Analytical solution for 1D heat equation

Hello,

I need help with analytical solution for following heat equation.

I already have numerical solution using euler(explicit and implicit).

Adding code for explicit solution...

clear all;
close all;
clc

%% zadani uhloy
%def oblasti
a=0; %x nalezi {a,b}
b=2*pi;
d=1; %okrajova podminka t nalezi {0,d}

h=2*pi/100; %volba kroku x


p=100; %koeficient pred druhou parcialni derivaci U dle x
tau=(h^2*1/2)/p % výpočet časového kroku 
%% priprava na vypocet
%vypocet mnozstvi kroku
stx=round(1+(2*pi-a)/h);
stt=1+d/tau;
x(1)=a;
t(1)=0;

%vypocet sigma

sig=p*tau/h^2


%% sestavovani site 
%parametrizace delky x
for j=2:stx
    x(j)=a+(j-1)*h;
end
%parametrizace casu t
for k=2:stt
    t(k)=(k-1)*tau;
end
%preddefinovani matice vyslednych U
%U=zeros(stx,stt);

%funkce f
%for j=1:stx
%    for k=1:stt
%        f(j,k)=exp(-t(k));
%    end
%end

% pocatecni podminky
for j=1:stx
U(j,1)=sin(x(j))+(1/10)*sin(10*x(j));
end
% okrajove podminky
for k=1:stt
U(1,k)=0;
U(stx,k)=0;
end

%% vypocet
%vypocet jednotlivych hodnot U uvnitr mnoziny
%tj bez OP a PP ktere byly jizvypocteny
for k=2:stt
for j=2:(stx-1)
    U(j,k)=sig*(U(j+1,k-1)+U(j-1,k-1))+(1-2*sig)*U(j,k-1);%+tau*f(j,k-1);
end
end

figure(1)
sur=surf(t, x, U);
set(sur,'LineStyle','none')

figure(2)
plot(x, U(:,1))

%U1_p1=U;,    t1_p1=t;
%U1_p10=U;   t1_p10=t;
U1_p100=U;  t1_p100=t;
%save U1_p1 , save t1_p1
%save U1_p10, save t1_p10
save U1_p100, save t1_p100

I need the same solution as my Figure 1 for different inputs of the constant 'a' in my code it is p, because I need to compare it with explicit and implicit solution.

FIgure (1) a=100

FIgure (1) a=1

I don't know how to do analytical solution in MATLAB for it.

Could someone help me please?

Thank you very much for your time and help! It will really help me.

Answer 1

[David Goodmanson]

Hi Dominick,

you have the sum of two sine terms for the boundary condition, sin(x) and (1/10) sin(10x). So consider a general sine term, A*sin(n*x). Taking the second derivative of this, you get -n^2*A*sin(n*x), same function as before, times a constant. This suggests separation of variables and a product-type solution for all of x and t. If you try

U(x,t) = A*sin(n*x)*exp(-b*t)

then the time derivative of the exponential is proportional to itself, which looks favorable. After taking the derivatives in the heat equation

a d^2U/dx^2 = dU/dt

you can solve for b in terms of a and n. For a particular choice of A and n you can satisfy the first boundary condition term, and for another choice you can satisfy the second boundary condition term. Since the heat equation is a linear partial differential equation, and since the two boundary terms are additive, the sum of the two solution satisfies the total boundary condition and you have the result.