Please modify this code to solve the discretised equations developed in $($i$).$You will need to update the parameters involved in the calculations based on the provided information. These locations are indicated by $"..."$ in the code.There are also bugs in the code which need to be corrected.$($HINT: There are three bugs. One bug is in the Part A boundary point definition. Another bug is in the calculation of the residual for the interface node. The third bug is in the calculation of the analytical temperature distribution for Part B$).$ Provide the results where deltax$=0\mathrm{.}05$m and deltax$=0\mathrm{.}02$clcclose all$\%$ Firstly, let us define our parametersxmin$\_$A $=...$; $\%...$m in our casexmax$\_$A $=...$; $\%...$m in our casexmin$\_$B $=$ xmax$\_$A; $\%...$m in our casexmax$\_$B $=...$; $\%...$m in our casedx $=...$; $\%$ Grid spacing sizenx$\_$A $=$ int$16(($xmax$\_$A$-$xmin$\_$A$)/($dx$))$; $\%$ Number of cells across Part Anx$\_$B $=$ int$16(($xmax$\_$B$-$xmin$\_$B$)/($dx$))$; $\%$ Number of cellsl$\_$A $=$ xmax$\_$A $-$ xmin$\_$A; $\%$ Length of Part Al$\_$B $=$ xmax$\_$B $-$ xmin$\_$B; $\%$ Length of Part BT$\_$inf $=...$; $\%...$K in our caseh $=...$; $\%...$ W$.$m$^-2.$K$^-1$qdot $=...$; $\%...$ W$/$m$^3$ in this case k$\_$A $=...$; $\%...$ W$.$m$^-1.$K$^-1$ in our casek$\_$B $=...$; $\%...$ W$.$m$^-1.$K$^-1$ in our caseit$\_$max $=...$; $\%$ Let us use $1000$ to begin withTol $=0\mathrm{.}000001$; $\%$ Let us use $1$E$-6$ to begin withit$\_$count $=0$; $\%$ Start the iteration counterTA $=$ zeros$(1,($nx$\_$A$+1))+425$; $\%$ Initialise temperature T$\_$A array for all points in Part A $($our initial guess is $425)$TA$\_$d $=$ zeros$(1,($nx$\_$A$+1))+425$; $\%$ Initialise duplicated temperature array for all points in Part ATB $=$ zeros$(1,($nx$\_$B$+1))+425$; $\%$ Initialise temperature T array for all points in Part B $($our initial guess is $425)$TB$\_$d $=$ zeros$(1,($nx$\_$B$+1))+425$; $\%$ Initialise duplicated temperature array for all points in Part Bres$\_$it$\_$max $=$ zeros$($it$\_$max,$1)$; $\%$ Initialise residual arraytic; $\%$ This starts our timer for it $=1$:it$\_$max $\%$ Start our loop through iterations for the internal points it$\_$count $=$ it$\_$count $+1$; $\%$ Update iteration counter$\%$ Firstly, let us take a duplicate of TA and TB$.$ This will be the data from our$\%$ previous iteration. For i$=1,$ this will be our initial guess. Our initial$\%$ guess is $0$ here.TA$\_$d $=$ TA; $\%$ Duplicate TA array and store as TA$\_$d arrayTB$\_$d $=$ TB; $\%$ Duplicate TB array and store as TB$\_$d array$\%$ Part A $-$ Boundary pointTA$(1)=($TA$\_$d$(2)+(($qdot$.*$dx$^2)./(2*$k$\_$A$))$; $\%$ First cell in T array updated to be T$\_$L$\%$ Part B $-$ Boundary TB$($nx$\_$B$+1)=($TB$\_$d$($nx$\_$B$)+((($h$*$dx$)/$k$\_$B$)*$T$\_$inf$))/(1+(($h$*$dx$)/$k$\_$B$))$; $\%$ Final cell in T array updated to be T$\_$R $\%$ Part A$/$B Interface nodeTA$($nx$\_$A$+1)=($k$\_$A$/($k$\_$A$+$k$\_$B$))*$TA$\_$d$($nx$\_$A$)+($k$\_$B$/($k$\_$A$+$k$\_$B$))*$TB$\_$d$(2)+(($qdot$.*$dx$^2)./(2*($k$\_$A$+$k$\_$B$)))$; $\%$ Final point in TA array $-$ First Point in TB array TB$(1)=$ TA$($nx$\_$A$+1)$;$\%$ Calculate new internal point values for Part Afor i $=2$:nx$\_$A TA$($i$)=($TA$\_$d$($i$-1)+$ TA$\_$d$($i$+1)+(($qdot$.*$dx$^2)./$k$\_$A$))/2$;end$\%$ Calculate new internal point values for Part Bfor i $=2$:nx$\_$B TB$($i$)=($TB$\_$d$($i$-1)+$ TB$\_$d$($i$+1))/2$;end$\%$ Calculate residualres$\_$it $=0\mathrm{.}0$; $\%$ Initialise our residual value for this iteration as $0.\%$ Firstly, determine the residual for the boundary nodesres$\_$it $=$ res$\_$it $+$ abs$($TA$(1)-(($TA$(2)+(($qdot$.*$dx$^2)./(2*$k$\_$A$)))))$;res$\_$it $=$ res$\_$it $+$ abs$($TB$($nx$\_$B$+1)-(($TB$($nx$\_$B$)+((($h$*$dx$)/$k$\_$B$*$T$\_$inf$)))/(1+(($h$*$dx$)/$k$\_$B$))))$;$\%$ Secondly, determine the residual for the interface noderes$\_$it $=$ res$\_$it $+$ abs$($TA$($nx$\_$A$+1)-(($k$/($k$\_$A$+$k$\_$B$))*$TA$($nx$\_$A$)+($k$\_$B$/($k$\_$A$+$k$\_$B$))*$TB$(2)+(($qdot$.*$dx$^2)./(2*($k$\_$A$+$k$\_$B$)))))$;$\%$ Now, determine the residuals for the internal nodes $-$ first Part A$,$ then Part Bfor i $=2$:nx$\_$A res$\_$it $=$ res$\_$it $+$ abs$($TA$($i$)-(($TA$($i$-1)+$ TA$($i$+1)+(($qdot$.*$dx$^2)./$k$\_$A$))/2))$;endfor i $=2$:nx$\_$B res$\_$it $=$ res$\_$it $+$ abs$($TB$($i$)-(($TB$($i$-1)+$ TB$($i$+1))/2))$;endif res$\_$it $<$ Tol fprintf$("$Converged$!")$ fprintf$("$Residual $=\%20\mathrm{.}15$f$",$ res$\_$it$)$ fprintf$("$Number of iterations for Convergence $=\%$d$",$ it$\_$count$)$breakendres$\_$it$\_$max$($it$)=$ res$\_$it;end $\%$ End of our iteration loop starting on line $43$toc; $\%$ This gives us our elapsed time to either complete the $\%$ maximum number of iterations or obtain a converged solution.$\%$ We can now plot our results.$\%$ Firstly, let us see our residual behaviourfigure$(1)$plot$($res$\_$it$\_$max,$\text{'}$r$\text{'},$'LineWidth',$1)\%$ Plots the range of approximate values on the graphhold on $\%$ Retains plots in the current axes so that new plots addedset$($gca$,$'FontSize',$15)$; $\%$ Sets the axis font sizexlim$([0$ it$\_$count$+1])$; $\%$ Sets the x axis limitsxlabel$(\text{'}$Iterations$\text{'},$'interpreter','latex','FontSize',$10)$; $\%$ Adds the x labelylabel$(\text{'}$Residual$\text{'},$'interpreter','latex','FontSize',$10)$; $\%$ Adds the y labellegend$(\text{'}$Point$-$Jacobi','interpreter','latex','FontSize',$10)\%$ Adds legendgrid onhold offfigure$(2)$semilogy$($res$\_$it$\_$max,$\text{'}$r$\text{'},$'LineWidth',$1)$hold on $\%$ Retains plots in the current axes so that new plots addedset$($gca$,$'FontSize',$15)$; $\%$ Sets the axis font sizexlim$([0$ it$\_$count$+1])$; $\%$ Sets the x axis limitsxlabel$(\text{'}$Iterations$\text{'},$'interpreter','latex','FontSize',$10)$; $\%$ Adds the x labelylabel$(\text{'}$Residual$\text{'},$'interpreter','latex','FontSize',$10)$; $\%$ Adds the y labellegend$(\text{'}$Point$-$Jacobi','interpreter','latex','FontSize',$10)\%$ Adds legendgrid onhold off$\%$ Now, let us plot our temperature across x plotting both our analytical$\%$ and numerical answerxA $=$ zeros$(($nx$\_$A$+1),1)$;xA$(1)=$ xmin$\_$A;xA$($nx$\_$A$+1)=$ xmax$\_$A;for i $=2$:nx$\_$A xA$($i$)=$ xA$($i$-1)+$ dx; endxB $=$ zeros$(($nx$\_$B$+1),1)$;xB$(1)=$ xmin$\_$B;xB$($nx$\_$B$+1)=$ xmax$\_$B;for i $=2$:nx$\_$B xB$($i$)=$ xB$($i$-1)+$ dx; end$\%$ Analytical resultsxxA $=$ xmin$\_$A:$0\mathrm{.}001$:xmax$\_$A; $\%$ Specify the range of x for our function $\%$ Steps of $0\mathrm{.}001$ are used.xxB $=$ xmin$\_$B:$0\mathrm{.}001$:xmax$\_$B; $\%$ Specify the range of x for our function $\%$ Steps of $0\mathrm{.}001$ are used.TA$\_$an $=$ @$($xxA$)-($qdot$/(2*$k$\_$A$)).*$xxA$.^2+($qdot$*($l$\_$A$^2)/(2*$k$\_$A$))+($qdot$*$l$\_$A$*($l$\_$B$/$k$\_$B $+1/$h$))+$T$\_$inf; $\%$ This is our analytical function for Part ATB$\_$an $=$ @$($xxB$)-($qdot$*$l$\_$A