Solving Multi-variable Partial Differential Equation

boundary conditions;

Please help me to solve above differential eqation. I want P = P(x,y) and also generate a 3D plot for the same.

I am a new to MATLAB, any help shall be highly appreciated.

Thanks in advance.

Answer 1

[Torsten]

The only thing you can get with the "boundary condition" you gave is the solution on the characteristic curve of the PDE through (x0,y0).

Look up "method of characteristics" for more details.

And I don't understand why you write

I am a new to MATLAB, any help shall be highly appreciated.

It's not a MATLAB problem.

[Captain Rituraj Singh]

I wrote this because I don't know which MATLAB subroutine or function is suitable to solve such an equation. I want to write a MATLAB code for above equation but not certain where to start. I won't post anything here if it is not related to MATLAB.

[Torsten]

Do you know how the method of characteristics works and how it is to be applied to your problem ?

If you understand this method, you will see that your problem is equivalent to solving three ordinary differential equations. The MATLAB solver to be used for this system is ode45.

[Captain Rituraj Singh]

I am not aware of "method of characteristics". But I have used ode45 earlier and sample program can help me better, though I fill try to look into "method of characteristics".

Thanks

[Torsten]

Set x = x(t) and y = y(t) (x(t) and y(t) will describe the characteristic curve through (x0,y0)) and P = P(x(t),y(t)).

Then

dP/dt = dP/dx * dx/dt + dP/dy * dy/dt 

So if you define the curve (x(t),y(t)) by

dx/dt = cosh(y)   , x(0) = 0.1
dy/dt = -sinh(y)/x, y(0) = 0.1,

then

dP/dt = -P/(3*x), P(0) = P(x0,y0) = 0.5.

on this curve.

Summarizing: you get a solution for P on the curve (x(t),y(t)) by solving the three ordinary differential equations

dx/dt = cosh(y) , x(0) = 0.1,
dy/dt = -sinh(y)/x, y(0) = 0.1,
dP/dt = -P/(3*x), P(0) = P(x0,y0) = 0.5

More is not possible with the condition P(0.1,0.1) = 0.5.

It's like solving an initial value problem: Given the condition in only one point gives a 1d solution curve, not a 2d function P(x,y).

Do you see what I mean: it's not a problem with MATLAB ?

[Torsten]

fun = @(t,u)[cosh(u(2));-sinh(u(2))/u(1);-u(3)/(3*u(1))];
u0 = [0.1;0.1;0.5];
tstart = -10;
tend = 10;
tspan1 = [0 tstart];
[T1,U1] = ode45(fun,tspan1,u0);
tspan2 = [0 tend];
[T2,U2] = ode45(fun,tspan2,u0);
T = [flipud(T1);T2(2:end)];
U = [flipud(U1);U2(2:end,:)];
plot3(U(:,1),U(:,2),U(:,3))
xlabel('x')
ylabel('y')
zlabel('P')

[Captain Rituraj Singh]

Thank you so much @Torsten, provided solution was very helpful.