almost-autnomous differential equation
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y' = y^2 − t. This differential equation has “stationary” solutions, but unlike with an autonomous equation, those stationary solutions are not horizontal. Vary the initial condition y(0) = c for a bit to try to get a sense of what the solutions look like. (Picking values between −3 and +3 should be good enough.)
Part A: There’s a value P such that, if y(0) < P, then the solution to the initial value problem decreases, while if y(0) > P, the solution to the initial value problem increases. Figure out what P is to two decimal places.
Attempted code:
syms y(t);
for c = -3:1:3
fc = dsolve('Dy = y^2 - t' , y(0) == c);
end
Not sure how I get it to print out an answer for p and get it to be for p to two decimal places.
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Star Strider
2016-7-23
编辑：Star Strider
2016-7-23
You need to solve it symbolically, but then you can use the matlabFunction function to create an anonymous function from it:
syms y(t) c
fc(t,c) = dsolve(diff(y) == y^2 - t, y(0) == c);
fc_fcn = matlabFunction(fc)
fc_fcn = @(t,c) -(airy(3,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(1,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0))./(airy(2,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(0,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0));
You also need to decide on an appropriate range for ‘t’ if you haven’t already been given one. It might be easier to use a ribbon plot for this until you home in on the correct value for ‘c’.
8 个评论
Bob
2016-7-23
Star Strider
2016-7-23
I’m not certain what you did when you put ‘[my] code into [your] code’. First, see if you can run my code alone as I posted it.
Otherwise, there could be version differences. I’m using R2016a, and the code I posted ran for me without error.
You can always just use the anonymous function form of the solution:
fc_fcn = @(t,c) -(airy(3,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(1,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0))./(airy(2,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(0,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0));
Bob
2016-7-23
Star Strider
2016-7-23
Apparently, ‘fc’ has some sort of interesting behaviour such that it neither increases nor decreases with time for some value of ‘c’. My guess is that you need to put ‘fc_fcn’ into a loop and find the value of ‘c’ that most closely produces the desired behaviour.
I’ll think about this to see if I can come up with a more efficient way to approach it, but ‘brute-forcing’ it may be the only way.
Bob
2016-7-23
Star Strider
2016-7-23
I don’t know what your function is supposed to do, only that you’re supposed to find a value of ‘c’ that does it.
This may give you some insight into the function’s behaviour:
t = linspace(0, 5, 50);
c = linspace(-3, 3, 50);
[T,C] = meshgrid(t,c);
fc_fcn = @(t,c) -(airy(3,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(1,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0))./(airy(2,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(0,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0));
figure(1)
meshc(T, C, fc_fcn(T,C))
grid on
axis([xlim ylim [-1 1]*1E+1])
xlabel('t')
ylabel('c')
zlabel('fc')
Bob
2016-7-24
Star Strider
2016-7-24
Not directly, but since I don’t understand what behaviour the particular value of ‘c’ is supposed to do, it should give you a way of determining the behaviour you’re looking for.
You can also plot the derivative (Jacobian) with meshc or contour if that would help:
t = linspace(0, 5, 50);
c = linspace(-3, 3, 50);
[T,C] = meshgrid(t,c);
fc_fcn = @(t,c) -(airy(3,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(1,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0))./(airy(2,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(0,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0));
dfc_fcn = gradient(fc_fcn(T, C));
figure(1)
contour(T, C, dfc_fcn, 50)
grid on
xlabel('t')
ylabel('c')
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