The exact solution to a partial differential equation
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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.
采纳的回答
Stephan
2018-8-8
Hi,
% 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
6 个评论
Brian May
2018-8-10
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
2022-2-4
@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
2024-11-13
% define equation
eqn = u == k* diff(u,2); is wrong
@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.
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