The exact solution to a partial differential equation

I know that MATLAB Has a PDE solver, but I wonder if it is possible to obtain the exact solution. For example, consider the heat equation

u_t = k u{xx}_

Is it possible to solve it with a set of initial and boundary conditions to calculate the exact equation of u

u = f(x,t)

and

u(x=0) = f(t)

I don't need numeral solution or the graph but the general equations.

Answer 1

[Stephan]

Hi,

this works by using the symbolic math toolbox :

% define symbolic variables
syms u(t) k
% define equation
eqn = u == k* diff(u,2);
% Solution without initial conditions
sol = dsolve(eqn);
% Define first derivative from u(t) for initial conditions
Du = diff(u);
% Define initial conditions
conds = [u(0) == 1, Du(0) == 0];
% Get solution with initial conditions
sol_conds = dsolve(eqn, conds);

will give:

>> pretty(sol)
      /      t    \         /    t    \
C1 exp| - ------- | + C2 exp| ------- |
      \   sqrt(k) /         \ sqrt(k) /

>> pretty(sol_conds)
   /    t    \      /      t    \
exp| ------- |   exp| - ------- |
   \ sqrt(k) /      \   sqrt(k) /
-------------- + ----------------
       2                 2

or if you use a live script it looks like this:

Best regards

Stephan

[Brian May]

Thanks for the detailed answer, Stephan. It answered my question, but still, I am not quite sure how to apply the boundary conditions. I am asking about the line of

conds = [u(0) == 1, Du(0) == 0];

Imagine a simple case of

IC: u(x,0) = A // Initial temperature across the slab

BC: u(0,t) = B // One side is kep at temperature B

BC: u_x(L,t) = 0 // The other side is insulated

How can I apply these three conditions to conds?

[Stephan]

If you have an analytical description of this case (an ode or a sytem of odes) you could also try solving it with symbolic toolbox. I never tried. So i can not definitly say, if this will work.

Note that we have not modelled a location x in the equation above as a variable.

[Darshan Soni]

@Stephan Can you tell me how can I solve 2 dimensional PDE in MATLAB and get analytical solution for that?

diff(F(x, y), x, 4)/E2 + diff(F(x, y), y, 4)/E1 + diff(diff(F(x, y), x, 2), y, 2)/G12 - (v*diff(diff(F(x, y), x, 2), y, 2))/E1 - (v*diff(diff(F(x, y), x, 2), y, 2))/E2 - (ay*h*sym(pi)^2*sin((sym(pi)*x)/a)*sin((sym(pi)*y)/b)*C(t))/a^2 - (ax*h*sym(pi)^2*sin((sym(pi)*x)/a)*sin((sym(pi)*y)/b)*C(t))/b^2 + (h*sym(pi)^4*sin((sym(pi)*x)/a)^2*sin((sym(pi)*y)/b)^2*A(t)^2)/(a^2*b^2) - (4*h*sym(pi)^2*cos((sym(pi)*x)/a)*cos((sym(pi)*y)/b)*sin((sym(pi)*x)/a)*sin((sym(pi)*y)/b)*A(t)^2)/(a*b)

[MINATI PATRA]

@Stephan u_t = diff(u, t). So considered equation

% define equation

eqn = u == k* diff(u,2); is wrong

[Peter]

The question was about the second derivative (u{xx}_) so I believe @Stephan's answer is correct

[Torsten]

@Stephan 's answer doesn't help to answer the original question. The requested u is a function of x and t following the partial differential equation

du/dt = k * d^2u/dx^2

The code

% define symbolic variables
syms u(t) k
% define equation
eqn = u == k* diff(u,2);
% Solution without initial conditions
sol = dsolve(eqn);

assumes that u is a function of t alone and solves k*u''(t) = u(t) which is an ordinary differential equation.