About Solving nonlinear ODE

I want to solve under nonlinear ODE.

A1*diff(y,t) - A2*(2*g*y + diff(y,t)^2)^(1/2) = K1

and constants are A1 = 78, A2 = 0.0081, g = 9.8, K1 = 0.025 and initial condition is y(0) == 9

I coded

syms y(t)
ode = A1*diff(y,t) - A2*(2*g*y + diff(y,t)^2)^(1/2) == K1 ;
cond = y(0) == 9 ;
ySol(t) = dsolve (ode, cond)

but mathlab couldn't solve this problem, so I transformed equation,

((A1^2) - (A2^2))*(diff(y,t))^2 - 2*A1*K1*diff(y,t) - (A2^2)*2*g*y + K1^2 =0

that I shifted K1 to the left side and A2*(2*g*y + diff(y,t)^2)^(1/2) to the right side, and squared them.

then I coded

syms y(t)
ode = ((A1^2) - (A2^2))*(diff(y,t))^2 - 2*A1*K1*diff(y,t) - (A2^2)*2*g*y + K1^2 == 0 ;
cond = y(0) == 9 ;
ySol(t) = dsolve (ode, cond)

matlab solved the problem, but solution was not that i wanted. (solution equation included lambert function etc)

I thought the solution would be quadratic function.

Help me, please and thak you for your answer.

[darova]

quadratic equation of this type has 2 roots:

Maybe you need to solve it using discriminant and choose some root

Answer 1

[Animesh]

Hey @WooJin Choi

Seems like you are trying to solve nonlinear ODE given by “A1*diff(y,t) - A2*(2*g*y + diff(y,t)^2)^(1/2) = K1” for a quadratic solution.

The presence of Lambert function in the solution indicates that the solution to this equation is more complex than a simple quadratic equation.

In the given ODE, the term “dy/dt” appears both linearly and inside a square root, and moreover, it is not the only term inside the square root. The presence of “y” inside the square root along with “dy/dt” significantly complicates the solution. This means that we cannot move the terms around to obtain a quadratic equation in “dy/dt”.

Therefore, the nonlinearity and the structure of the ODE prevent us from reducing it to a quadratic form that can be solved analytically using the quadratic formula.

Although we can consider numerical methods to approximate the solution for the given ODE. To solve this equation numerically, we need to reformulate it so that it can be handled by MATLAB's numerical solvers, which require the derivative to be isolated on one side of the equation. Since we cannot isolate “dy/dt” due to the square root, we can instead use a root-finding approach at each time step to determine “dy/dt”

Here is a MATLAB script that uses “fsolve” to find “dy/dt” and then employs a simple numerical integration technique to approximate the solution over a time span:

function dydt = solveForDerivative(t, y, A1, A2, g, K1, dydt_guess) 
% Define the function for which we want to find the root 
func = @(dydt) A1*dydt - A2*sqrt(2*g*y + dydt^2) - K1; 
% Use a root-finding method like fsolve to find dydt 
options = optimoptions('fsolve', 'Display', 'none'); % Suppress fsolve output 
dydt = fsolve(func, dydt_guess, options); 
end 

A1 = 78; 
A2 = 0.0081; 
g = 9.8; 
K1 = 0.025; 
% Define initial conditions and time span 
y0 = 9; 
dydt_guess = 0; % Initial guess for dy/dt 
% Define time points where we want the solution 
t_points = linspace(0, 10, 100); % 100 time points from 0 to 10 
y_values = zeros(size(t_points)); 
y_values(1) = y0; % Initial condition 
% Perform a custom numerical integration 
for i = 2:length(t_points) 
    % Current time 
    t = t_points(i); 
    y_prev = y_values(i-1);  
    % Calculate dy/dt using the custom ODE function 
    dydt = solveForDerivative(t, y_prev, A1, A2, g, K1, dydt_guess);  
    % Update the guess for the next iteration 
    dydt_guess = dydt;  
    % Estimate the new value of y using Euler's method (simplest integration method) 
    dt = t_points(i) - t_points(i-1); 
    y_values(i) = y_prev + dydt * dt; 
end 
plot(t_points, y_values); 
xlabel('Time'); 
ylabel('y(t)'); 
title('Approximate numerical solution of the ODE'); 

Here, we have used Euler’s method for numerical integration, but you can go for more sophisticated methods like “runga-kutta” or MATLAB functions like “ode45” with custom iteration to handle non linearity.

Moreover, you can refer the following MathWorks documentations for more information:

• Solving a non linear ODE: https://www.mathworks.com/help/curvefit/solving-a-nonlinear-ode.html
• “fsolve” function: https://www.mathworks.com/help/optim/ug/fsolve.html
• Basic Solver Selection: https://www.mathworks.com/help/matlab/math/choose-an-ode-solver.html#:~:text=i*yv(%3A%2C2)%3B-,Basic%20Solver%20Selection,-ode45%20performs%20well