General solutions and boundary conditions

Adding solutions

Lesson 6.2.1 found two basic solutions $y_1$ and $y_2$ of each equation $ay'' + by' + cy = 0$. For $y'' + 5y' + 4y = 0$ they are $e^{-t}$ and $e^{-4t}$.

Any combination of them is also a solution. Substitute $y = c_1y_1 + c_2y_2$: each derivative splits the same way, so the left side becomes $c_1$ times what $y_1$ gives plus $c_2$ times what $y_2$ gives.

Both are 0, so the sum is 0. For example, $2e^{-t} - 3e^{-4t}$ solves $y'' + 5y' + 4y = 0$.

In the words of lesson 3.5.1, the solutions are the span of $y_1$ and $y_2$: a set with two dimensions, one for each constant. The pair must be independent. $e^{-t}$ and $3e^{-t}$ are multiples of each other, and their combinations never give $e^{-4t}$.

For each case of lesson 6.2.1, the general solution reads:

Initial conditions: a 2 × 2 system

An initial-value problem gives the starting value and slope, $y(0) = y_0$ and $y'(0) = v_0$. Each condition is one linear equation in $c_1$ and $c_2$.

Example 1. Solve $y'' + 5y' + 4y = 0$ with $y(0) = 1$ and $y'(0) = 0$. The general solution is $y = c_1e^{-t} + c_2e^{-4t}$, with slope $y' = -c_1e^{-t} - 4c_2e^{-4t}$. At $t = 0$:

Adding the two equations gives $-3c_2 = 1$, so $c_2 = -\frac{1}{3}$ and $c_1 = \frac{4}{3}$. The solution is $y = \frac{4}{3}e^{-t} - \frac{1}{3}e^{-4t}$. It starts at 1, flat, and falls towards 0:

In Example 1 the matrix is $\begin{pmatrix} 1 & 1 \\ -1 & -4 \end{pmatrix}$, with determinant $-4 + 1 = -3$ (lesson 3.4.1). That the solution exists and is unique is a theorem about these equations; this lesson takes it as given.

Example 2. Solve $y'' + 2y' + 5y = 0$ with $y(0) = 1$ and $y'(0) = 0$. The general solution is $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$, so $y(0) = c_1 = 1$.

The slope needs the product rule (lesson 2.3.2): at $t = 0$ it is $y'(0) = -c_1 + 2c_2$, where $-c_1$ comes from the factor $e^{-t}$. Setting it to 0 gives $c_2 = \frac{1}{2}$, so

This is the curve lesson 6.2.1's explorer draws for $b = 2$ and $c = 5$. At $t = 1$ it is 0.014164, just above 0, and at $t = 2$ it is $-0.139672$.

Boundary conditions: two points

A boundary-value problem gives conditions at two different points instead, such as $y(0) = A$ and $y(L) = B$. They still give two linear equations in $c_1$ and $c_2$. The difference is that the system can fail.

Example 3: one solution. Solve $y'' - y = 0$ with $y(0) = 0$ and $y(1) = 1$. The roots are $\pm 1$, so $y = c_1e^{t} + c_2e^{-t}$, and the conditions read

So $c_2 = -c_1$ and $c_1(e - e^{-1}) = 1$. Since $e - e^{-1} = 2.350402$, $c_1 = 0.425459$. The solution passes $y(0.5) = 0.443409$ and starts with slope $y'(0) = 0.850918$.

Example 4: none, or infinitely many. Take $y'' + y = 0$ with $y(0) = 0$. The general solution is $c_1\cos t + c_2\sin t$, and $y(0) = 0$ forces $c_1 = 0$. So the candidates are $c_2\sin t$, and the second condition decides.

• $y(\frac{\pi}{2}) = 1$: $c_2\sin\frac{\pi}{2} = c_2 = 1$. One solution, $y = \sin t$.
• $y(\pi) = 1$: $c_2\sin\pi = 0$ for every $c_2$, never 1. No solution.
• $y(\pi) = 0$: every $c_2$ works. Infinitely many solutions, $y = c_2\sin t$.

The 2 × 2 matrix explains it. For $y(0) = A$ and $y(L) = B$ it is $\begin{pmatrix} 1 & 0 \\ \cos L & \sin L \end{pmatrix}$, with determinant $\sin L$. When $L$ is a multiple of $\pi$, the determinant is 0: lesson 3.3.2's rank test then gives none or infinitely many, depending on $B$.

With $B = 1$ the ranks are 1 and 2, so there is no solution. With $B = 0$ both are 1, below the two unknowns, so there are infinitely many.

A condition at infinity

On a half-line $t \ge 0$, the second condition can be about the far end. Solve $y'' - y = 0$ with $y(0) = 1$ and $y$ bounded: it stays between two fixed numbers as $t \to \infty$.

The general solution is $c_1e^{t} + c_2e^{-t}$. Any $c_1 \neq 0$ makes $y$ grow without bound, so $c_1 = 0$, and $y(0) = 1$ gives $c_2 = 1$: the solution is $e^{-t}$.

The equidimensional equation

Some equations have coefficients that change, in a pattern that keeps them easy. In the equidimensional equation

each derivative is multiplied by the power of $x$ that matches its order.

If $x$ is in metres, $y''$ is in units of $y$ per metre squared. So $x^2y''$ is in units of $y$, like $xy'$ and $y$. Every term has the same units: hence the name.

The exponential trial of lesson 6.2.1 fails here, but a power works. For $y = x^p$, the power rule (lesson 2.3.1) gives $xy' = px^p$ and $x^2y'' = p(p - 1)x^p$. Every term carries $x^p$:

Example 5. For $x^2y'' + 2xy' - 2y = 0$, the equation for $p$ is $p(p - 1) + 2p - 2 = p^2 + p - 2 = 0$. A table of trial powers finds its roots:

The algebra agrees: $p^2 + p - 2 = (p - 1)(p + 2)$, so $p = 1$ or $p = -2$. The general solution is $y = c_1x + c_2x^{-2}$.

Now add conditions: $y(1) = 3$, and $y$ bounded as $x \to \infty$. The term $c_1x$ grows, so $c_1 = 0$. Then $y(1) = c_2 = 3$, and $y = 3x^{-2}$, so $y(2) = 0.75$.

Repeated and complex powers

Put $x = e^s$, so that $s = \ln x$. By the chain rule (lesson 2.3.3), the equation becomes a constant-coefficient equation in $s$, with the equation for $p$ as its characteristic equation. Then $x^p = e^{ps}$, and lesson 6.2.1's other two cases carry over:

• A repeated root $p_0$ gives $x^{p_0}$ and $x^{p_0}\ln x$. For $x^2y'' - xy' + y = 0$, $p(p - 1) - p + 1 = (p - 1)^2$, so the solutions are $x$ and $x\ln x$.
• Complex roots $\alpha \pm i\omega$ give $x^{\alpha}\cos(\omega\ln x)$ and $x^{\alpha}\sin(\omega\ln x)$. For $x^2y'' + xy' + y = 0$, $p^2 + 1 = 0$, and the solutions are $\cos(\ln x)$ and $\sin(\ln x)$.

Case study 6.4.4 uses the same change of variable on a harder equation.

Solving a boundary-value problem numerically

solve_ivp needs a starting value and a starting slope. A boundary-value problem gives a starting value and an end value instead. Shooting turns one into the other: guess the slope, solve forward, and see where the end lands.

For Example 3, start at $y(0) = 0$ with slope $s$. A slope of 0.4 lands at $y(1) = 0.47$, and a slope of 1.3 at 1.53: too low, then too high. Between them, brentq (lesson 1.3.2) finds the slope that lands on 1.

SciPy also has a ready-made solver, solve_bvp. It takes the equation as a pair $(y, v)$ with $v = y'$, the two conditions as a function, and a first guess on a grid of points.

Lesson 6.2.3 practises both lessons of this module. Case study 6.2.4 puts the equidimensional equation and a condition at infinity together to value a put.

Common mistakes

1. Two conditions need both constants. $y = c_1e^{-t}$ alone cannot meet $y(0) = 1$ and $y'(0) = 0$: it would need $c_1 = 1$ and $c_1 = 0$.
2. Differentiate the whole combination. For $e^{-t}(c_1\cos 2t + c_2\sin 2t)$, $y'(0) = -c_1 + 2c_2$, not $2c_2$.
3. The pair must be independent. $c_1e^{-t} + c_2\cdot 3e^{-t}$ is only a multiple of $e^{-t}$; its matrix has determinant 0.
4. A boundary-value problem may have no solution, or many. $y'' + y = 0$ with $y(0) = 0$ and $y(\pi) = 1$ has none.
5. Bounded removes the growing part. On $t \ge 0$ it removes $e^{t}$; on $x \ge 1$ it removes positive powers such as $x$.
6. Use $p(p - 1)$, not $p^2$. For $x^2y'' - 6y = 0$ the equation is $p(p - 1) - 6 = 0$, with roots 3 and $-2$, not $p^2 - 6 = 0$.
7. Powers need $x > 0$. $x^{-2}$ and $x\ln x$ are solutions on $x > 0$; at $x = 0$ the equation's leading coefficient $ax^2$ is 0.