Boundary Value Problems
Discussion
For a second-order linear equation, the general solution will have two arbitrary constants, and you will need two additional pieces of information to determine those constants. So far we have considered initial value problems, where those additional pieces of information give the value of the function and its derivative at a single point. In the theory section, we stated a theorem that for reasonable problems, such an initial value problem always had a unique solution. But sometimes the two data points we know give values at different points. These are called boundary value problems. As noted in the theory section, with boundary values sometimes even nice problems may have no solution, or many solutions. In this section we will explore these issues. We will start with a paradigm and then some examples of the types of results that can occur.Paradigm
$$ y'' + 4y' + 3y = 0, \qquad y(0)=1, y(1)=1$$ Step 1: Find the general solution, the same way as always. I won't bother to show all the sub-steps since I think you are good with those by now. $$ y = C_1 e^{-3x} + C_2 e^{-x} $$ Step 2: Plug in the boundary values to get two equations in two unknowns (or $n$ equations in $n$ unknowns if you have an $n^{th}$-order equation). $$\begin{align} y(0)&=C_1 + C_2 \buildrel{\text{set}}\over{=} 1 \\ y(1)&=C_1 e^{-3} + C_2 e^{-1} \buildrel{\text{set}}\over{=} 1 \end{align} $$ Step 3: Solve the system of equations for the values of the constants. If we multiply the second equation by $e$ we get $C_1 e^{-2} + C_2 = e$. Subtracting this from the first equation we get $(1-e^{-2})C_1 = 1-e$ and hence $C_1 = (1-e)/(1-e^{-2})$. Plugging this into the first equation then gives us $C_2=(e-e^{-2})/(1-e^{-2})$ and so the solution to the initial value problem is $$ y(x) = \frac{1-e}{1-e^{-2}}e^{-3x} + \frac{e-e^{-2}}{1-e^{-2}}e^{-x}. $$Examples
EXAMPLE 1: An inhomogeneous equation $$ y'' - 4y = 6e^x, \qquad y(0)=1, y(\ln(2))=2 $$ Step 1: $$ y = C_1 e^{2x} + C_2e^{-2x} - 2e^{x}$$ Step 2: $$ \begin{align} y(0) &= C_1 + C_2 - 2 \buildrel{\text{set}}\over{=} 1 \\ y(\ln(2)) &= 4C_1 + \frac{C_2}{4} - 4 \buildrel{\text{set}}\over{=} 2 \end{align}$$ Step 3: Moving all the constants to the right, we get $$\begin{align} C_1 + C_2 &= 3 \\ C_1 + \frac{C_2}{4} &= 6 \end{align}$$ Subtracting the second equation from the first we get $(3/4)C_2 = -3$, which gives us $C_2 = -4$. Plugging this into the first equation we then find $C_1 = 7$ and so the solution to the boundary value problem is $$ y = 7e^{2x} - 4e^{-2x} - 2e^x. $$ Note that the only contribution of the inhomogeneous equation is that we have to move the values of the forcing function to the right when we solve for the constants. Because of that, all the qualitative features of boundary value problems show up in homogeneous equations. So we will concentrate on homogeneous equations to keep the algebra simpler.The next three examples will all feature the same equation, but with different boundary values leading to different numbers of solutions. EXAMPLE 2: One Solution $$ y'' + y = 0, \qquad y(0)=0, y(1)=1 $$ Step 1: $$ y = C_1 \cos(x) + C_2 \sin(x) $$ Step 2: $$\begin{align} y(0) &= C_1 \cos(0) + C_2 \sin(0) \buildrel{\text{set}}\over{=} 0 \\ y(1) &= C_1 \cos(1) + C_2 \sin(1) \buildrel{\text{set}}\over{=} 1 \end{align}$$ Step 3: From the first equation we know $C_1=0$ and then the second equation tells us $C_2=1/\sin(1)$. So the solution to the boundary value problem is $$ y = \frac{\sin(x)}{\sin(1)}. $$ EXAMPLE 3: No Solutions $$ y'' + y = 0, \qquad y(0)=0, y(\pi)=1 $$ Step 1: $$ y = C_1 \cos(x) + C_2 \sin(x) $$ Step 2: $$\begin{align} y(0) &= C_1 \cos(0) + C_2 sin(0) \buildrel{\text{set}}\over{=} 0 \\ y(\pi) &= C_1\cos(\pi) + C_2 \sin(\pi) \buildrel{\text{set}}\over{=} 1 \end{align}$$ Step 3: From the first equation we get $C_1 = 0$, but the second equation requires $C_1 = -1$. This is impossible, so there are no solutions to this boundary value problem. EXAMPLE 4: Infinitely Many Solutions $$ y'' + y = 0, \qquad y(0)=0, y(\pi)=0 $$ Step 1: $$ y = C_1 \cos(x) + C_2 \sin(x) $$ Step 2: $$\begin{align} y(0) &= C_1 \cos(0) + C_2 \sin(0) \buildrel{\text{set}}\over{=} 0 \\ y(\pi) &= C_1 \cos(\pi) + C_2 \sin(\pi) \buildrel{\text{set}}\over{=} 0 \end{align}$$ Step 3: Both the first and second equations turn into $C_1=0$, so definitely we need $C_1=0$. But neither equation says anything about $C_2$, so we can choose $C_2$ to be any value we wish. The solutions to the boundary value problem are all the functions $$ y = C_2 \sin(x). $$
So even for a very simple equation like $y'' + y = 0$, the boundary value problem is not guaranteed to have a unique solution like an initial value problem would. You might get a unique solution, but you might also get no solution or infinitely many solutions. This need not be a disaster, in fact in many applications it is a blessing. If you take Partial Differential Equations, you will often reduce a partial differential equation to a collection of ordinary differential equations, solve the ordinary differential equations, and then add up the solutions in the appropriate way to get the solution to the original partial differential equation. Typically the ordinary differential equations are going to be boundary value problems. Suppose the ordinary differential equation is $L(y)=0$ with boundary values $y(a)=0$ and $y(b)=0$. Then one solution must be $y=0$. If that were the only solution, you wouldn't have anything interesting to add up, just a bunch of zeros. But if you have a situation like Example 4 above, then you can still get interesting functions that enable you to build your solution to the partial differential equation. In these sorts of situations, you may need to choose the parameters of your ordinary differential equation carefully to be sure you get interesting solutions.
EXAMPLE 5: Choosing Boundary Values With Interesting Solutions Find all values of $a$ where the following boundary value problem has non-zero solutions. $$ y'' + 4y = 0, \qquad y(0)=0, y(a)=0 $$ Step 1: $$ y = C_1 \cos(2x) + C_2 \sin(2x) $$ Step 2: $$\begin{align} y(0) &= C_1 \cos(0) + C_2 \sin(0) \buildrel{\text{set}}\over{=} 0 \\ y(a) &= C_1 \cos(a) + C_2 \sin(a) \buildrel{\text{set}}\over{=} 0 \end{align}$$ Step 3: From the first equation we see $C_1=0$. Then the second equation becomes $C_2 \sin(a) = 0$. So either $C_2=0$ or $\sin(a)=0$. If $C_2=0$, then our solution becomes $y = 0 \cos(x) + 0 \sin(x) = 0$. Since we are looking for non-zero solutions, that won't work. But if $\sin(a)=0$, then any function of the form $y = C_2 \sin(x)$ is a solution and we have non-zero solutions. So the condition we need is that $\sin(a)=0$, or $a = n\pi$ for some integer $n$. EXAMPLE 6: Choosing an Equation With Interesting Solutions Find all values of $\lambda$ where the following boundary value problem has non-zero solutions. $$ y'' + \lambda y = 0, \qquad y(0)=0, y(1)=0 $$ We will have to consider three cases depending on the sign of $\lambda$. Case 1: $\lambda<0$. We let $\mu=\sqrt{-\lambda}$. Step 1: $$ y = C_1 \exp\left(\mu x\right) + C_2 \exp\left(-\mu x\right) $$ Step 2: $$\begin{align} y(0)&= C_1 + C_2 \buildrel{\text{set}}\over{=} 0 \\ y(1)&= C_1 \exp\left(\mu\right) + C_2 \exp\left(-\mu\right) \buildrel{\text{set}}\over{=} 0 \end{align} $$ Step 3: Multiplying the second equation by $\exp(\mu)$ and subtracting it from the first equation we get $\left(1-\exp(2\mu)\right)C_1 = 0$. Since $\lambda$ is strictly less than 0, $1-\exp(2\mu) \ne 0$, and so $C_1=0$. Then we must also have $C_2=0$, so there are no non-zero solutions if $\lambda>0$. Case 2: $\lambda=0$. Step 1: $$ y = C_1 x + C_2 $$ Step 2: $$\begin{align} y(0) &= C_2 \buildrel{\text{set}}\over{=} 0 \\ y(1) &= C_1+C_2 \buildrel{\text{set}}\over{=} 0 \end{align} $$ Again, the only solution is $C_1=0$ and $C_2=0$ so there are no non-zero solutions. Case 3: $\lambda>0$. We let $\mu=\sqrt{\lambda}$. Step 1: $$ y = C_1 \cos\left(\mu x\right) + C_2 \sin\left(\mu x\right) $$ Step 2: $$\begin{align} y(0) &= C_1 \cos(0) + C_2 \sin(0) \buildrel{\text{set}}\over{=} 0 \\ y(1) &= C_1 \cos\left(\mu\right) + C_2\sin\left(\mu\right) \buildrel{\text{set}}\over{=} 0 \end{align}$$ Step 3: From the first equation we have $C_1 = 0$, and so the second equation becomes $C_2\sin\left(\mu x\right)=0$. Then either $C_2=0$, which makes our solution $y=0$, or $\sin\left(\mu x\right)=0$, in which case we have infinitely many non-zero solutions $y = C_2\sin\left(\mu x\right)$. That means $\mu = n\pi$ for some integer $n$. So the values of $\lambda$ for which we have non-zero solutions are $\lambda=n^2\pi^2$ for some integer $n$. This last example was rather involved. But if you have to work with partial differential equations (which show up in many applications), it is by far the most important example in this section.
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