
## 4.5.7 Quasi-linearization

n previous examples, we have always linearized the discretization of the original problem. However, it is possible to first linearize the differential equation and only successively discretize it. This method is normally called quasi-linearization. Let us now look at an example given by a second order non linear equation: $$$$\tag{4.128} y''(x) = f(x,y,y')$$$$

We now write (4.128) as: $$$$\tag{4.129} g(x,y,y',y'') \equiv y''(x) - f(x,y,y') = 0$$$$

and setting $$$$\tag{4.130} \delta y = y_{m+1} - y_m \ , \ \delta y' = y'_{m+1} -y'_m\ , \ \delta y'' = y''_{m+1} - y''_m$$$$

where $$m$$ stands for the iteration number (in this section we will be using sub-scripts for iteration number)

By expanding (4.129) around iteration $$m$$ we get: $$$$\tag{4.131} g(x,y_{m+1},y'_{m+1},y{''}_{m+1}) \approx g(x,y_m,y'_m,y{''}_m) + \left( \dfrac{\partial g}{\partial y}\right)_m \delta y + \left( \dfrac{\partial g}{\partial y'}\right)_m \delta y' + \left( \dfrac{\partial g}{\partial y^{''}}\right)_m \delta y{''}\\$$$$

Suppose that we have iterated so many times that: $$\begin{equation*} g(x,y_{m+1},y'_{m+1},y^{''}_{m+1}) \approx g(x,y_m,y'_m,y^{''}_m)\approx 0 \end{equation*}$$

which inserted in (4.131) gives: $$$$\tag{4.132} \left( \frac{\partial g}{\partial y}\right)_m \delta y + \left( \frac{\partial g}{\partial y'}\right)_m \delta y' + \left( \frac{\partial g}{\partial y^{''}}\right)_m \delta y^{''} = 0$$$$

After derivation of (4.129): $$$$\tag{4.133} \frac{\partial g}{\partial y} = -\frac{\partial f}{\partial y} \ , \ \frac{\partial g}{\partial y'} = -\frac{\partial f}{\partial y'} \ , \ \frac{\partial g}{\partial y^{''}} = 1$$$$

which inserted in (4.132) gives: $$$$\tag{4.134} \delta y'' = \left(\frac{\partial f}{\partial y}\right)_m\delta y + \left(\frac{\partial f}{\partial y'}\right)_m\delta y'$$$$

By inserting (4.130) in (4.134) and making use of (4.128), we get: $$$$\tag{4.135} \begin{array}{c} y''_{m+1}- \left(\dfrac{\partial f}{\partial y'}\right)_m \; y'_{m+1} - \left(\dfrac{\partial f}{\partial y}\right)_m \; y_{m+1}\\\\ = f(x,y_m,y'_m)-\left(\dfrac{\partial f}{\partial y}\right)_m \; y_m - \left(\dfrac{\partial f}{\partial y'}\right)_m \; y'_m \end{array}$$$$

Finally we re-write (4.135) with our regular notation, i.e. using super-scripts for iteration number: $$$$\tag{4.136} \begin{array}{c} (y'')^{m+1}- \left(\dfrac{\partial f}{\partial y'}\right)_m \; (y')^{m+1} - \left(\dfrac{\partial f}{\partial y}\right)_m \; y^{m+1}\\\\ = f(x,y^m,(y')^m)-\left(\dfrac{\partial f}{\partial y'}\right)_m \; (y')^m - \left(\dfrac{\partial f}{\partial y}\right)_m \; y^m \end{array}$$$$

We have used a second order equation as an example, but (4.134) - (4.136) allows for generalization to n-th order equations. Take for example a third order equation: $$$$\tag{4.137} \begin{array}{c} (y^{'''})^{m+1} - \left(\dfrac{\partial f}{\partial y''}\right) \; (y'')^{m+1} - \left(\dfrac{\partial f}{\partial y'}\right)_m \; (y')^{m+1} - \left(\dfrac{\partial f}{\partial y}\right)_m \; y^{m+1} \\\\ = f(x,y^m,(y')^m,(y'')^m)-\left(\dfrac{\partial f}{\partial y''}\right)_m \; (y'')^{m}\\\\ - \left(\dfrac{\partial f}{\partial y'}\right)_m \; (y')^{m}-\left(\dfrac{\partial f}{\partial y}\right)_m \; (y)^{m} \end{array}$$$$

## 4.5.8 Example:

1)

Consider the following equation $$\begin{equation*} y''(x) = \frac{3}{2}y^2 \end{equation*}$$

One can easily check that $$\frac{\partial f}{\partial y'} = 0$$ and $$\frac{\partial f}{\partial y} = 3y$$. Inserting these terms in (4.136) gives: $$\begin{equation*} (y'')^{m+1} - 3y^m \; y^{m+1} = -\frac{3}{2}(y^m)^2 \end{equation*}$$

By discretizing with central differences $$y_i'' \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2}$$ we get the following algebraic relation: $$\begin{equation*} y_{i-1}^{m+1} - (2+3h^2y_i^m)y_i^{m+1} + y_{i+1}^{m+1} = -\frac{3}{2}(hy_i^m)^2 \end{equation*}$$

which is in agreement with (4.102).

2)

Falkner-Skan equation reads: $$\begin{equation*} y'' + y \; y'' + \beta \; [1-(y')^2] = 0 \end{equation*}$$

We re-write the equation as: $$\begin{equation*} y'' = -\left(y\; y'' + \beta \; [1 - (y')^2]\right) = f(x,y,y',y'') \end{equation*}$$

Noting that in this case $$\frac{\partial f}{\partial y''} = -y$$ , $$\frac{\partial f}{\partial y'}=2\beta y'$$ and $$\frac{\partial f}{\partial y} = -y''$$, equation (4.137) gives: $$\begin{equation*} \begin{array}{c} (y''')^{m+1} + y^m \; (y'')^{m+1} -2\beta(y')^m \; (y')^{m+1} + (y'')^m \; y^{m+1}\\ = -\beta \left( 1 + \left[(y')^m\right]^2\right) + y^m \; (y'')^m \end{array} \end{equation*}$$

Using central differences one can verify that the result is in agreement with equation (F.O.16) in Appendix F of the Numeriske Beregninger.