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

Both conditions sit at the same point $x=1$ (one is the value, one is the slope — both at $x=1$). $\Rightarrow$ IVP. Key idea: "initial" is about one location, not about the number $0$.

Conditions at two different points $x=0$ and $x=\pi /2$. $\Rightarrow$ BVP (a two-point problem).

One condition at $x=2$, the other at $x=5$ — different points. $\Rightarrow$ BVP (a mixed one, since it uses a value at one end and a slope at the other, but still a BVP because the locations differ).

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

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

Unique: $y(x)=2\cos x-3\sin x.$ Why unique? Both conditions at one point $\Rightarrow$ we read off $c_1,c_2$ directly; no matrix can be singular here.

$y^{\prime \prime }=0$ integrates twice to a straight line: $y=c_{1}x+c_2$.

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

Unique: $y(x)=10-2x.$ Check via the determinant with $y_1=x,y_2=1$: $\det M=y_1(0)y_2(4)-y_2(0)y_1(4)=0\cdot 1-1\cdot 4=-4\ne 0$. Nonzero $\Rightarrow$ exactly one 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: $y(x)=\cos x+5\sin x.$ Here $\det M=y_1(0)y_2(\pi /2)-y_2(0)y_1(\pi /2)=1\cdot 1-0\cdot 0=1\ne 0$, so the endpoints are not a bad pair — one clean solution.

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

• $y(0)=c_1=0$.
• $y(\pi )=c_1\cos \pi +c_2\sin \pi =-c_1+0=0$ — this just repeats $c_1=0$ and says nothing about $c_2$.

$\det M=\cos 0\cdot \sin \pi -\sin 0\cdot \cos \pi =1\cdot 0-0\cdot (-1)=0$. Because the right-hand side $(0,0)$ is compatible, we get infinitely many solutions: $y(x)=c_2\sin x,c_2\text{arbitrary}.$ Picture: look at the orange sine curves in the figure — all pass through both endpoints $x=0$ and $x=\pi$.

• $y(0)=c_1=0$.
• $y(\pi )=-c_1=0$, but we demanded $y(\pi )=1$. So $0=1$ — impossible.

Same singular $\det M=0$, but now the target is incompatible. $\Rightarrow$ No solution. Why: when $\det M=0$, the second equation lives entirely inside the first's shadow; if the requested value doesn't land in that shadow, nothing satisfies both.

Characteristic behaviour gives $y=c_1\cos 2x+c_2\sin 2x$.

• $y(0)=c_1=0\Rightarrow y=c_2\sin 2x$.
• $y(\pi )=c_2\sin (2\pi )=c_2\cdot 0=0$ — automatically satisfied for any $c_2$.

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

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

• $y(0)=c_1=0\Rightarrow y=c_2\sin kx$.
• $y(L)=c_2\sin kL=0$. For $c_2\ne 0$ we need $\sin kL=0\Rightarrow kL=n\pi ,n=1,2,3,\dots$

So $k=\frac{n\pi }{L}$ and the eigenvalues are ${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2,n=1,2,3,\dots$ with eigenfunctions $y_n=\sin \left(\frac{n\pi }{L}x\right)$. Why $n\ge 1$? $n=0$ gives $k=0$, i.e. $y\equiv 0$ — the trivial solution we're excluding.

From 4.1 with $L=2$: allowed eigenvalues are ${\lambda }_n={\left(\frac{n\pi }{2}\right)}^2=\frac{n^2{\pi }^2}{4}$.

• $n=1:\frac{{\pi }^2}{4}\approx 2.467$ — allowed ✔
• $n=2:{\pi }^2\approx 9.870$ — allowed ✔
• $\lambda =1$: is $1=\frac{n^2{\pi }^2}{4}$? That needs $n=\frac{2}{\pi }\approx 0.637$, not an integer — not allowed (only trivial solution).
• $\lambda =4$: needs $n=\frac{4}{\pi }\approx 1.273$, not an integer — not allowed.

So the nontrivial ones are $\boxed{\lambda =\frac{{\pi }^2}{4}}$ and $\boxed{\lambda ={\pi }^2}$.

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

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

Both constants forced to $0$ $\Rightarrow$ only the trivial solution $y\equiv 0$. That counts as a unique solution (the trivial one). No free constant survives, so $\det M\ne 0$ here.

Integrate twice: $y^{\prime }=3x^2+c_1$, then $y=x^3+c_{1}x+c_2$.

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

Unique: $y(x)=x^3-x.$ Check: $y^{\prime \prime }=6x$ ✔; $y(0)=0$ ✔; $y^{\prime }(x)=3x^2-1$, $y^{\prime }(1)=2$ ✔. Because one condition is a value at $x=0$ and the other a slope at $x=1$ (different points), this is a BVP — but a well-posed one.

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

• $y(0)=c_1=1$.
• $y(\pi )=-c_1+0=-1$. We need $-c_1=-1$, i.e. $c_1=1$ — consistent with the first condition.

$\det M=\cos 0\sin \pi -\sin 0\cos \pi =0$: singular, but the target is compatible. $\Rightarrow$ Infinitely many solutions: $y=\cos x+c_2\sin x$ for any $c_2$. Contrast with 3.2: same singular matrix, but there the target ($0=1$) was incompatible $\Rightarrow$ none. Here $c_1=1$ agrees with both rows $\Rightarrow$ a whole family.

All conditions at one point? ::: IVP — solve by reading off $c_1,c_2$; always uniquely solvable (Wronskian $\ne 0$). Conditions at two points and $\det M\ne 0$? ::: Unique BVP solution. $\det M=0$ with compatible data? ::: Infinitely many solutions. $\det M=0$ with incompatible data? ::: No solution. Where do the special ${\lambda }_n=(n\pi /L{)}^2$ come from? ::: The $\det M=0$ condition of the BVP — Sturm-Liouville Theory eigenvalues.