The superposition principle

Let $ (n, m) \in \hspace{0.05em} \mathbb{N}^2 $ be two natural numbers and:

- a series of continuous functions $ a_1(x), a_2(x), \hspace{0.2em} ... \hspace{0.2em}, a_n(x) $
- a series of functions $ f_1(x), f_2(x), \hspace{0.2em} ... \hspace{0.2em}, f_m(x) $

Let $ y(x) $ be a function of class $ \mathbb{C}^{n}$ on an interval $I$. We note $y^{(n)}$ its $n$-th derivative.

In the context of solving a linear differential equation of order $n$ where the right hand side is a linear combination of functions such that $ (E)$:

$$ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = \lambda_1 f_1(x) + \lambda_2 f_2(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \lambda_m f_m(x) \qquad (E) $$

with for all $ k \in [\![ 1, m ]\!] $, the function $ (y_k) $ as a specific solution of the equation $ (E_k) $ :

$$ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_k(x) \qquad (\tilde E_k) \\ $$

The superposition principle tells us that:

$$ \forall k \in [\![ 1, m ]\!], $$

$$ y_k \enspace \underline{specific \ solution} \ of \ (\tilde E_k) \Longleftrightarrow (\lambda_1 y_1 + \lambda_2 y_2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \lambda_m y_m) \ \underline{total \ specific \ solution} \ of \ (E) $$

Demonstration

Let $ (n, m) \in \hspace{0.05em} \mathbb{N}^2 $ be two natural numbers and:

- a series of continuous functions $ a_1(x), a_2(x), \hspace{0.2em} ... \hspace{0.2em}, a_n(x) $
- a series of functions $ f_1(x), f_2(x), \hspace{0.2em} ... \hspace{0.2em}, f_m(x) $

Let $ y(x) $ be a function of class $ \mathbb{C}^{n}$ on an interval $I$. We note $y^{(n)}$ its $n$-th derivative.

We start from the equation $ (E) $, linear differential of order $ n$ where the right hand side is a linear combination of functions.

Specific solution for each function $ f_k(x)$

For all $ k \in [\![ 1, m ]\!] $, we then have a series of equations $ (\tilde E_k) $ to solve:

$$ \left \{ \begin{gather*} a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_1(x) \qquad (\tilde E_1) \\ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_2(x) \qquad (\tilde E_2) \\ ........................ \\ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_k(x) \qquad (\tilde E_k) \\ ........................ \\ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_m(x) \qquad (\tilde E_m) \\ \end{gather*} \right \}$$

In this series of equations, we then notice that:

$$ \forall k \in [\![ 1, m ]\!], \enspace y_k \enspace solution \enspace of \enspace (\tilde E_k) $$

If each $ y_k $ is a solution for $ (\tilde E_k) $, then each $ y_k $ verifies:

$$ a_n(x) y_k^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y_k'(x) + a_0(x) y_k(x) = f_k(x) \qquad (\tilde E_k)$$

Thereby, by multiplying $ (\tilde E_k) $ by $ \lambda_k $:

$$ a_n(x) \lambda_k y_k^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) \lambda_k y_k'(x) + a_0(x) \lambda_k y_k(x) = \lambda_k f_k(x) \qquad ( \lambda_k \tilde E_k)$$

Now, thanks to linearity of the derivative, we know that:

$$ \bigl( \lambda f \bigr)' = \lambda f' $$

So in our case,

$$ \forall k \in [\![ 1, m]\!], \enspace \Bigl( \lambda_k f_k \Bigl)' = \lambda_k f'_k \qquad (1) $$

Thanks to$ (1) $, we can rearrange it and see that:

$$ a_n(x) (\lambda_k y_k)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_k y_k \bigr)'(x) + a_0(x) \bigl(\lambda_k y_k\bigr)(x) = \lambda_k f_k(x) \qquad ( \lambda_k \tilde E_k)$$

This equation shows that:

$$ \forall k \in [\![ 1, m]\!], \enspace \lambda_k y_k \enspace solution \enspace of \enspace (\lambda_k \tilde E_k) $$

Total specific solution from the aggregation of functions $ \lambda_k f_k $

Previously, we were able to see that each $ \lambda_k y_k $ is solution for $ (\lambda_k \tilde E_k) $:

$$ \left \{ \begin{gather*} a_n(x) (\lambda_1 y_1)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_1 y_1 \bigr)(x)' + a_0(x) \bigl(\lambda_1 y_1\bigr)(x) = \lambda_1 f_1(x) \qquad (\lambda_1 \tilde E_1) \\ a_n(x) (\lambda_2 y_2)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_2 y_2\bigr)(x) ' + a_0(x) \bigl(\lambda_2 y_2\bigr)(x) = \lambda_2 f_2(x) \qquad (\lambda_2 \tilde E_2) \\ ........................ \\ a_n(x) (\lambda_k y_k)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_k y_k \bigr)(x)' + a_0(x) \bigl(\lambda_k y_k \bigr)(x) = \lambda_k f_k(x) \qquad (\lambda_k \tilde E_k) \\ ........................ \\ a_n(x) (\lambda_m y_m)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_m y_m\bigr)(x) ' + a_0(x) \bigl(\lambda_m y_m \bigr)(x) = \lambda_m f_m(x) \qquad (\lambda_m \tilde E_m) \\ \end{gather*} \right \} $$

By adding up the $ (\lambda_k \tilde E_k) $ from $1 $ to $m $:

$$ a_n(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k^{(n)} \Biggr]} _\text{ $ y_s^{(n)} $} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k' \Biggr]} _\text{ $ y_s' $} \hspace{0.2em} + \hspace{0.2em} a_0(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k \Biggr]} _\text{ $ y_s$} \hspace{0.2em} = \hspace{0.2em} \sum_{k=1}^m \lambda_k f_k \qquad (E-bis) $$

Thus, a total specific solution $ y_s $ which will be the addition of all the specific solutions $ \lambda_k y_k $:

$$ y_s(x) = \lambda_1 y_1(x) + \lambda_2 y_2(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \lambda_m y_m(x) $$

In the end, thanks to $ (E-bis) $, we see that this function is indeed a solution of $ ( E) $:

$$ a_n(x) y_s^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) y_s'(x) + a_0(x) y_s(x) = \lambda_1 f_1(x) + \lambda_2 f_2(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em}+ \hspace{0.2em} \lambda_m f_m(x) \qquad (E) $$

And as a result,

$$ \forall k \in [\![ 1, m ]\!], $$

Example of superposition of solutions

Let $ (E) $ be a first order linear differential equation $ LDE_1 $ with a constant coefficient $ (H) $ its associated homogeneous equation:

$$ \Biggl \{ \begin{align*} y'(x) + 2 y(x) = 2x^2 + 3cos(x) + 1 \qquad (E) \\ y'(x) + 2y(x) = 0 \qquad (H) \end{align*} $$

We then have a series of equations $ (\tilde E_1), (\tilde E_2), (\tilde E_3) $ to solve:

$$ \left \{ \begin{gather*} y'(x) + 2 y(x) = x^2 \qquad \qquad (\tilde E_1) \\ y'(x) + 2 y(x) = cos(x) \qquad (\tilde E_2) \\ y'(x) + 2 y(x) = 1 \qquad \qquad (\tilde E_3) \\ \end{gather*} \right \} $$

Solving the first equation $ (\tilde E_1)$

This equation was solved in the example of solving $ LDE_1 $ with constant coefficient.

The specific solution $ y_1 $ of $ (\tilde E_1) $ is:

$$ y_1(x)= \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} $$

Solving the second equation $ (\tilde E_2)$

We have a solution to the homogeneous equation $ (H) $ (see the example of solving $ EDL_1 $ with constant coefficient):

$$ y_h(x) = Ke^{-2x} $$

We then look for a particular solution $ y_s $ of type:

$$ y_2(x) = K(x) e^{-2x} \qquad (y_2) $$

After having performed a variation of parameters , we seek to determine the function $ K(x) $ such as:

$$ K'(x) = cos(x)e^{2x} $$

Now, taking its antiderivative, we do have:

$$ K(x) = \int^x cos(t)e^{2t} \hspace{0.2em} dt$$

We perform an integration by parts with the following choice for $ u $ and $ v' $:

$$ \Biggl \{ \begin{align*} u(t) = cos(t) \\ v'(t) = e^{2t } \hspace{0.1em} dt \end{align*} $$

$$ \Biggl \{ \begin{align*} u'(t) = -sin(t) dt \\ v(t) = \frac{1}{2} e^{2t } \end{align*} $$

$$K(x) = \frac{1}{2} \Biggl[cos(t) e^{2t }\Biggr]^x - \int^x \frac{1}{2} (-sin(t)) \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$

$$K(x) = \frac{1}{2} \Biggl[cos(t)e^{2t }\Biggr]^x + \int^x \frac{1}{2} sin(t) \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$

$$K(x) = \frac{1}{2} \Biggl[cos(t) e^{2t }\Biggr]^x + \frac{1}{2} \Biggl( \frac{1}{2} \Biggl[sin(t) e^{2t }\Biggr]^x - \frac{1}{2}\int^x cos(t) e^{2t} \hspace{0.2em} dt \Biggr) $$

We note that $ K(x) $ reappears, so we replace it:

$$K(x) = \frac{1}{2} \Biggl[cos(t) e^{2t }\Biggr]^x + \frac{1}{4} \Biggl[sin(t) e^{2t }\Biggr]^x - \frac{1}{4} K(x) $$

$$ \frac{5}{4} K(x) = \frac{1}{2} cos(x)e^{2x } + \frac{1}{4} sin(x) e^{2x } $$

$$ K(x) =\frac{8}{20} cos(x) e^{2x } + \frac{4}{20} sin(x) e^{2x } $$

$$ K(x) = e^{2x } \biggl( \frac{2}{5}cos(x) + \frac{1}{5} sin(x) \biggr) \qquad (K) $$

Injecting $ K $ into $ y_s $, the exponentials annihilate:

$$ y_2(x) = e^{2x } \biggl( \frac{2}{5} cos(x) + \frac{1}{5} sin(x) \biggr) e^{-2x} $$

The specific solution $ y_2 $ of $ (\tilde E_2) $ is:

$$ y_2(x)= \frac{2 }{5}cos(x) +\frac{1}{5}sin(x) $$

Solving the third equation $ (\tilde E_3)$

$$ y'(x) + 2 y(x) = 1 \qquad \qquad (\tilde E_3) $$

Here, $\frac{1 }{2} $ is an obvious solution.

Thus, the specific solution $ y_3 $ of $ (\tilde E_3) $ is:

$$ y_3(x)= \frac{1 }{2} $$

Superposition of solutions$: \sum \lambda_k y_k $

We saw in the demonstration above that:

$$ \forall k \in [\![ 1, m ]\!], $$

$$ y_k \ \underline{specific \ solution} \ of \ (\tilde E_k) \Longleftrightarrow (\lambda_1 y_1 + \lambda_2 y_2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \lambda_m y_m) \ \underline{total \ specific \ solution} \ of \ (E) $$

So in our case:

$$ 2 y_1(x) + 3 y_2(x) + y_3(x) \ \underline{total \ specific \ solution} \ of \ ( E) $$

Let us then calculate this particular total solution $y_s$:

$$ y_s(x) = 2 y_1(x) + 3 y_2(x) + y_3(x) $$

$$ y_s(x) = 2 \Biggl( \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} \Biggr) + 3 \Biggl(\frac{2 }{5}cos(x) +\frac{1}{5}sin(x) \Biggr) + \frac{1}{2} $$

$$ y_s(x) = x^2 - x + \frac{1}{2} + \frac{6 }{5}cos(x) +\frac{3}{5}sin(x) + \frac{1}{2} $$

$$ y_s(x) = x^2 - x + 1 + \frac{6 }{5}cos(x) +\frac{3}{5}sin(x) $$

Verification of the total specific solution $ y_s$

If $ y_s $ is solution for $ (E) $, then:

$$ y_s'(x) + 2 y_s(x) = 2x^2 + 3cos(x) + 1 $$

Let us check it.

$$ y_s'(x) + 2 y_s(x) = \Biggl( 2x -1 -\frac{6 }{5} sin(x) +\frac{3}{5} cos(x)\Biggr) + 2\Biggl( x^2 - x + 1 + \frac{6 }{5}cos(x) +\frac{3}{5}sin(x) \Biggr) $$

$$ y_s'(x) + 2 y_s(x) = 2x -1 -\frac{6 }{5} sin(x) +\frac{3}{5} cos(x) + 2x^2 - 2x +2 + \frac{12 }{5}cos(x) + +\frac{6}{5}sin(x)$$

By tidying up a little:

$$ y_s'(x) + 2 y_s(x) = 2x^2 + \hspace{0.2em} \underbrace{ 2x - 2x} _\text{$=0$} \hspace{0.2em} + \hspace{0.2em} 2 -1 + \hspace{0.2em} \underbrace{\frac{6}{5}sin(x) -\frac{6 }{5} sin(x)} _\text{$=0$} \hspace{0.2em} + \frac{12 }{5}cos(x) +\frac{3}{5} cos(x) $$

$$ y_s'(x) + 2 y_s(x) = 2x^2 +1 + \frac{15}{5}cos(x) $$

$$ y_s'(x) + 2 y_s(x) = 2x^2 + 3cos(x) +1 $$

We definitely verified that $ y_s $ was a solution for $ (E) $.

The superposition principle

Demonstration

Specific solution for each function \( f_k(x)\)

Total specific solution from the aggregation of functions \( \lambda_k f_k \)

Example of superposition of solutions

Solving the first equation \( (\tilde E_1)\)

Solving the second equation \( (\tilde E_2)\)

Solving the third equation \( (\tilde E_3)\)

Superposition of solutions\(: \sum \lambda_k y_k \)

Verification of the total specific solution \( y_s\)