# Analytical solution for 1D heat equation

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Dominik Horcicka on 6 May 2022
Edited: David Goodmanson on 12 May 2022
Hello,
I need help with analytical solution for following heat equation.
I already have numerical solution using euler(explicit and implicit).
clear all;
close all;
clc
%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.
Thank you very much for your time and help! It will really help me.

David Goodmanson on 6 May 2022
Edited: David Goodmanson on 12 May 2022
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.

R2021b

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