Pari/GP, 86 bytes

f(a,b,c,d)=[print(concat(r~))|r<-matrix(a,b,i,j,[c,d][vecmin([i-1,j-1,a-i,b-j])%2+1])] 

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R, 85 bytes

function(a,b,s,t,m=matrix(t,b,a),r=row(m),c=col(m)){m[!pmin(b-r,a-c,r-1,c-1)%%2]=s;m} 

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Idea of pmin inspired by @alephalpha's vecmin.

J, ¹⁸ 17 bytes

{~2|<./&(<.&i.-)/ 

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Note: Explanation needs slight update but idea is the same as current solution

Consider 7 5:

• <./&i./ First uses &i. to generate both 0 1 2 3 4 5 6 and 0 1 2 3 4, and the creates a "mininum" function table:

  0 0 0 0 0 0 1 1 1 1 0 1 2 2 2 0 1 2 3 3 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 

• - Note using i. on a negative number counts down, so i. _3 is 2 1 0.

• -<.&(...)] Is a fork which applies the verb from step to both 7 5 and _7 _5, and then takes the min. So min of:

   0 0 0 0 0 4 3 2 1 0 0 1 1 1 1 4 3 2 1 0 0 1 2 2 2 min 4 3 2 1 0 0 1 2 3 3 3 3 2 1 0 0 1 2 3 4 2 2 2 1 0 0 1 2 3 4 1 1 1 1 0 0 1 2 3 4 0 0 0 0 0 

  which gives:

  0 0 0 0 0 0 1 1 1 0 0 1 2 1 0 0 1 2 1 0 0 1 2 1 0 0 1 1 1 0 0 0 0 0 0 

• 2| Then we mod by 2:

  0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 

• {~ And map to chars:

  ----- -@@@- -@-@- -@-@- -@-@- -@@@- ----- 

ayr -0, 17 14 bytes

Since the 05Ab1e answer takes the input [row, column], I can get -3 bytes. Images and try it link outdated, just replace the code with this snippet.

~2||:v:\/@`v:~ 

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Explained

-0 makes ~ and |: (descending range) 0-indexed

The program A (~2||:v:\/@`v:~) B is equivalent to A ~ 2 | (v:\/ |: B) v: v:\/ ~ B

How to get there:

Expansion of the train (lowercase = fn, uppercase = var):

• B (f A g h i j) C
• B (f (A g h i j)) C
• B (f (A g (h i j))) C
• B f (A g (h i j)) C
• B f (A g ((h i j) C))
• B f (A g ((h C) i (j C)))

Substituting in the symbols:
B ~ (2 | ((|: C) v:\/@`v: (~ C)))

Focusing on the fork (h i j):
v:\/@`v: could be written as f \ / @` g
which is g @ (f \ /)
A g @ f B is (f A) g f B
so (|: C) v:\/@`v: (~ C) is (v:\/ |: C) v: (v:\/ ~ C)

Therefore
A (~2||:v:\/@`v:~) B is equivalent to A ~ 2 | (v:\/ |: B) v: v:\/ ~ B

Which leads to the explanation:

A is the string, B is the dims

• |: ... ~ Take the descending and ascending 0-ranges of each element of B

• v:\/ Convert these each separately to a minimum matrix of shape B

• v: Take the minimum of these two matrices on an element-wise basis

• 2| Mod 2 each element

• ~ Use the matrix as indices into string A

Haskell, 64 bytes

(x!y)q=[[cycle q!!minimum[k,1+x-k,j,1+y-j]|j<-[1..y]]|k<-[1..x]] 

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Also works with even sized mats and more than two character patterns.

Charcoal, 20 bytes

ＮθＥ⊘⊕Ｎ⭆⊘⊕θ§ζ⌊⟦ιλ⟧‖Ｏ⌈ 

Try it online! Link is to verbose version of code. Takes the two symbols as a single argument. Explanation:

Ｎθ 

Input the width.

Ｅ⊘⊕Ｎ⭆⊘⊕θ§ζ⌊⟦ιλ⟧ 

Input the height and loop over the top left quarter of the mat, outputting the symbols cyclically indexed by the minimum of the row and column.

‖Ｏ⌈ 

Reflect with overlap to complete the map.

Alternative approach, also 20 bytes:

ＮθＥ⮌…⁰Ｎ⭆⮌…⁰θ§ζ⌊⟦ικλμ 

Try it online! Link is to verbose version of code. Takes the two symbols as a single argument. Explanation:

Ｎθ 

Input the width.

Ｅ⮌…⁰Ｎ⭆⮌…⁰θ§ζ⌊⟦ικλμ 

Input the height and loop over the mat, reversing the ranges of the row and column, outputting the symbols cyclically indexed by the minimum of the row and column and reversed row and column.

05AB1E, 13 bytes

>;L`â€ßZä€ûûè 

First input is a pair of \$[row,column]\$, second input is a pair of \$[symbol2,symbol1]\$. Output is a matrix of characters.

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Explanation:

>; # Increase both values in the first (implicit) input-pair by 1, then halve # e.g. [5,13] → [6,14] → [3,7] L # Map both to a list in the range [1,n] # → [[1,2,3],[1,2,3,4,5,6,7]] ` # Pop and push both separated to the stack # → [1,2,3] and [1,2,3,4,5,6,7] â # Get the cartesian product of these two lists # → [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[2,1],[2,2],[2,3],[2,4],[2,5], # [2,6],[2,7],[3,1],[3,2],[3,3],[3,4],[3,5],[3,6],[3,7]] €ß # Get the minimum of each inner pair # → [1,1,1,1,1,1,1,1,2,2,2,2,2,2,1,2,3,3,3,3,3] Z # Push the flattened maximum (without popping the list) # → [1,1,1,1,1,1,1,1,2,2,2,2,2,2,1,2,3,3,3,3,3] and 3 ä # Split the list into that many equal-sized parts so we have a matrix # → [[1,1,1,1,1,1,1],[1,2,2,2,2,2,2],[1,2,3,3,3,3,3]] €û # Palindromize each row # → [[1,1,1,1,1,1,1,1,1,1,1,1,1], # [1,2,2,2,2,2,2,2,2,2,2,2,1], # [1,2,3,3,3,3,3,3,3,3,3,2,1]] û # Palindromize the entire matrix # → [[1,1,1,1,1,1,1,1,1,1,1,1,1], # [1,2,2,2,2,2,2,2,2,2,2,2,1], # [1,2,3,3,3,3,3,3,3,3,3,2,1], # [1,2,2,2,2,2,2,2,2,2,2,2,1], # [1,1,1,1,1,1,1,1,1,1,1,1,1]] è # 0-based modulair index each into the (implicit) second input-string # e.g. "-@" → # [["@","@","@","@","@","@","@","@","@","@","@","@","@"], # ["@","-","-","-","-","-","-","-","-","-","-","-","@"], # ["@","-","@","@","@","@","@","@","@","@","@","-","@"], # ["@","-","-","-","-","-","-","-","-","-","-","-","@"], # ["@","@","@","@","@","@","@","@","@","@","@","@","@"]] # (after which the matrix is output implicitly) 

Python3, 240 bytes

b,a,c,d=input().split() g=range a,b=int(a),int(b) r=[[d]*b for i in g(a)] x=y=0 while 1: for i in g(x,a-x):r[i][y]=r[i][b-1-y]=c for j in g(y,b-y):r[x][j]=r[a-1-x][j]=c x+=2;y+=2 if x>=a/2or y>=b/2:break print('\n'.join(map(''.join,r))) 

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APL+WIN, 22 bytes

Prompts for column then row as integers and the symbols as string. Index origin = 0

⎕[2|n⌊⌽⍉⌽⍉n←(⍳⎕)∘.⌊⍳⎕] 

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Python 3, 174 bytes

w,h,x,y=input().split() n='\n' f=lambda x,y,w,h:x*w+n+x+f(y,x,w-2,h-2).replace(n,x+n+x)+x+n+x*w if w*h-w-h+1else[n.join(x*(h-w+1)),x*(w-h+1)][w>h] print(f(x,y,int(w),int(h))) 

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Charcoal, 36 28 bytes

ＮθＮηＷ∧›θ⁰›η⁰«Ｂθη§ζⅈ↘≧⁻²θ≧⁻²η 

-10 bytes thanks to @Neil and +2 bytes for a bug-fix.

Try it online (verbose) or try it online (pure).

Explanation:

Get the first two inputs as integers:

InputNumber(q);InputNumber(h); ＮθＮη 

Continue looping while both q and h are still larger than 0:

While(And(Greater(q,0),Greater(h,0)){ ... } Ｗ∧›θ⁰›η⁰« ... 

Print a box with dimensions q by h, with the X'th character of the third input-string z as border, where X is the current x-position of the cursor:

 Box(q,h,AtIndex(z,X())); Ｂθη§ζⅈ 

Move the cursor once towards the bottom-right:

 Move(:DownRight); ↘ 

Decrement both q and h by 2:

 MapAssignRight(Minus,2,q);MapAssignRight(Minus,2,h); ≧⁻²θ≧⁻²η 

Vyxal, 18 bytes

›½ɾΠvg:G$ṅ/vf⌊v∞∞İ 

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How?

›½ɾΠvg:G$ṅ/vf⌊v∞∞İ › # Increment both of the (implicit) first input ½ # Halve both ɾ # Generate a [1, n] range for both Π # Cartesian product of the two vg # Minimum of each :G # Duplicate and get the maximum $ # Swap ṅ # Join by nothing / # Split this into that many equal parts vf # Convert each back to a list ⌊ # Convert each to an integer v∞ # Palindromise each ∞ # Palindromise the list İ # Index these into the (implicit) second input