Mathematics-III (BT-401) - Unit 3 Short Notes

How unit 3 is examined

This unit solves ordinary differential equations step by step (Taylor, Euler, Runge-Kutta, Milne, Adams) and partial differential equations by finite differences; the marks sit in modified Euler, Runge-Kutta, Taylor and Milne, with one paper each on Poisson and Crank-Nicholson.

Ordinary differential equations: Taylor's series

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Definition. For $y'=f(x,y)$, $y(x_0)=y_0$, Taylor's series method expands the solution about $x_0$ and evaluates it at $x_0+h$.

Formula. $$y(x_0+h)=y_0+hy_0'+\frac{h^2}{2!}y_0''+\frac{h^3}{3!}y_0'''+\cdots$$

Key points.

1. The derivatives $y'',y''',\dots$ are found by differentiating $y'=f(x,y)$ repeatedly and substituting $x_0,y_0$.
2. Each term is smaller than the last because $h$ is small, so the series is cut when the next term no longer changes the required decimal place.
3. The method needs $f$ to be easily differentiable, which is why it is used only for simple equations.
4. Picard's method instead iterates $y_{n+1}=y_0+\int_{x_0}^{x}f(x,y_n)\,dx$ starting from $y_1=y_0+\int f(x,y_0)dx$.

Example. $y'=x^2y-1$, $y(0)=1$, find $y(0.1)$. Then $y''=2xy+x^2y'$, $y'''=2y+4xy'+x^2y''$, $y^{iv}=6y'+6xy''+x^2y'''$.

$y(0.1)=1-0.1+0+\frac{2(0.1)^3}{6}-\frac{6(0.1)^4}{24}=$ 0.90031.

Example. $xy'=x-y$, $y(2)=2$, $y(2.1)$: $y'=(x-y)/x=0$; $xy''=1-2y'$ gives $y''=0.5$; $xy'''=-3y''$ gives $y'''=-0.75$; $xy^{iv}=-4y'''$ gives $y^{iv}=1.5$. So $y(2.1)=2+0+\frac{0.01}{2}(0.5)-\frac{0.001}{6}(0.75)+\frac{0.0001}{24}(1.5)=$ 2.00238.

Example. Picard, $y'=1+xy$, $y(0)=1$: $y_1=1+x+\frac{x^2}{2}$, $y_2=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{8}$, $y_3=y_2+\frac{x^5}{15}+\frac{x^6}{48}$. This gives $y(0.1)=$ 1.105, $y(0.2)=$ 1.223, $y(0.3)=$ 1.355.

<mark>Taylor's method writes $y(x_0+h)$ as a power series in $h$ whose coefficients are the successive derivatives of $y$ at $(x_0,y_0)$.</mark>

Answer frame. Open with the series formula; find $y',y'',y'''$ at the starting point in a small table; substitute $h$ term by term; close with the value rounded as asked. For Picard write the iteration, integrate twice or thrice, then substitute each $x$.

Asked: [7 marks] (Dec 2020, Jun 2022) Find $y(0.1)$ for $\frac{dy}{dx}=x^2y-1$, $y(0)=1$ using Taylor's series method. Asked: [7 marks] (Dec 2020) Using Taylor's series solve $xy'=x-y$, $y(2)=2$ at $x=2.1$, correct to five decimal places. Asked: [7 marks] (May 2019) Use Picard's method to approximate $y$ at $x=0.1,0.2,0.3$, given $y=1$ at $x=0$ and $\frac{dy}{dx}=1+xy$, correct to three decimal places.

Euler and modified Euler's methods

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Definition. Euler's method advances the solution of $y'=f(x,y)$ by moving along the tangent at each point for one step $h$; the modified method corrects that step with the average of the slopes at both ends.

Formula. $$y_{n+1}=y_n+hf(x_n,y_n)\quad\text{(Euler)}$$ $$y_{n+1}^{(0)}=y_n+hf(x_n,y_n),\qquad y_{n+1}^{(k)}=y_n+\frac h2\left[f(x_n,y_n)+f(x_{n+1},y_{n+1}^{(k-1)})\right]$$

Key points.

1. Euler's method is a first-order method, so its error per step is of order $h^2$ and it needs a very small $h$ for good accuracy.
2. It uses the slope only at the left end of each interval, so the error builds up as the tangent leaves the true curve.
3. Modified Euler first predicts $y_{n+1}^{(0)}$ by Euler's formula and then corrects it using the trapezoidal average of two slopes.
4. The corrector is repeated until two successive values agree to the required decimals, and only then is the next step begun.
5. Modified Euler is second order (error per step of order $h^3$), so it is far more accurate than Euler for the same $h$.
6. Every step needs $f(x_n,y_n)$ with the latest $y_n$, so the steps cannot be skipped.

Example. $y'=x+y$, $y(0)=1$, $h=0.1$, find $y(0.3)$ by modified Euler.

Answers of the other papers (same method):

Example (plain Euler). $y'=\frac{y^2-x}{y^2+x}$, $y(0)=1$, $h=0.1$: $f(0,1)=1$ so $y(0.1)=1+0.1(1)=$ 1.1; $f(0.1,1.1)=\frac{1.11}{1.31}=0.84733$ so $y(0.2)=1.1+0.084733=$ 1.18473. For $y'=y^2-x^2$, $y(0)=1$: $y(0.1)=1+0.1(1)=$ 1.1.

==Modified Euler predicts with $y^{(0)}_{n+1}=y_n+hf(x_n,y_n)$ and corrects with $y_{n+1}=y_n+\frac h2[f(x_n,y_n)+f(x_{n+1},y_{n+1}^{(k-1)})]$ until two iterates agree.==

Pitfall: Stopping after one correction, or restarting the predictor from the old $y_n$ instead of the converged value, gives a wrong last decimal.

Answer frame. Open with both formulas and $f(x,y)$, $h$, $x_0$, $y_0$; tabulate predictor and corrector iterates for each step; give the converged value at each $x$; close with a bold answer line. For plain Euler write one line per step: $x_n$, $y_n$, $f$, $y_{n+1}$.

Asked: [7 marks] (May 2019, Jun 2020, Jun 2023) Using modified Euler's method find $y$ at $x=0.3$ given $\frac{dy}{dx}=x+y$, $y(0)=1$; also $h=0.2$, compute $y(1)$. Asked: [7 marks] (Jun 2020) Solve $\frac{dy}{dx}=x+y$ at $x=0.3$ with modified Euler's method, $y(0)=1$, $h=0.1$. Asked: [7 marks] (Nov 2022) Find $y(4.2)$ by Euler's modified method with $h=0.1$ from $\frac{dy}{dx}=\frac{2-y}{4x}$, $y=1$ at $x=4$. Asked: [7 marks] (Nov 2022) Use Euler's method for $\frac{dy}{dx}=\frac{y^2-x}{y^2+x}$, $x=0$, $y=1$; compute $y(0.1)$, $y(0.2)$. Asked: [7 marks] (Nov 2022) Use Euler's modified method for $\frac{dy}{dx}=x+\sqrt y$, $y(2)=4$, $h=0.2$; compute $y(2.4)$. Asked: [7 marks] (Jun 2023) Use Euler's method to compute the solution of $\frac{dy}{dx}=y^2-x^2$, $y(0)=1$ at $x=0.1$.

Runge-Kutta method of fourth order

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Definition. The Runge-Kutta method of fourth order (RK4) finds $y_{n+1}$ from $y_n$ using a weighted average of four slope estimates taken inside the step, without any derivatives of $f$.

Formula. $$k_1=hf(x_n,y_n),\quad k_2=hf\left(x_n+\tfrac h2,y_n+\tfrac{k_1}2\right)$$ $$k_3=hf\left(x_n+\tfrac h2,y_n+\tfrac{k_2}2\right),\quad k_4=hf(x_n+h,y_n+k_3)$$ $$y_{n+1}=y_n+\tfrac16(k_1+2k_2+2k_3+k_4)$$

Key points.

1. RK4 agrees with the Taylor series up to $h^4$, so its error per step is of order $h^5$.
2. $k_1$ is the slope at the start, $k_2$ and $k_3$ are slopes at the midpoint, and $k_4$ is the slope at the end of the step.
3. The two midpoint slopes get weight 2 each, as in Simpson's rule, which gives the $\frac16$ average.
4. It is self-starting, needing only $y_0$, and it is used to supply starting values for Milne and Adams methods.
5. It needs four evaluations of $f$ per step, which is its cost.
6. For $y''=g(x,y,z)$ put $z=y'$ and step $y$ and $z$ together with $k_i$ and $l_i$ as below.

Second order equation. $k_1=hz_n$, $l_1=hg(x_n,y_n,z_n)$; $k_2=h(z_n+\frac{l_1}2)$, $l_2=hg(x_n+\frac h2,y_n+\frac{k_1}2,z_n+\frac{l_1}2)$; $k_3=h(z_n+\frac{l_2}2)$, $l_3=hg(x_n+\frac h2,y_n+\frac{k_2}2,z_n+\frac{l_2}2)$; $k_4=h(z_n+l_3)$, $l_4=hg(x_n+h,y_n+k_3,z_n+l_3)$; then $y_{n+1}=y_n+\frac16(k_1+2k_2+2k_3+k_4)$, $z_{n+1}=z_n+\frac16(l_1+2l_2+2l_3+l_4)$.

Example. $y'=x+y$, $y(0)=1$, $h=0.1$, find $y(0.1)$, $y(0.2)$.

Results: $y(0.1)=$ 1.1103, $y(0.2)=$ 1.2428.

Other papers ($h$ as stated, one step unless shown):

Second order paper. $xy''+y'+xy=0$ is Bessel's equation of order zero. Put $y=\sum a_nx^n$: the $x^0$ term gives $a_1=0$, and $a_n=-a_{n-2}/n^2$, so with $y(0)=1$ the series is $y=1-\frac{x^2}{4}+\frac{x^4}{64}-\frac{x^6}{2304}+\cdots$. It gives $y(0.5)=0.93847$, $y(1)=0.76520$, $y(1.5)=0.51183$.

Pitfall: As printed, $y'(0)=1$ is impossible for this equation, because $x=0$ forces $y'(0)=0$; state this and use $y'(0)=0$, since RK4 also cannot start at the singular point $x=0$.

==$y_{n+1}=y_n+\frac16(k_1+2k_2+2k_3+k_4)$, where $k_1,k_2,k_3,k_4$ are $h$ times the slope at the start, twice at the midpoint, and at the end.==

Answer frame. Open with $f(x,y)$, $x_0$, $y_0$, $h$ and the four $k$ formulas; compute $k_1$ to $k_4$ in a table; average and add to $y_0$; restart from the new $(x,y)$ for the next step; close with a bold answer line. For a second order equation write $z=y'$ first.

Asked: [7 marks] (Dec 2020, Jun 2020, Jun 2022, Nov 2022, Jun 2026) Use the Runge-Kutta method to approximate $y$ at $x=0.1$ and $0.2$ given $y=1$ at $x=0$ and $\frac{dy}{dx}=x+y$; also $y(1.05)$ from $\frac{dy}{dx}=x^2+y^2$, $y(1)=1.2$, $h=0.05$; also $y(0.1)$ from $\frac{dy}{dx}=x^2+y$, $y(0)=-1$. Asked: [7 marks] (Dec 2020) Solve $y'=x^2-y$, $y(0)=1$ at $x=0.1,0.2$ by fourth-order Runge-Kutta. Asked: [7 marks] (Jun 2020) Apply Runge-Kutta to $10\frac{dy}{dx}=x^2+y^2$, $y(0)=1$ for $x=0.1$. Asked: [7 marks] (Jun 2022) Solve $xy''+y'+xy=0$, $y(0)=1$, $y'(0)=1$, for $x=0$ to $x=1.5$.

Milne's predictor-corrector methods

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Definition. Milne's method is a multistep predictor-corrector method that predicts $y_{n+1}$ from four earlier points and corrects it with Simpson's rule.

Formula. $$y_{n+1}^{p}=y_{n-3}+\frac{4h}{3}\left(2f_{n-2}-f_{n-1}+2f_n\right)$$ $$y_{n+1}^{c}=y_{n-1}+\frac h3\left(f_{n-1}+4f_n+f_{n+1}\right)$$

Key points.

1. The predictor is an open formula and the corrector is Simpson's rule, and the corrector is repeated until $y^c_{n+1}$ stops changing.
2. Four starting values $y_{n-3},\dots,y_n$ are required, found by Taylor, Picard or Runge-Kutta.
3. Only one new value of $f$ is needed per correction, so it is cheaper than RK4 per step.
4. It is not self-starting, and it can become unstable for some equations.

Example. $y'=x+y$, $y(0)=1$, $h=0.1$, find $y(0.3)$. RK4 starting values: $y(-0.1)=0.909675$, $y(0)=1$, $y(0.1)=1.110342$, $y(0.2)=1.242805$, so $f=0.809675,\ 1,\ 1.210342,\ 1.442805$.

Predictor: $y^p(0.3)=0.909675+\frac{0.4}{3}(2(1)-1.210342+2(1.442805))=$ 1.39971. Corrector: $f(0.3,1.39971)=1.69971$, so $y^c=1.110342+\frac{0.1}3(1.210342+4(1.442805)+1.69971)=$ 1.39972, unchanged on repeating: $y(0.3)=$ 1.3997.

==Milne's predictor $y_{n+1}=y_{n-3}+\frac{4h}3(2f_{n-2}-f_{n-1}+2f_n)$ is corrected by Simpson's rule $y_{n+1}=y_{n-1}+\frac h3(f_{n-1}+4f_n+f_{n+1})$.==

Answer frame. Open with both formulas; list the four starting values and their $f$ in a table; compute the predictor, then the corrector until it repeats; close with the bold value.

Asked: [7 marks] (Jun 2022, Jun 2026) Use Milne's method to solve $\frac{dy}{dx}=x+y$, $y(0)=1$, from $x=0.20$ to $x=0.30$.

Adams' predictor-corrector methods

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Definition. The Adams-Bashforth predictor and Adams-Moulton corrector are multistep formulas that use four earlier slopes.

Formula. $$y_{n+1}^{p}=y_n+\frac h{24}(55f_n-59f_{n-1}+37f_{n-2}-9f_{n-3})$$ $$y_{n+1}^{c}=y_n+\frac h{24}(9f_{n+1}+19f_n-5f_{n-1}+f_{n-2})$$

Key points.

1. Four starting values are needed, taken from Runge-Kutta.
2. The corrector is repeated until it converges.
3. Both formulas are of fourth order, error of order $h^5$.

Partial differential equations: finite difference solution of the two dimensional Laplace equation

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Definition. Laplace's equation $u_{xx}+u_{yy}=0$ is solved by replacing the derivatives with central differences on a square grid of side $h$.

Formula. $$u_{i,j}=\tfrac14\left(u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}\right)$$

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Key points.

1. Each interior value is the average of its four neighbours, and boundary values are given.
2. The unknowns are found by solving the linear system, or by Liebmann's iteration starting from a guess.
3. The diagonal formula $u=\frac14(\text{four diagonal neighbours})$ gives a first guess.

Poisson equation

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Definition. Poisson's equation $\nabla^2u=f(x,y)$ is Laplace's equation with a source term, with the same five-point formula.

Formula. $$u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}-4u_{i,j}=h^2f(x_i,y_j)$$

Key points.

1. The right side is $h^2f$ at the node, so the mesh needs $f$ at every interior point.
2. Each interior node gives one linear equation, and boundary values move to the right side.
3. For $h=\frac13$ there are four interior nodes $u_1,u_2$ (bottom, $y=\frac13$) and $u_3,u_4$ (top, $y=\frac23$).

Example. $\nabla^2u=-81xy$, $h=\frac13$, so $h^2f=-9xy$, with $u=0$ on $x=0,y=0$ and $u=100$ on $x=1,y=1$.

By symmetry $u_2=u_3$, so $u_1=25.79$, $u_2=u_3=51.08$, $u_4=76.54$: $u_1=25.79,\ u_2=u_3=51.08,\ u_4=76.54$.

Asked: [7 marks] (Jun 2026) Solve $\nabla^2u=-81xy$, $0<x,y<1$, $h=\frac13$, with $u(0,y)=u(x,0)=0$, $u(1,y)=u(x,1)=100$.

Implicit and explicit methods for the one dimensional heat equation (Bender-Schmidt and Crank-Nicholson)

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Definition. The heat equation $u_t=c^2u_{xx}$ is solved on a grid with step $h$ in $x$ and $k$ in $t$, with $\lambda=\frac{c^2k}{h^2}$.

Formula. $$u_{i,j+1}=\lambda u_{i-1,j}+(1-2\lambda)u_{i,j}+\lambda u_{i+1,j}\quad\text{(explicit)}$$

Key points.

1. The explicit method is stable only for $\lambda\le\frac12$, and Bender-Schmidt takes $\lambda=\frac12$: $u_{i,j+1}=\frac12(u_{i-1,j}+u_{i+1,j})$.
2. The implicit Crank-Nicholson method is stable for every $\lambda$ but needs a linear system at each time level.
3. The explicit method gives each new value directly, so it is simpler.

Crank-Nicholson methods

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Definition. Crank-Nicholson averages the space differences at levels $j$ and $j+1$, giving a tridiagonal system for each new time level.

Formula. $$-\lambda u_{i-1,j+1}+2(1+\lambda)u_{i,j+1}-\lambda u_{i+1,j+1}=\lambda u_{i-1,j}+2(1-\lambda)u_{i,j}+\lambda u_{i+1,j}$$

Key points.

1. It is of second order in $h$ and $k$ and is stable for all $\lambda$.
2. The paper gives $h=1$ but no $k$, so choose $\lambda=1$ ($k=1$), which turns the formula into $-u_{i-1,j+1}+4u_{i,j+1}-u_{i+1,j+1}=u_{i-1,j}+u_{i+1,j}$.
3. Boundary values are added to the right side of the first and last equations at both levels.

Example. $u_t=u_{xx}$, $h=1$, $u(x,0)=20$, $u(0,t)=0$, $u(5,t)=100$, nodes $x=1,2,3,4$. Level 1: $4u_1-u_2=20$, $-u_1+4u_2-u_3=40$, $-u_2+4u_3-u_4=40$, $-u_3+4u_4=220$, giving $u=(10.05,\ 20.19,\ 30.72,\ 62.68)$. Level 2 uses these as the right side: $u=(11.03,\ 23.91,\ 43.86,\ 68.64)$.

Asked: [7 marks] (Jun 2026) Solve $\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}$ in $0<x<5$ by Crank-Nicholson, $u(x,0)=20$, $u(0,t)=0$, $u(5,t)=100$, $h=1$.

Finite difference explicit method for the wave equation

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Definition. The wave equation $u_{tt}=c^2u_{xx}$ is solved by central differences in both $x$ and $t$, with $r=\frac{ck}{h}$.

Formula. $$u_{i,j+1}=2(1-r^2)u_{i,j}+r^2(u_{i+1,j}+u_{i-1,j})-u_{i,j-1}$$

Key points.

1. The scheme is stable for $r\le1$, and $r=1$ reduces it to $u_{i,j+1}=u_{i+1,j}+u_{i-1,j}-u_{i,j-1}$.
2. The first level comes from $u_t(x,0)=0$: $u_{i,1}=\frac12(u_{i+1,0}+u_{i-1,0})$ when $r=1$.

Last-minute revision

• Euler: $y_{n+1}=y_n+hf(x_n,y_n)$; modified Euler corrects with $\frac h2[f_n+f_{n+1}]$ until convergence.
• $y'=x+y$, $y(0)=1$: modified Euler $y(0.3)=1.4004$ ($h=0.1$), $y(1)=3.4548$ ($h=0.2$).
• RK4: $y_{n+1}=y_n+\frac16(k_1+2k_2+2k_3+k_4)$; $y'=x+y$ gives $y(0.1)=1.1103$, $y(0.2)=1.2428$.
• Taylor: $y(x_0+h)=y_0+hy_0'+\frac{h^2}2y_0''+\cdots$; $y'=x^2y-1$ gives $y(0.1)=0.90031$.
• Picard: $y_{n+1}=y_0+\int f(x,y_n)dx$; $y'=1+xy$ gives 1.105, 1.223, 1.355.
• Milne predictor $y_{n-3}+\frac{4h}3(2f_{n-2}-f_{n-1}+2f_n)$, corrector is Simpson; $y'=x+y$ gives $y(0.3)=1.3997$.
• Adams predictor uses $55,-59,37,-9$ over 24; corrector $9,19,-5,1$ over 24.
• Laplace: $u_{i,j}$ is the average of four neighbours; Poisson: neighbours $-4u=h^2f$.
• Poisson paper: $u_1=25.79$, $u_2=u_3=51.08$, $u_4=76.54$.
• Heat explicit stable for $\lambda\le\frac12$; Crank-Nicholson stable for all $\lambda$, tridiagonal system.
• Wave explicit stable for $r=\frac{ck}h\le1$.

Memory hooks

• Euler = tangent step; modified Euler = tangent, then average the two slopes.
• RK4 weights $1,2,2,1$ over 6, like Simpson's rule.
• Milne: 4 old points, Simpson corrects; Adams: 55, 59, 37, 9 then 9, 19, 5, 1.
• Poisson is Laplace with a source: $h^2f$ on the right.
• Explicit needs $\lambda\le\frac12$, Crank-Nicholson has no limit.

Coverage checklist

• Ordinary differential equations: Taylor's series: Dec 2020 and Jun 2022 ($y'=x^2y-1$), Dec 2020 ($xy'=x-y$), Picard May 2019.
• Euler and modified Euler's methods: modified Euler ($x+y$, $\frac{2-y}{4x}$, $x+\sqrt y$), Euler ($\frac{y^2-x}{y^2+x}$, $y^2-x^2$).
• Runge-Kutta method of fourth order for solving first and second order equations: five first-order papers and $xy''+y'+xy=0$.
• Milne's predicator-corrector methods: Jun 2022, Jun 2026.
• Adam's predicator-corrector methods: formulas only, unasked.
• Partial differential equations: Finite difference solution two dimensional Laplace equation: formula only, unasked.
• Poission equation: Jun 2026.
• Implicit and explicit methods for one dimensional heat equation (Bender-Schmidt and Crank-Nicholson methods): formulas only, unasked.
• Crank-Nicholson methods: Jun 2026.
• Finite difference explicit method for wave equation: formula only, unasked.