how to change U(0,t) = 1 to Ux(0,t) = 1

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Omar Benavente
Omar Benavente 2022-2-10
回答： Shushant 2023-9-26
a = @(t) 1;
b = @(t) 1;
xArray = dx*(0:J);
tArray = dt*(0:N);
for n = 1 : N
u(1, n+1)=a(tArray(n+1)); % Boundary Conditions at x=0
u(J+1, n+1)=b(tArray(n+1)); % Boundary Conditions at x=1
for j = 2 : J
u(j, n+1) = u(j,n) + mu*(u(j+1,n) - 2*u(j,n) + u(j-1,n));
end
end
I have the two boundaries of u(0,t) = 1and u(1,t), What if I change the boundary condition u(0,t) =1 to ux(0,t) = 1(differentiate with respect to x). How do I change my boundary condition to ux(0,t) = 1 according to my code?
Sincerely yours,
Omar

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回答（1 个）

Shushant
Shushant 2023-9-26
Hi Omar Benavente,
I understand that you have an equation "u(0,t) = 1" and you want to differentiate this equation with respect to "x" to obtain "ux(0,t)".
To define an equation in terms of "x", you need to first make "x" a symbolic object. Refer to the following documentation for more information on "syms" and how to make a variable into a symbolic object-
https://in.mathworks.com/help/symbolic/syms.html
Then you have to define "u(0,t)" in terms of the symbolic object "x". Afterwards, you can use the "diff" function to differentiate "u(0,t)" to obtain "ux(0,t)". Refer to the following documentation for more information on how to use "diff" function with symbolic objects-
https://in.mathworks.com/help/symbolic/diff.html
I hope this helped in solving the issue you were facing.
Thank you,
Shushant

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