Can someone please assist me in plotting a 3-D surface?

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Omorodion Solomon
Omorodion Solomon 2021-8-19
评论： Walter Roberson 2021-8-21
Ran in:
I want to plot a 3D graph of the series when alpha=0.25 and and k=1,2,...10.
This is my code but im getting an error :
alpha=0.25;
[x,t] = meshgrid(-4:.1:4, 0.1:.2:0.4);
k=[1:10];
A=exp(x);
B=(t.^(k.*alpha))./factorial(k).*alpha. ^k;
Invalid use of operator.

Error in connector.internal.fevalMatlab

Error in connector.internal.fevalJSON
C=sum(B);
D=A.*C;
mesh(t,x,D)
xlabel('t')
ylabel('x')
zlebel('u_10')

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采纳的回答

Omorodion Solomon
Omorodion Solomon 2021-8-21

The approximate and exact solution are not corresponding. Please assist further. Here is my code:

alpha=0.25; r=5; t=0.002; x=[-5:0.2:5]; k = reshape(0:100, 1, 1, []); A=exp(x); B = (t.^(k.*alpha))./factorial(k).*alpha.^k; C = sum(B,3); Numerical=A.*C; Exact=A.*exp((r-4).*(t.^alpha)./(alpha)); Z1=Numerical; Z2=Exact; hZ1 = plot(x,Z1,'-r'); hold on hZ2 = plot(x,Z2,'-g'); hold off grid on xlabel('x') ylabel('u') legend([hZ1(1),hZ2(1)], 'CFRDTM','EXACT', 'Location','NE')

更多回答（1 个）

Walter Roberson
Walter Roberson 2021-8-19
Ran in:
alpha=0.25;
[x,t] = meshgrid(-4:.1:4, 0.1:.2:0.4);
k = reshape(1:10, 1, 1, []);
A=exp(x);
B = (t.^(k.*alpha))./factorial(k).*alpha.^k;
C = sum(B,3);
D=A.*C;
mesh(t,x,D)
xlabel('t')
ylabel('x')
zlabel('u_10')
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2021-8-20
You had
k=[1:10];
B=(t.^(k.*alpha))./factorial(k).*alpha. ^k;
so you obviously wanted to create length(k) = 10 outputs for each input value.
But you have an array of inputs -- t is a 2D array, 81 x 2, and you want to produce length(k) = 10 outputs for each of those 81 x 2 inputs. How do you want to arrange that? Do you want the output to be 10 x 81 x 2? Do you want it to be 81 x 10 x 2 ? Do you want it to be 81 x 2 x 10? Do you want it to be 810 x 2? Do you want it to be 81 x 20?
You followed that with
C=sum(B);
which values are you wanting to sum?
mesh(t,x,D)
as the results are being sent to mesh() with t and x both being 2D arrays, we can see that D is expected to be the same size as t and x are. So with that 81 x 2 input, after the sum of the 10 values per entry, you want an 81 x 2 output. The only way this can make sense is if for each input location, you want to sum the values calculated from the 10 different k entries.
You can do some complicated memory arrangements, but by far the easiest way to handle all of this is to arrange that the output, B, is one of:
  • 10 x 81 x 2 -- in which case you want to sum() along the first dimension to get 1 x 81 x 2 and then collapse the first dimension to get 81 x 2
  • 81 x 10 x 2 -- in which case you want to sum() along the second dimension to get 81 x 1 x 2 and then collapse the second dimension to get 81 x 2
  • 81 x 2 x 10 -- in which case you want to sum() along the third dimension to get 81 x 2 x 1 which will automatically collapse to 81 x 2
sum along the first dimension is sum(B,1) . sum along the second dimension is sum(B,2) . sum along the third dimension is sum(B,3) . sum(B) by itself sums along the first non-singular dimension.
I decided that it was easiest to arrange to sum along the third dimension, as afterwards you do not need to rearrange to get to the 81 x 2 result you need.
The easiest way to get the 10 different k-based results for each location, to get 81 x 2 x 10 from an 81 x 2 input, is to arrange that k is 1 x 1 x 10 instead of the 1 x 10 that you had coded before. After that, the code takes advantage of "implicit expansion"

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