c_2\text{arbitrary}.$ Picture: figure mein orange sine curves dekho — saari dono endpoints $x=0$ aur $x=\pi$ se pass karti hain.

• $y(0)=c_1=0$.
• $y(\pi )=-c_1=0$, lekin humne demand kiya tha $y(\pi )=1$. Toh $0=1$ — impossible.

Same singular $\det M=0$, lekin ab target incompatible hai. $\Rightarrow$ Koi solution nahi. Kyun: jab $\det M=0$ hota hai, doosri equation pehli ki shadow mein hi rehti hai; agar requested value us shadow mein nahi padti, toh kuch bhi dono satisfy nahi karta.

Characteristic behaviour se $y=c_1\cos 2x+c_2\sin 2x$ milta hai.

• $y(0)=c_1=0\Rightarrow y=c_2\sin 2x$.
• $y(\pi )=c_2\sin (2\pi )=c_2\cdot 0=0$ — kisi bhi $c_2$ ke liye automatically satisfy hota hai.

$\det M=\cos 0\cdot \sin 2\pi -\sin 0\cdot \cos 2\pi =0$. $\Rightarrow$ Infinitely many: $y=c_2\sin 2x$.

$\lambda =k^2$ likho with $k>0$, toh $y=c_1\cos kx+c_2\sin kx$.

• $y(0)=c_1=0\Rightarrow y=c_2\sin kx$.
• $y(L)=c_2\sin kL=0$. $c_2\ne 0$ ke liye humhe $\sin kL=0\Rightarrow kL=n\pi ,n=1,2,3,\dots$ chahiye.

Toh $k=\frac{n\pi }{L}$ aur eigenvalues hain ${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2,n=1,2,3,\dots$ with eigenfunctions $y_n=\sin \left(\frac{n\pi }{L}x\right)$. $n\ge 1$ kyun? $n=0$ se $k=0$ milta hai, yaani $y\equiv 0$ — woh trivial solution hai jise hum exclude kar rahe hain.

4.1 se $L=2$ ke saath: allowed eigenvalues hain ${\lambda }_n={\left(\frac{n\pi }{2}\right)}^2=\frac{n^2{\pi }^2}{4}$.

• $n=1:\frac{{\pi }^2}{4}\approx 2.467$ — allowed ✔
• $n=2:{\pi }^2\approx 9.870$ — allowed ✔
• $\lambda =1$: kya $1=\frac{n^2{\pi }^2}{4}$? Uske liye $n=\frac{2}{\pi }\approx 0.637$ chahiye, jo integer nahi hai — allowed nahi (sirf trivial solution).
• $\lambda =4$: $n=\frac{4}{\pi }\approx 1.273$ chahiye, jo integer nahi hai — allowed nahi.

Toh nontrivial wale hain $\boxed{\lambda =\frac{{\pi }^2}{4}}$ aur $\boxed{\lambda ={\pi }^2}$.

$y=c_1\cos x+c_2\sin x$, toh $y^{\prime }=-c_1\sin x+c_2\cos x$.

• $y^{\prime }(0)=-c_1\cdot 0+c_2\cdot 1=c_2=0$.
• $y^{\prime }(\pi /2)=-c_1\sin (\pi /2)+c_2\cos (\pi /2)=-c_1\cdot 1+0=-c_1=0\Rightarrow c_1=0$.

Dono constants $0$ par force ho gaye $\Rightarrow$ sirf trivial solution $y\equiv 0$. Yeh ek unique solution count hota hai (trivial wala). Koi free constant nahi bachta, toh yahan $\det M\ne 0$ hai.

Do baar integrate karo: $y^{\prime }=3x^2+c_1$, phir $y=x^3+c_{1}x+c_2$.

• $y(0)=c_2=0$.
• $y^{\prime }(1)=3(1{)}^2+c_1=3+c_1=2\Rightarrow c_1=-1$.

Unique: $y(x)=x^3-x.$ Check: $y^{\prime \prime }=6x$ ✔; $y(0)=0$ ✔; $y^{\prime }(x)=3x^2-1$, $y^{\prime }(1)=2$ ✔. Kyunki ek condition $x=0$ par value hai aur doosri $x=1$ par slope hai (alag points), yeh hai ek BVP — lekin ek well-posed wala.

$y=c_1\cos x+c_2\sin x$.

• $y(0)=c_1=1$.
• $y(\pi )=-c_1+0=-1$. Humhe $-c_1=-1$ chahiye, yaani $c_1=1$ — pehli condition ke saath consistent hai.

$\det M=\cos 0\sin \pi -\sin 0\cos \pi =0$: singular, lekin target compatible hai. $\Rightarrow$ Infinitely many solutions: $y=\cos x+c_2\sin x$ kisi bhi $c_2$ ke liye. 3.2 se contrast: same singular matrix, lekin wahan target ($0=1$) incompatible tha $\Rightarrow$ koi nahi. Yahan $c_1=1$ dono rows se agree karta hai $\Rightarrow$ ek poora family.

Saari conditions ek hi point par hain? ::: IVP — $c_1,c_2$ read off karke solve karo; hamesha uniquely solvable (Wronskian $\ne 0$). Conditions do points par hain aur $\det M\ne 0$? ::: Unique BVP solution. $\det M=0$ with compatible data? ::: Infinitely many solutions. $\det M=0$ with incompatible data? ::: Koi solution nahi. Special ${\lambda }_n=(n\pi /L{)}^2$ kahan se aate hain? ::: BVP ki $\det M=0$ condition se — Sturm-Liouville Theory eigenvalues.