Help with heat equation

Question

Janvier Solaris 2018-6-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405779-help-with-heat-equation

评论： Janvier Solaris 2018-6-16

image.png

Hello I have the following question

I am trying to understand how to do from the heat equation part Can anyone help?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2018-6-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405779-help-with-heat-equation#answer_324770

在 MATLAB Online 中打开

Try

sigma = 0.2;
r = 0.05;
S0 = 10 : 10 : 100;
K = 50
T = 2
xmax = 10
xmin = -10
N = [10, 100, 200, 400, 800];
M = N
for k = 1 : length(N)
    thisM = M(k);
    thisN = N(k);
    blsprice(................
    absoluteError = .........
    plot(S0, absoluteError, 'b*-');
    hold on;
    grid on;
end

I don't know what "the Call Option" or the heat equation or blsprice() is so I can't help with that, but at least this is a start. Presumably blsprice() uses those parameters as inputs. I'm also not sure how to calculate the absolute error.

3 个评论
显示 1更早的评论隐藏 1更早的评论

Image Analyst 2018-6-15

在 MATLAB Online 中打开

You can't do

for n = 2:N

because N is an array. You have to think what you really want. You need to extract the values of M and N from the array and use them somehow.

Even if you did

for n = N

which is allowed, n would take on the value of N(1) first, which is 10. So then you say

v(1, 10) = v(1,1);

well, what about the rest of the array? What about columns 2 through 9?

You also shouldn't do

v = zeros(M,N);

because they're arrays. You can if you use thisM and thisN and put that inside the loop as I showed you. Take another crack at it.

Janvier Solaris 2018-6-16

在 MATLAB Online 中打开

Hi @Image Analyst, I was able to modify my code as such which works similarly to your modifications;

    S0 = 10:10:100;
  K = 50;
  r = .05;
  T = 2;
  sigma = .2;
  price = blsprice(S0,K,r,T,sigma)
amax = 10;
amin = -10;
M = [10 100 200 400 800];
N = [10 100 200 400 800];
for k = 1:5
      dx = (amax - amin)/(M(k)-1);
      dt = T/(N(k)-1);                         
      v = zeros(M(k), N(k));                       
      r= dt/dx^2;                          
      x = linspace(amin,amax,M(k));
      v(:,1) = max(exp(x)-1,0);               
         for n = 2:N(k)                      
           %2 lines below are initial and boundary conditions
             v(1,n) = v(1, 1);            
              v(N(k),n) = v(M(k),1);
              v(2:M(k)-1,n) = (1-2*r)*v(2:M(k)-1,n-1)+r*v(1:M(k)-2,n-1)+r*v(3:M(k),n-1);
         end
         approx = v(2:M(k)-1,n)
  end
      error = abs(approx - price)  
     B= table(approx')
     G=table(price')
surf(v)

the loop grabs every item within the array as thisN and thisM would have .

Now the only issue I am getting is that I am getting the error, the limits are too large and it is not plotting for my 3d plot. when I change the array to for k = 1: 3 then it works but not for 400 and higher would you know anything about "limits are too large" error?

请先登录，再进行评论。